Maharashtra Board Class 10 Science Chapter 1 Gravitation Solutions

Std 10 Science Part 1 Chapter 1 Gravitation Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 1 Gravitation Question Answer Maharashtra Board

Class 10 Science Chapter 1 Gravitation Exercise Question 1.
Study the entries in the following table and rewrite them putting the connected items in a single row :

I	II	III
Mass	\(m/s^2\)	Zero at the centre of the earth
Weight	kg	Measure of inertia
Acceleration due to gravity	\(N.m^2/kg^2\)	Same in the entire universe
Gravitational constant	N	Depends on height

Answer:

I	II	III
Mass	kg	Measure of inertia
Weight	N	Depends on height
Acceleration due to gravity	\(m/s^2\)	Zero at the centre of the earth
Gravitational constant	\(N.m^2/kg^2\)	Same in the entire universe

In simple words: This exercise requires matching related physical quantities and their properties/units into coherent rows, indicating a fundamental understanding of gravitation concepts. Each row correctly links a property like 'Mass' to its unit 'kg' and characteristic 'Measure of inertia'.

🎯 Exam Tip: Pay close attention to the units and characteristics of physical quantities as they are frequently tested. Understanding the relationships between mass, weight, acceleration due to gravity, and the gravitational constant is crucial for scoring well.

 

Gravitation Class 10 Maharashtra Board Question 2.
Answer the following questions.
(a) What is the difference between mass and weight of an object? Will the mass and weight of an object on the earth be the same as their values on Mars? Why?
Answer:
The mass of an object is the amount of matter present in it. It is same everywhere in the Universe and is never zero. It is a scalar quantity and its SI unit is kg. The weight of an object is the force with which the earth (or any other planet/ moon/star) attracts it. It is directed towards the centre of the earth. The weight of an object is different at different places on the earth. It is zero at the earth's centre. It is a vector quantity and its SI unit is the newton (N). The magnitude of weight = mg.

The mass of an object will be the same on the earth and Mars, but the weight will not be the same because the value of g on Mars is different from that on the earth.
(b) what are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force?
Answer:
(i) Free fall:
Whenever an object moves under the influence of the force of gravity alone, it is said to be falling freely.
(ii) Acceleration due to gravity:
The acceleration produced in a body due to the gravitational force of the earth is called the acceleration due to gravity.
[Note: On the earth's surface, the value of the acceleration due to gravity is almost uniform. If a body falls from a low altitude, the value of the acceleration due to gravity is almost the same.]
(iii) Escape velocity:
When a body is thrown vertically upward from the surface of the earth, the minimum initial velocity of the body for which the body is able to overcome the downward pull by the earth and can escape the earth forever is called the escape velocity.
(iv) Centripetal force:
In uniform circular motion of a body, the force acting on the body is directed towards the centre of the circle. This force is called centripetal force.
(c) Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?
Kepler's first law :
The orbit of a planet is an ellipse with the Sun at one of the foci.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सूर्य के चारों ओर एक ग्रह की दीर्घवृत्तीय कक्षा को दर्शाता है। सूर्य 'S' पर स्थित है और ग्रह 'A', 'B', 'C', 'D', 'E', 'F' बिंदुओं पर दिखाया गया है, जो विभिन्न समय अंतरालों में ग्रह की स्थिति को इंगित करता है। यह केप्लर के पहले नियम को स्पष्ट करता है, जिसमें बताया गया है कि ग्रह सूर्य के चारों ओर एक दीर्घवृत्तीय कक्षा में घूमते हैं, जिसमें सूर्य एक फोकस पर होता है।

Figure 1.5 shows the elliptical orbit of a planet revolving around the Sun (S).

Kepler's second law :
The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
A → B, C → D and E → F are the displacements of the planet in equal intervals of time.
The straight lines AS, CS and ES sweep equal areas in equal intervals of time.
Area ASB = area CSD = area ESF.

Kepler's third law :
The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.
Thus, if r is the average distance of the planet from the Sun and T is its period of revolution, then,
\(T^2 \alpha r^2\), i.e., \(T^2/r^3 = constant = K\)

For simplicity, we shall assume the orbit to be a circle.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सूर्य (S) के चारों ओर एक ग्रह (P) की गोलाकार गति को दर्शाता है। 'r' ग्रह की कक्षा की त्रिज्या है, जो सूर्य से उसकी दूरी को इंगित करती है। यह सरल आरेख बताता है कि ग्रह सूर्य के चारों ओर एक वृत्ताकार पथ पर गति कर रहा है।

In Fig. 1.6,
S denotes the position of the Sun, P denotes the position of a planet at a given instant and r denotes the radius of the orbit (= the distance of the planet from the Sun). Here, the speed of the planet is uniform.
\[v = \frac{\text{circumference of the circle}}{\text{period of revolution of the planet}}\]
\[= \frac{2\pi r}{T}\]
If m is the mass of the planet, the centripetal force exerted on the planet by the Sun (= gravitational force),
\[F = \frac{mv^2}{r}\]
\[\implies F = \frac{m(2\pi r/T)^2}{r} = \frac{4\pi^2mr^2}{T^2r} = \frac{4\pi^2mr}{T^2}\]
According to Kepler's third law,
\[T^2 = Kr^3\]
\[\implies F = \frac{4\pi^2mr}{Kr^3} = \frac{4\pi^2m}{K} \left(\frac{1}{r^2}\right)\]
Thus, \(F \alpha \frac{1}{r^2}\) as \(\frac{4\pi^2m}{K}\) is constant in a particular case.
(d) A stone thrown vertically upwards with initial velocity u reaches a height 'h' before coming down. Show that the time taken to go up is same as the time taken to come down.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पत्थर की ऊर्ध्वगामी और अधोगामी गति को दर्शाता है। 'A' प्रारंभिक बिंदु है, 'B' अधिकतम ऊंचाई है, और 'h' तय की गई कुल ऊंचाई है। 't1' ऊपर जाने में लगा समय है और 't2' नीचे आने में लगा समय है। यह आरेख गति के नियमों को समझने में मदद करता है।

We have, \(v = u + at\) .....(1)
and \(s = ut + \frac{1}{2} at^2\) .....(2)
\[\implies s = (v - at) t + \frac{1}{2} at^2\]
\[= vt - at^2 + \frac{1}{2} at^2\]
\[\implies s = vt - \frac{1}{2} at^2\] ....(3)
As the stone moves upward from A → B,
\(s = AB = h, t = t_1\),
\(a = -g\) (retardation),
\(u = u\) and \(v = 0\)
\[\implies \text{From Eq. (3), } h = 0 - \frac{1}{2} (-g)t_1^2\]
\[\implies h = \frac{1}{2} gt_1^2\].....(4)
As the stone moves downward from B → A,
\(t = t_2, u = 0, s = h\) and \(a = g\)
\[\implies \text{from Eq. (2), } h = \frac{1}{2} gt_2^2\].....(5)
From Eqs. (4) and (5), \(t_1^2 = t_2^2\)
\[\implies t_1 = t_2\] (\(t_1\) and \(t_2\) are positive)
(e) If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?
Answer:
To pull an object along the floor, it is necessary to do work against the force of friction between the object and the surface of the floor. This force of friction is proportional to the weight, mg, of the object. If the value of g becomes twice its value, the weight of the object and hence the force of friction will become double. Therefore, it will become two times more difficult to pull a heavy object along the floor.
In simple words: Mass is the amount of matter, constant everywhere, a scalar quantity (SI unit: kg). Weight is the gravitational force on an object, varies with 'g', a vector quantity (SI unit: N). The mass of an object will remain the same on Earth and Mars, but its weight will differ because the gravitational acceleration 'g' is different on each planet. Free fall is motion under gravity alone; acceleration due to gravity is the acceleration caused by Earth's gravitational force; escape velocity is the minimum speed needed to escape a planet's gravity; and centripetal force is the inward-directed force keeping an object in circular motion.

🎯 Exam Tip: Distinguishing between mass and weight, understanding Kepler's laws, and defining key gravitational terms are fundamental for conceptual clarity and high scores. Remember the direct and inverse proportionality in Kepler's third law and Newton's law of gravitation for problem-solving.

 

10th Gravitation Chapter Exercise Question 3.
Explain why the value of g is zero at the centre of the earth.
Answer:
The value of g changes while going deep inside the earth. It goes on decreasing as we go from the earth's surface towards the earth's centre.

We shall treat the earth as a sphere of uniform density. If we consider a particle of mass m at point P at a distance (R – d) from the earth's centre, where R is the radius of the earth and d is the depth below the earth's surface, the gravitational force on the particle due to the earth is

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पृथ्वी के अंदर गुरुत्वाकर्षण त्वरण (g) को दर्शाता है। 'A' पृथ्वी की सतह पर एक बिंदु है, 'P' पृथ्वी के भीतर एक कण का स्थान है, और 'R' पृथ्वी की त्रिज्या है। 'd' सतह से गहराई है, और 'R-d' केंद्र से कण की दूरी है। छायांकित भाग 'गोलाकार खोल' को दर्शाता है जो इस मामले में गुरुत्वाकर्षण बल में प्रभावी नहीं है।

\[F = \frac{GmM'}{(R-d)^2}\]
where 'M' is the mass of the sphere of radius \((R – d)\).
\[M' = \frac{4}{3} \pi (R-d)^3 \times \frac{M}{\frac{4}{3} \pi R^3} = \frac{M(R-d)^3}{R^3}\]
because the outer spherical shell is not effective (Fig. 1.10). In this case, the acceleration due to gravity is
\[g = \frac{F}{m} = \frac{G}{m} \frac{M(R-d)^3}{(R-d)^2 R^3} = \frac{GM(R-d)}{R^3}\]
where M is the mass of the earth. Thus, g decreases as d increases. It is less than that at the earth's surface \(\left(\frac{GM}{R^2}\right)\) At the earth's centre, \(d = R\)
\[\implies g = 0\]
[Note: The formulae given in the answer are not given in the textbook. The formula \(\text{density} = \frac{\text{mass}}{\text{volume}}\) is used to find M'.]
In simple words: The value of 'g' decreases as you go deeper into the Earth. At the center, the gravitational forces from all directions cancel out, effectively making the net gravitational force, and thus 'g', zero.

🎯 Exam Tip: Remember that 'g' is maximum at the surface and decreases with both altitude and depth. The concept of gravitational forces cancelling out at the Earth's center is a key point for explaining why 'g' is zero there.

 

Std 10 Science Chapter 1 Gravitation Question Answer Question 4.
Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be \(\sqrt{8}T\).
Answer:
\[T = \frac{2\pi}{\sqrt{GM}} r^{3/2}\]
where T = period of revolution of a planet around the Sun, M = mass of the Sun, G = gravitational constant and r = radius of the orbit assumed to be circular = distance of the planet from the Sun.
For \(r = R, T = T_1\).
\[\implies T_1 = \frac{2\pi}{\sqrt{GM}} R^{3/2}\]
For \(r = 2R, T = T_2\).
\[T_2 = \frac{2\pi}{\sqrt{GM}} (2R)^{3/2}\]
\[T_2 = \frac{2\pi}{\sqrt{GM}} R^{3/2} \times 2^{3/2}\]
\[T_2 = T_1 2^{3/2}\]
\[\implies T_2 = T_1 \sqrt{8}\]
\[T_2 = \sqrt{8}T\]
In simple words: Using Kepler's Third Law (\(T^2 \alpha r^3\)), if the orbital radius is doubled (from R to 2R), the square of the new period will be \( (2R)^3 = 8R^3 \), meaning the new period will be \(\sqrt{8}\) times the original period.

🎯 Exam Tip: This question tests your understanding of Kepler's Third Law. Ensure you can manipulate the proportionality \(T^2 \alpha r^3\) correctly to solve problems involving changes in orbital radius and period.

 

Class 10 Science 1 Chapter 1 Gravitation Question 5.
Solve the following examples.
(a) An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?
Solution:
Data: \(u = 0 \, m/s\), \(s = 5m\), \(t = 5s\), \(g = ?\)
\[s = \frac{1}{2} gt^2\]
\[\implies 5 \, m = \frac{1}{2} g (5 \, s)^2\]
\[5 \, m = \frac{1}{2} g \times 25 \, s^2\]
\[\implies g = \frac{2 \times 5}{25} \, m/s^2\]
\[\implies g = 0.4 \, m/s^2 \text{ (on the planet).}\]
(b) The radius of planet A is half the radius of planet B. If the mass of A is \(M_A\), what must be the mass of B so that the value of g on B is half that of its value on A?
(Practice Activity Sheet – 4)
Solution:
Data : \(R_A = R_B/2\), \(g_B = \frac{1}{2} g_A\), \(M_B = ?\)
\[g = \frac{GM}{R^2}\]
\[g_A = \frac{GM_A}{R_A^2} \text{ and } g_B = \frac{GM_B}{R_B^2}\]
\[\implies \frac{g_B}{g_A} = \left(\frac{M_B}{M_A}\right) \left(\frac{R_A}{R_B}\right)^2\]
\[\implies \frac{1}{2} = \left(\frac{M_B}{M_A}\right) \left(\frac{1}{2}\right)^2\]
\[\frac{1}{2} = \frac{M_B}{M_A} \times \frac{1}{4}\]
\[\frac{M_B}{M_A} = \frac{4}{2}\]
\[\implies M_B = 2M_A\]
(c) The mass and weight of an object on the earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.
Solution:
Data: \(m = 5 \, kg\), \(W = 49 \, N\),
\(g_M = \frac{g_E}{6}\) (on the moon) \(=\) ?, \(W\) (on the moon) \(=\) ?
(i) The mass of the object on the moon = the mass of the object on the earth = 5 kg
(ii) \(W = mg\)
\[\frac{W_M}{W_E} = \frac{mg_M}{mg_E} = \frac{g_M}{g_E} = \frac{1}{6}\]
\[\implies W_M = \frac{W_E}{6} = \frac{49 \, N}{6} = 8.167 \, N\]
(weight of the object on the moon).
(d) An object thrown vertically upwards reaches a height of 500 m. what was its initial velocity? How long will the object take to come back to the earth? Assume \(g = 10 \, m/s^2\)
Solution:
100 m/s and 20 s
(e) A ball falls off a table and reaches the ground in 1 s. Assuming \(g = 10 \, m/s^2\), calculate its speed on reaching the ground and the height of the table.
Solution:
Data: \(t = 1s\), \(g = 10 \, m/s^2\), \(u = 0 \, m/s\),
\(s = ?\), \(v = ?\)
(i) \(s = ut + \frac{1}{2} gt^2\)
\[s = \frac{1}{2} gt^2 \text{ for } u = 0 \, m/s\]
\[\implies s = \frac{1}{2} \times 10 \, m/s^2 \times (1s)^2\]
\[= 5 \, m\]
\[\implies \text{The height of the table } = 5 \, m.\]
(ii) \(v = u + at = u + gt\)
\[= 0 \, m/s + 10 \, m/s^2 \times 1 \, s\]
\[= 10m/s\]
\[\implies \text{The velocity of the ball on reaching the ground } = 10 \, m/s.\]
(f) The masses of the earth and moon are \(6 \times 10^{24} \, kg\) and \(7.4 \times 10^{22} \, kg\), respectively. The distance between them is \(3.84 \times 10^5 \, km\). Calculate the gravitational force of attraction between the two. Use \(G = 6.7 \times 10^{-11} \, N.m^2 \, kg^{-2}\).
Solution:
Data: \(m_1 = 6 \times 10^{24} \, kg\),
\(m_2 = 7.4 \times 10^{22} \, kg\),
\(r = 3.84 \times 10^5 \, km = 3.84 \times 10^8 \, m\),
\(G = 6.7 \times 10^{-11} \, N.m^2 \, kg^{-2}, F = ?\)
\[F = \frac{Gm_1m_2}{r^2}\]
\[F = \frac{6.7 \times 10^{-11} \, N.m^2 \, kg^{-2} \times 6 \times 10^{24} \, kg \times 7.4 \times 10^{22} \, kg}{(3.84 \times 10^8 \, m)^2}\]
\[F = \frac{6.7 \times 6 \times 7.4 \times 10^{35}}{3.84 \times 3.84 \times 10^{16}} \, N = 2.017 \times 10^{20} \, N\]
This is (the magnitude of) the gravitational force between the earth and the moon.
(g) The mass of the earth is \(6 \times 10^{24} \, kg\). The distance between the earth and the Sun is \(1.5 \times 10^{11} \, m\). If the gravitational force between the two is \(3.5 \times 10^{22} \, N\), what is the mass of the Sun? (Use \(G = 6.7 \times 10^{-11} \, N.m^2 \, kg^{-2}\))
Solution:
Data: \(m_1 = 6 \times 10^{24} \, kg\),
\(r = 1.5 \times 10^{11} \, m, F = 3.5 \times 10^{22} \, N\),
\(G = 6.7 \times 10^{-11} \, N.m^2 \, kg^{-2}, m_2 = ?\)
\[F = \frac{Gm_1m_2}{r^2}\]
\[\implies m_2 = \frac{Fr^2}{Gm_1}\]
\[m_2 = \frac{3.5 \times 10^{22} \, N \times (1.5 \times 10^{11} \, m)^2}{6.7 \times 10^{-11} \, N.m^2 \, kg^{-2} \times 6 \times 10^{24} \, kg}\]
\[m_2 = \frac{3.5 \times 1.5 \times 1.5 \times 10^{44}}{6.7 \times 6 \times 10^{13}} \, kg\]
\[= 1.96 \times 10^{30} \, kg \text{ (mass of the sun)}\]
In simple words: These problems involve applying kinematic equations for free fall and Newton's Law of Universal Gravitation to calculate 'g', mass, velocity, distance, and gravitational force. It's essential to correctly identify given values and use the appropriate formulas, paying attention to units and conversions, especially for large numbers in scientific notation.

🎯 Exam Tip: For numerical problems, always write down the given data and the formula to be used. Show all calculation steps clearly. Pay attention to units and scientific notation, as small errors can lead to large discrepancies in the final answer.

 

Gravitation Class 10 Exercise Answers Project:
Take weights of five of your friends. Find out what their weights will be on the moon and the Mars.
Answer:
Help: The weight of a body
(i) On the earth. \(W_1 = mg_1\)
(ii) on the moon, \(W_2 = mg_2\)
(iii) on Mars, \(W_3 = mg_3\)
\[\implies W_2 = W_1 \times \frac{g_2}{g_1} \text{ and } W_3 = W_1 \times \frac{g_3}{g_1}\]
Now, \(g_1 = 9.81 \, m/s^2\), \(g_2 = 1.67 \, m/s^2\) and \(g_3 = 3.72 \, m/s^2\)
If \(W_1 = 500 \, N\),
\[W_2 = 500 \times \frac{1.67}{9.81} \, N = 85.12N \text{(approx.)}\]
\[\text{and } W_3 = 500 \times \frac{3.72}{9.81} \, N = 189.6 \, N \text{(approx.)}\]
In simple words: To find a person's weight on the Moon or Mars, you multiply their Earth weight by the ratio of the gravitational acceleration on that body to Earth's gravitational acceleration. Since 'g' is different, weight changes, but mass remains constant.

🎯 Exam Tip: This project highlights the difference between mass (constant) and weight (varies with 'g'). Understanding the formula \(W = mg\) and how to apply it for different planetary bodies is crucial. Ensure correct conversion of gravitational acceleration values for accurate calculations.

 

Can you recall? (Text Book Page No. 1)

 

10th Class Science Part 1 Chapter 1 Gravitation Exercise Question 1.
What are the effects of a force acting on an object? (Note: a body = an object)
Answer:
- A force can set a body in motion. For example, if a ball at rest on the floor is pushed, it rolls on the floor.
- A force can stop a moving body. For example, a moving bicycle can be brought to rest by application of brakes.
- A force acting on a body can change the speed of the body. For example, when brakes are applied to a moving bicycle, its speed decreases due to the friction between the brake shoes and the rim of the tire.
- A force can change the direction of motion of the body. For example, in a uniform circular motion of a body, the direction of motion of the body keeps on changing due to the applied force.
- A force can change the speed as well as the direction of motion of the body. For example, when a ball bowled by a bowler is hit by a batsman, there occurs a-change in the speed as well as the direction of motion of the ball.
- A force can change the shape and size of the body on which it acts. For example, when a rubber ball is pressed, it gets deformed and hence no longer remains spherical. Also, there can be a decrease in its volume.
In simple words: A force can cause an object to start moving, stop moving, change its speed, alter its direction of motion, or even deform its shape and size. These are the fundamental ways forces interact with objects.

🎯 Exam Tip: When asked about the effects of force, remember the five key changes: initiating motion, stopping motion, changing speed, changing direction, and changing shape/size. Providing simple, clear examples for each effect can earn full marks.

 

Question 1. Gravitation Exercise Question 2.
What types of forces are you familiar with?
Answer:
The gravitational force between the earth and the moon, the electromagnetic force between two charged particles in motion, the nuclear force between a proton and a neutron in the nucleus of an atom.
In simple words: Common forces include gravitational force (attraction between masses), electromagnetic force (between charged particles), and nuclear forces (holding atomic nuclei together).

🎯 Exam Tip: List at least three fundamental forces (gravitational, electromagnetic, nuclear strong/weak) and give a brief, clear example for each to show understanding.

 

Gravitation 10th Class Exercise Question 3.
What do you know about the gravitational force?
Answer:
The gravitational force is a universal force, i.e., it acts between any two objects in the universe.
In simple words: Gravitational force is a universal attractive force that exists between any two objects possessing mass in the entire universe.

🎯 Exam Tip: Emphasize "universal" and "attractive" when defining gravitational force. Mentioning its dependence on mass and distance (even if not explicitly stating the inverse square law here) adds value.

 

Can you recall? (Text Book Page No. 1)

 

Science Part 1 Gravitation Exercise Question 1.
What are Newton's laws of motion?
Answer:
(1) Newton's first law of motion: An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.
(2) Newton's second law of motion: The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.
(3) Newton's third law of motion: Every action force has an equal and opposite reaction force that acts simultaneously.
[Note: Equal in magnitude and opposite in direction.]
In simple words: Newton's first law states that an object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. The second law relates force, mass, and acceleration (\(F=ma\)) or change in momentum. The third law states that for every action, there is an equal and opposite reaction.

🎯 Exam Tip: Clearly state all three laws. For the second law, mentioning either \(F=ma\) or the rate of change of momentum is acceptable. For the third law, highlight "equal and opposite" and "simultaneous action."

 

Use your brainpower! (Text Book Page No. 4)

 

10th Science Part 1 Gravitation Exercise Question 1.
If area ESF in figure 1.5 is equal to area ASB, what will you infer about EF?
Answer:
The time taken by the planet to move from E to F equals the time taken by the planet to move from A to B.
In simple words: This inference is based on Kepler's Second Law, which states that a line joining a planet and the Sun sweeps out equal areas in equal intervals of time. Therefore, equal areas imply equal time intervals.

🎯 Exam Tip: This question directly tests Kepler's Second Law. State the law clearly and apply it to the given scenario. Understanding that equal areas correspond to equal time intervals is key.

 

Use your brainpower (Text Book Page No. 7)

 

Gravitation Exercise 10th Class Question 1.
According to Newton's law of gravitation, every object attracts every other object. Thus, if the earth attracts an apple towards itself, the apple also attracts the earth towards itself with the same force. Why then does the apple fall towards the earth, but the earth does not move towards the apple?
Answer:
The earth and the apple move towards each other, but the magnitude of the displacement of the earth is negligible relative to that of the apple. Also the observer is located on the earth.
[Note: The mass of the earth is far greater than that of an apple. Hence, the magnitude of the acceleration of the earth is negligible relative to that of the apple.]
In simple words: While the apple exerts an equal and opposite force on the Earth, the Earth's enormous mass means its acceleration (and thus displacement) due to this force is extremely small and unnoticeable compared to the apple's acceleration.

🎯 Exam Tip: This question requires applying Newton's Third Law (equal and opposite forces) along with Newton's Second Law (\(F=ma\)). Emphasize the inverse relationship between acceleration and mass to explain the negligible movement of the Earth.

 

Gravitation Class 10 Question And Answer Question 2.
The gravitational force due to the earth also acts on the moon because of which it revolves around the earth. Similar situation exists for the artificial satellites orbiting the earth. The moon and the artificial satellites orbit the earth. The earth attracts them towards itself but unlike the falling apple, they do not fall on the earth, why?
Answer:
This is because of the velocity of the moon and the satellites along their orbits. If this velocity was not there, they would have fallen on the earth.
In simple words: The Moon and satellites orbit the Earth due to their tangential velocity, which continuously causes them to "fall around" the Earth rather than directly into it. Without this orbital velocity, they would indeed fall.

🎯 Exam Tip: The key concept here is orbital velocity. Explain that the gravitational force provides the necessary centripetal force for circular motion, and the tangential velocity prevents them from falling directly to Earth. Compare it to a horizontally launched projectile that falls a great distance. This is important to score well.

 

Think about it (Text Book Page No. 8)

 

Std 10 Science Chapter 1 Gravitation Exercise Question 1.
What would happen if there were no gravity?
Answer:
There would be no gravitational attraction between any two particles and hence no formation of the solar system, galaxy, etc.
In simple words: Without gravity, objects would float freely, planets and stars wouldn't form, and the universe as we know it—with its structured solar systems and galaxies—would not exist.

🎯 Exam Tip: Focus on the fundamental role of gravity in celestial mechanics and the formation of large structures (planets, stars, galaxies). A complete absence of gravity implies no attractive force between masses.

 

Science 1 Gravitation Question 2.
What would happen if the value of G was twice as large?
Answer:
The gravitational force between any two particles would become double, also the value of g would become double.
In simple words: If 'G' were twice as large, gravitational forces between all objects would double, leading to stronger attractions and causing the acceleration due to gravity ('g') to also double.

🎯 Exam Tip: Understand that the universal gravitational constant 'G' directly influences the strength of gravitational force. If 'G' changes, all gravitational phenomena (like 'g' and orbital stability) would be affected proportionally.

 

Can you tell? (Text Book Page No. 8)

 

Gravitation Class 10 Exercise Question 1.
What would be the value of g on the surface of the earth if its mass was twice as large and its radius half of what it is now?
Answer:
\[g = \frac{GM}{R^2}\]
\[\implies g_1 = \frac{GM_1}{R_1^2} \text{ and } g_2 = \frac{GM_2}{R_2^2}\]
\[\frac{g_2}{g_1} = \frac{M_2}{M_1} \left(\frac{R_1}{R_2}\right)^2\]
Given \(M_2 = 2M_1\) and \(R_2 = \frac{R_1}{2}\)
\[\frac{g_2}{g_1} = (2) \left(\frac{R_1}{R_1/2}\right)^2 = 2(2)^2 = 8\]
\[\implies g_2 = 8g_1\]
Thus, the value of g on the surface of the earth would be eight times the present value.
In simple words: If Earth's mass doubled and its radius halved, the acceleration due to gravity ('g') on its surface would become eight times greater.

🎯 Exam Tip: This problem requires understanding the formula for 'g' and how it scales with mass and radius. Remember to square the inverse radius ratio in the proportionality. This type of problem is common for testing analytical skills.

 

Think about it (Text Book Page No. 9)

 

Std 10 Science Chapter 1 Gravitation Answers Question 1.
Will the direction of the gravitational force change as we go inside the earth?
Answer:
No.
In simple words: No, the direction of gravitational force always points towards the center of the Earth, regardless of whether you are on the surface or inside it.

🎯 Exam Tip: The direction of gravitational force is always towards the center of the mass exerting the force. This is a simple but important conceptual point often overlooked.

Use Your Brainpower! (Text Book Page No. 10)

Question 1. Will your weight remain constant as you go above the surface of the earth?
Answer: No. As we go above the surface of the earth, our weight will go on decreasing.
In simple words: Weight depends on the distance from the Earth's center; as you go higher, this distance increases, causing your weight to decrease.

🎯 Exam Tip: Understand the inverse relationship between altitude and gravitational acceleration 'g' to correctly explain changes in weight.

Question 2. Suppose you are standing on a tall ladder. If your distance from the centre of the earth is 2R, what will be your weight?
Answer: W = \( \frac{GMm}{r^2} = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2} = \frac{1}{4} (\frac{GMm}{R^2}) = \frac{1}{4} \) weight on the surface of the earth
In simple words: If your distance from the Earth's center doubles to 2R, your weight will become one-fourth of what it is on the surface because gravitational force is inversely proportional to the square of the distance.

🎯 Exam Tip: Remember that gravitational force (and thus weight) decreases quadratically with increasing distance from the center of the gravitating body.

Use Your Brain Power! (Text Book Page No. 12)

Question 1. According to Newton's law of gravitation, the earth's gravitational force is higher on an object of larger mass. Why doesn't that object fall down with higher velocity as compared to an object with lower mass?
Answer: F = ma and F = \( \frac{GMm}{r^2} \).
Acceleration, a = \( \frac{GM}{r^2} \). This is independent of the mass (m) of the object. Hence, an object of larger mass and an object of lower mass fall down with the same velocity.
In simple words: The acceleration due to gravity, 'g', is the same for all objects, regardless of their mass. This means both heavy and light objects fall at the same rate if air resistance is ignored.

🎯 Exam Tip: The acceleration due to gravity (g) is independent of the falling object's mass; it only depends on the mass of the planet and the distance from its center.

Use Your Brain Power! (Text Book Page No. 6)

Question 1. Assuming the acceleration in Example 2 above remains constant, how long will Mahendra take to move 1 cm towards Virat?
Answer: Here, u = 0
.. s = ut + \( \frac{1}{2} \) at² = 0 + \( \frac{1}{2} \) at²
.. t = \( \sqrt{\frac{2s}{a}} = \sqrt{\frac{2 \times 1 \times 10^{-2} \text{ m}}{5.34 \times 10^{-9} \text{ m/s}^2}} \)
= \( \sqrt{0.3745 \times 10^7 \text{ s}^2} = \sqrt{3.745 \times 10^6 \text{ s}} \)
= 1935 s = 32 minutes 15 seconds.
In simple words: If acceleration is constant, the time to cover a small distance can be calculated using kinematic equations, showing it takes a significant amount of time for a person to move 1 cm due to their mutual gravitational attraction.

🎯 Exam Tip: Be careful with unit conversions (cm to m) and scientific notation when performing calculations involving very small accelerations and forces.

Fill In The Blanks With Appropriate Words And Write The Completed Sentences:

Question 1. The ratio g(earth)/g(moon) is equal to........
Answer: The ratio g(earth)/g(moon) is equal to 6 (approximately.).
In simple words: The Earth's gravitational acceleration is roughly six times stronger than the Moon's, so the ratio of their 'g' values is approximately 6:1.

🎯 Exam Tip: Remember the approximate ratio of Earth's 'g' to Moon's 'g' (about 6:1) for quick estimations in problems.

Question 2. The value of the acceleration due to gravity........as we move from the equator to a pole.
Answer: The value of the acceleration due to gravity increases as we move from the equator to a pole.
In simple words: Due to Earth's slightly flattened shape, the poles are closer to the center, causing gravity to be stronger there than at the equator.

🎯 Exam Tip: Recognize that Earth's non-perfectly spherical shape causes 'g' to vary, being highest at the poles and lowest at the equator.

Question 3. If the earth shrinks to half of its radius, its mass remaining the same, the weight of an object on the earth will become........times.
Answer: If the earth shrinks to half of its radius, its mass remaining the same, the weight of an object on the earth will become four times.
In simple words: If Earth's radius is halved while its mass stays the same, the gravitational force (and thus weight) on its surface will increase fourfold because force is inversely proportional to the square of the radius.

🎯 Exam Tip: Apply the inverse square law for gravitational force and acceleration due to gravity; if radius is halved, force increases by (1/0.5)^2 = 4 times.

Question 4. The SI unit of weight is the........
Answer: The SI unit of weight is the newton.
In simple words: Weight is a force, and the standard international unit for force is the newton.

🎯 Exam Tip: Clearly distinguish between mass (kg) and weight (Newton) and their respective SI units.

Question 5. The CGS unit of weight is the........
Answer: The CGS unit of weight is the dyne
In simple words: In the CGS system, the unit for force, and thus weight, is the dyne.

🎯 Exam Tip: Be aware of both SI and CGS units for physical quantities, especially force.

Question 6. The weight of a body is ........at the poles.
Answer: The weight of a body is maximum at the poles.
In simple words: Objects weigh more at the poles because they are closer to the Earth's center compared to the equator, due to the Earth's oblate spheroid shape.

🎯 Exam Tip: Link Earth's shape (flattened poles, bulging equator) to the variation of 'g' and weight across its surface.

Question 7. Outside the earth, the weight of a body varies as........
Answer: Outside the earth, the weight of a body varies as \( \frac{1}{(R + h)^2} \).
In simple words: Above the Earth's surface, weight decreases with altitude, following an inverse square relationship with the total distance from the Earth's center.

🎯 Exam Tip: For objects outside the Earth, remember the distance 'r' in the gravitational force formula becomes (R+h), where R is Earth's radius and h is altitude.

Question 8. Due to the ........ force, the earth attracts all objects towards it.
Answer: Due to the gravitational force, the earth attracts all objects towards it. Gravitational
In simple words: The Earth pulls everything towards its center because of a fundamental interaction called gravitational force.

🎯 Exam Tip: Understand that gravity is the fundamental force responsible for objects falling towards Earth.

Question 9. The acceleration due to gravity does not depend on the ........ of the body.
Answer: The acceleration due to gravity does not depend on the mass of the body. Mass
In simple words: How fast something falls (its acceleration due to gravity) is not affected by how heavy it is, only by the mass of the planet it's falling towards.

🎯 Exam Tip: It's crucial to remember that 'g' is constant for all objects near Earth's surface, irrespective of their individual masses, assuming negligible air resistance.

Question 10. According to Kepler's first law, the orbit of a planet is ........ with the Sun at one of the........
Answer: According to Kepler's first law, the orbit of a planet is an ellipse with the Sun at one of the foci
In simple words: Kepler's first law states that planets move in oval-shaped paths (ellipses), with the Sun located at one of the two focal points of that ellipse.

🎯 Exam Tip: Memorize the three Kepler's Laws of Planetary Motion; the first law defines the shape of the orbit.

Question 11. According to Kepler's second law, the line joining the planet and the Sun ........ in equal intervals of time.
Answer: According to Kepler's second law, the line joining the planet and the Sun Sweeps equal areas in equal intervals of time.
In simple words: Kepler's second law means a planet moves faster when it's closer to the Sun and slower when it's farther away, ensuring the line connecting it to the Sun covers the same area in equal time periods.

🎯 Exam Tip: The second law implies conservation of angular momentum and explains why a planet's orbital speed varies.

Question 12. According to Kepler's third law T² \( \propto \) rⁿ, where n = ........
Answer: According to Kepler's third law T² \( \propto \) rⁿ, where n = 3
In simple words: Kepler's third law states that the square of a planet's orbital period is directly proportional to the cube of its average distance from the Sun.

🎯 Exam Tip: The relationship T² \( \propto \) r³ is fundamental for understanding orbital periods and distances in a solar system.

Question 13. For a freely falling object, we can write Newton's second equation of motion as ........
Answer: For a freely falling object, we can write Newton's second equation of motion as S = \( \frac{1}{2} \) gt²
In simple words: For an object falling freely from rest, the distance it falls can be calculated using the formula involving gravity and the square of the time taken.

🎯 Exam Tip: For free fall problems, remember the kinematic equations of motion with 'a' replaced by 'g', and 'u' often being zero for objects dropped from rest.

Question 1. (A) Write the proper answer in the square. (Practice Activity Sheet - 1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो द्रव्यमानों, m1 और m2, के बीच गुरुत्वाकर्षण बल को दर्शाता है। पहले सेट में, द्रव्यमान d दूरी पर हैं और बल F है। दूसरे सेट में, द्रव्यमान m1 और m2 हैं, लेकिन दूरी 2d है, और हमें नया बल F2 ज्ञात करना है।
If this F = x
Then F =
Answer: F = \( \frac{x}{4} \)
Explanation: F = \( G \frac{m_1 m_2}{r^2} \); \( F_1 = G \frac{m_1 m_2}{d^2} \) and
\( F_2 = G \frac{m_1 m_2}{(2d)^2} = \frac{1}{4} G \frac{m_1 m_2}{d^2} = \frac{1}{4} X \)
In simple words: Since gravitational force is inversely proportional to the square of the distance, doubling the distance reduces the force to one-fourth of its original value.

🎯 Exam Tip: Always apply the inverse square law carefully: if distance becomes 'n' times, the force becomes '1/n²' times.

Question 1. (B) Write the proper answer in the square. (March 2019)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो द्रव्यमानों, m1 और m2, के बीच गुरुत्वाकर्षण बल को दर्शाता है। पहले सेट में, द्रव्यमान d दूरी पर हैं। दूसरे सेट में, समान द्रव्यमान m1 और m2 हैं, लेकिन दूरी 3d है, और हमें नया बल F ज्ञात करना है।
If F = \( \frac{G m_1 m_2}{d^2} \)
then F =
Answer: F = \( \frac{G m_1 m_2}{9d^2} \)
In simple words: If the distance between two masses is tripled, the gravitational force between them becomes one-ninth of the original force, as per the inverse square law.

🎯 Exam Tip: Be precise with squaring the distance factor; if distance is 'n' times, the force is 1/n² times.

Choose The Correct Alternative And Rewrite The Statements:

Question 1. The gravitational force between two particles separated by a distance r varies as .......
(a) \( \frac{1}{r} \)
(b) r
(c) r²
(d) \( \frac{1}{r^2} \)
Answer: (d) \( \frac{1}{r^2} \)
In simple words: The gravitational force follows an inverse square law, meaning it is proportional to one divided by the square of the distance between the objects.

🎯 Exam Tip: The inverse square relationship (\(1/r^2\)) is a key characteristic of fundamental forces like gravity and electrostatic force; memorize it.

Question 2. In the usual notation, the acceleration due to gravity at a height h from the surface of the earth is ........
(a) \( g = \frac{GM}{(R+h)} \)
(b) \( g = \frac{GM}{\sqrt{R+h}} \)
(c) \( g = \frac{GM}{(R+h)^2} \)
(d) \( g = GM (R+h)^2 \)
Answer: (c) \( g = \frac{GM}{(R+h)^2} \)
In simple words: The acceleration due to gravity at an altitude 'h' above the Earth's surface is found by using the total distance from the Earth's center (R+h) in the inverse square law formula.

🎯 Exam Tip: When calculating 'g' at a height 'h', remember that the distance 'r' in the formula is 'R+h', not just 'h'.

Question 3. The SI unit of the universal constant of gravitation is ........
(a) N.m²/kg²
(b) N.kg²/m²
(c) m/s²
(d) kg.m/s²
Answer: (a) N.m²/kg²
In simple words: The SI unit for the gravitational constant (G) is derived from Newton's law of gravitation and is expressed as Newton-meter squared per kilogram squared.

🎯 Exam Tip: Derive the unit of 'G' from the formula F = \( G \frac{m_1 m_2}{r^2} \) to avoid memorization errors during exams.

Question 4. The escape velocity of a body from the earth's surface, Vsec = ........
(a) \( \sqrt{\frac{GM}{R}} \)
(b) \( 2 \sqrt{\frac{GM}{R}} \)
(c) \( \sqrt{\frac{2GM}{R}} \)
(d) \( \sqrt{\frac{GM}{2R}} \)
Answer: (c) \( \sqrt{\frac{2GM}{R}} \)
In simple words: Escape velocity is the minimum speed needed to break free from a planet's gravity, and its formula involves the square root of twice the gravitational constant times the planet's mass, divided by its radius.

🎯 Exam Tip: Memorize the formula for escape velocity and understand that it depends on the mass and radius of the celestial body, not the mass of the escaping object.

Question 5. How much will a person with 72 N weight on the earth, weigh on the moon? (Practice Activity Sheet-1)
(a) 12 N
(b) 36 N
(c) 21 N
(d) 63 N
Answer: (a) 12 N
In simple words: Since the Moon's gravity is about one-sixth of Earth's, a person weighing 72 N on Earth will weigh approximately 12 N on the Moon.

🎯 Exam Tip: For problems involving weight on different celestial bodies, remember the approximate ratio of gravitational acceleration (g) for the Earth and Moon (Earth g \( \approx \) 6 x Moon g).

Question 6. What will be the weight of a person on the earth, who weighs 9N on the moon? (Practice Activity Sheet – 2)
(a) 3 N
(b) 15 N
(c) 45 N
(d) 54 N
Answer: (d) 54 N
In simple words: If someone weighs 9 N on the Moon, where gravity is one-sixth of Earth's, they would weigh six times more, or 54 N, on Earth.

🎯 Exam Tip: Use the inverse of the gravitational ratio; if Moon's gravity is 1/6th of Earth's, then Earth's gravity is 6 times that of the Moon.

State Whether The Following Statements Are True Or False: (If A Statement Is False, Correct It And Rewrite It.)

Question 1. If the separation between two particles is doubled, the gravitational force between the particles becomes half the initial force.
Answer: False. (If the separation between two particles is doubled, the gravitational force between the particles becomes \( \frac{1}{4} \) times the initial force.)
In simple words: Doubling the distance between two objects reduces the gravitational force to one-fourth, not one-half, because gravity follows an inverse square law.

🎯 Exam Tip: Always remember the inverse square law for gravitational force (\(F \propto 1/r^2\)); if 'r' is doubled, 'F' becomes 1/4th.

Question 2. The CGS unit of the universal constant or gravitation is the dyne cm²/gram²?
Answer: True.
In simple words: Yes, the CGS unit for the universal gravitational constant is dyne-centimeter squared per gram squared, reflecting the force, distance, and mass units in the CGS system.

🎯 Exam Tip: Know both SI (N.m²/kg²) and CGS (dyne.cm²/g²) units for the universal gravitational constant 'G'.

Question 3. At the centre of the earth, the value of the acceleration due to gravity becomes zero.
Answer: True.
In simple words: At the very center of the Earth, the gravitational pulls from all directions cancel each other out, making the net acceleration due to gravity zero.

🎯 Exam Tip: Understand that inside a uniform spherical mass, the gravitational force (and 'g') decreases linearly towards the center, becoming zero right at the center.

Question 4. The weight of a body is minimum at the poles.
Answer: False. (The weight of a body is maximum at the poles.)
In simple words: This statement is false because a body's weight is actually maximum at the poles, not minimum, due to the Earth's flattened shape bringing the poles closer to the center.

🎯 Exam Tip: Connect the Earth's oblate spheroid shape (flattened at poles, bulging at equator) to the variation of 'g' and weight: 'g' is maximum at poles, minimum at equator.

Question 5. Mass is a vector quantity.
Answer: False. (Mass is a scalar quantity.)
In simple words: Mass only has a magnitude (how much "stuff"), not a direction, making it a scalar quantity.

🎯 Exam Tip: Differentiate between scalar quantities (magnitude only, e.g., mass, speed) and vector quantities (magnitude and direction, e.g., weight, velocity).

Question 6. weight is a vector quantity.
Answer: True.
In simple words: Weight is a force that acts in a specific direction (towards the center of gravity), so it is a vector quantity.

🎯 Exam Tip: Remember that weight is a force, and all forces are vector quantities, possessing both magnitude and direction.

Question 7. g has maximum value at the equator.
Answer: False. (g has maximum value at the poles.)
In simple words: This is false; the acceleration due to gravity ('g') is strongest at the poles and weakest at the equator because the poles are closer to the Earth's center.

🎯 Exam Tip: Reinforce the understanding that Earth's rotation and shape cause 'g' to be highest at poles and lowest at the equator.

Question 8. Outside the earth, g varies as 1/(R + h)².
Answer: True.
In simple words: Correct, outside Earth, the acceleration due to gravity ('g') decreases inversely with the square of the total distance from the Earth's center, which is the sum of Earth's radius (R) and the altitude (h).

🎯 Exam Tip: When dealing with 'g' at altitude, always use the total distance from the center, (R+h), and apply the inverse square law.

Question 9. The value of G changes from place to place.
Answer: False. (The value of G is the same throughout the universe.)
In simple words: This is false because the universal gravitational constant 'G' is a fundamental constant of nature and has the same value everywhere in the universe.

🎯 Exam Tip: Distinguish between 'G' (universal constant, always the same) and 'g' (acceleration due to gravity, varies by location and celestial body).

Question 10. The value of g increases with altitude.
Answer: False. (The value of g decreases with altitude.)
In simple words: This is false; as you go higher, the acceleration due to gravity ('g') actually decreases because you are moving further away from the Earth's center.

🎯 Exam Tip: Remember that 'g' decreases with altitude and depth (except for the surface value); it's highest on the surface.

Question 11. The escape velocity of a body does not depend on the mass of the body.
Answer: True.
In simple words: Yes, escape velocity only depends on the mass and radius of the planet, not on the mass of the object trying to escape its gravity.

🎯 Exam Tip: A common misconception is that heavier objects need more velocity to escape; clarify that escape velocity is independent of the escaping object's mass.

Question 12. The mass of a body is the amount of matter present in it.
Answer: True.
In simple words: Mass is a measure of the amount of substance in an object, a fundamental property that remains constant regardless of location.

🎯 Exam Tip: Define mass accurately as an intrinsic property of matter, distinct from weight which is a force.

Match The Following:

Question. Match the following :

Column A	Column B
(1) Escape velocity	(a) \( -\frac{GMm}{R+h} \)
(2) Gravitational acceleration	(b) \( \sqrt{\frac{2GM}{R}} \)
(3) Gravitational potential energy	(c) \( \frac{Gm_1m_2}{r^2} \)
(4) Gravitational force	(d) \( \frac{GM}{r^2} \) (r \( \ge \) R)
	(e) \( -\frac{GMm}{2(R+h)} \)

Answer:(1) Escape velocity : \( \sqrt{\frac{2GM}{R}} \)
(2) Gravitational acceleration : \( \frac{GM}{r^2} \) (r \( \ge \) R)
(3) Gravitational potential energy : \( -\frac{GMm}{R+h} \)
(4) Gravitational force : \( \frac{Gm_1m_2}{r^2} \)
In simple words: This matching exercise correctly pairs key gravitational concepts like escape velocity, gravitational acceleration, potential energy, and force with their corresponding mathematical formulas.

🎯 Exam Tip: Thoroughly understand and memorize the formulas for each gravitational concept as they are frequently tested in direct and problem-solving questions.

Answer The Following Questions In One Sentence Each:

Question 1. State the SI and CGS units of G.
Answer: The SI unit of G is N.m²/kg² and CGS unit is the dyne.cm²/g².
In simple words: The SI unit for the universal gravitational constant is Newton-meter squared per kilogram squared, while its CGS unit is dyne-centimeter squared per gram squared.

🎯 Exam Tip: Be prepared to state units for physical constants in both SI and CGS systems, ensuring correct notation.

Question 2. State any one characteristic of gravitational force.
Answer: Gravitational force between two particles does not depend on the nature of the medium between them.
In simple words: A key feature of gravity is that its strength between two objects isn't affected by what substance is in between them.

🎯 Exam Tip: Understand that gravitational force is a universal force that acts through any medium, unlike some other forces.

Question 3. Name the force that keeps a satellite in the orbit around the earth.
Answer: The gravitational force due to the earth keeps a satellite in the orbit around the earth.
In simple words: Earth's gravitational pull is the force that continuously keeps satellites orbiting around it.

🎯 Exam Tip: Recognize that gravity acts as the centripetal force required for orbital motion around any celestial body.

Question 4. Name the force due to which the earth revolves around the Sun.
Answer: The earth revolves around the Sun due to the gravitational force of attraction exerted on it by the Sun.
In simple words: The Sun's powerful gravitational pull is what keeps Earth, and all other planets, moving in their orbits around it.

🎯 Exam Tip: Understand that all orbital motions in a solar system are governed by the gravitational attraction between the central star and the orbiting bodies.

Question 5. What is the acceleration due to gravity at a height h ( = radius of the earth) from the surface of the earth? (g = 9.8 m/s²)
Answer: The acceleration due to gravity at a height h ( = radius of the earth) from the surface of the earth is 2.45 m/s².
[Explanation : g' = \( \frac{GM}{(R+h)^2} = \frac{GM}{4R^2} = \frac{g}{4} \)
= \( \frac{9.8}{4} \) m/s² = 2.45 m/s² for h = R ]
In simple words: When you go up to a height equal to Earth's radius, the acceleration due to gravity drops to one-fourth of its surface value, becoming 2.45 m/s².

🎯 Exam Tip: When h = R, the total distance from the center is 2R, so the gravitational acceleration becomes \(g/4\). Remember this specific case.

Question 6. What is the relation between the SI unit of weight and the CGS unit of weight?
Answer: The relation between the SI unit of weight (the newton) and the CGS unit of weight (the dyne) is 1 newton = 10\(^5\) dynes.
In simple words: One Newton, the standard unit of weight, is equivalent to 100,000 dynes, which is the weight unit in the CGS system.

🎯 Exam Tip: Know common unit conversions, especially between SI and CGS systems, like the Newton-dyne relationship.

Question 7. Write the formula for the centripetal force acting on a body performing circular motion.
Answer: F = \( \frac{mv^2}{r} \)
In simple words: Centripetal force, which keeps an object moving in a circle, is calculated by multiplying its mass by the square of its speed and then dividing by the radius of the circular path.

🎯 Exam Tip: The formula F = \( \frac{mv^2}{r} \) is crucial for any problem involving circular motion and centripetal force; commit it to memory.

Question 8. Write the formula for the escape velocity of a body from the earth's surface.
Answer: V\(_{esc}\) = \( \sqrt{\frac{2GM}{R}} \) or v\(_{esc}\) = \( \sqrt{2gR} \).
In simple words: The formula for escape velocity involves the square root of twice the gravitational constant times the Earth's mass divided by its radius, or simply the square root of twice 'g' times Earth's radius.

🎯 Exam Tip: Be familiar with both forms of the escape velocity formula, one using G and M, the other using g and R, and know when to apply each.

Question 9. What is the value of the acceleration due to gravity at the centre of the earth?
Answer: Zero.
In simple words: At Earth's core, the acceleration due to gravity is zero because gravitational forces from all directions perfectly balance each other out.

🎯 Exam Tip: Remember this specific condition: g = 0 at the Earth's center.

Question 10. What are the factors on which the maximum height attained by a body thrown upwards depends?
Answer: The initial velocity of the body, the acceleration due to gravity at that place, the buoyant force and frictional force due to air.
In simple words: The highest point an object reaches when thrown upwards is determined by how fast it's thrown, the local gravity, and any air resistance or buoyancy acting on it.

🎯 Exam Tip: For projectile motion, initial velocity and gravitational acceleration are primary factors, but acknowledge air resistance and buoyant force in real-world scenarios.

Some Of The Important Terms In Chapter Gravitation Are Given In The Following Box. Find Them:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक शब्द पहेली (word search) है जिसमें गुरुत्वाकर्षण अध्याय से संबंधित महत्वपूर्ण शब्द छिपे हुए हैं। छात्रों को इस बॉक्स में 'centripetal force', 'escape velocity', 'periodic time' और 'gravitational constant' जैसे शब्द ढूंढने हैं।
Answer:
(1) centripetal force
(2) escape velocity
In simple words: This is a word search puzzle where you need to find key terms related to gravitation, such as centripetal force and escape velocity, hidden within the grid of letters.

🎯 Exam Tip: Familiarize yourself with key terminology from each chapter to quickly identify and understand concepts.

Question. (Continued)
(3) periodic time
(4) gravitational constant.
In simple words: This completes the list of hidden terms from the word search, adding "periodic time" and "gravitational constant" to the found terms.

🎯 Exam Tip: Ensure you understand the definitions and significance of all key terms from the chapter, not just their names.

Answer The Following Questions:

Question 1. What is centripetal force?
(OR)
Define centripetal force.
Answer: In uniform circular motion of a body, the force acting on the body is directed towards the centre of the circle. This force is called centripetal force.
In simple words: Centripetal force is the inward-directed force that keeps an object moving in a circular path.

🎯 Exam Tip: Clearly state that centripetal force is always directed towards the center of the circular path and is essential for maintaining circular motion.

Question 2. Give one example of centripetal force.
Answer: The moon revolves around the earth due to the gravitational force exerted on it by the earth. This force is directed towards the centre of the earth and is thus a centripetal force.
In simple words: The Earth's gravity pulling on the Moon is an example of centripetal force, causing the Moon to orbit the Earth.

🎯 Exam Tip: Use real-world examples like planets orbiting the Sun, satellites orbiting Earth, or a stone whirled on a string to illustrate centripetal force.

Question 3. Name the source responsible for the motion of a planet around the Sun.
Answer: A planet revolves around the Sun due to the gravitational force exerted on it by the Sun.
In simple words: The Sun's gravitational force is what makes planets orbit around it.

🎯 Exam Tip: Gravitational force is the primary force responsible for celestial mechanics, including orbital motion.

Answer The Following Questions:

Question 1. In the following figure, an orbit of a planet around the Sun (S) has been shown. AB and CD are the distances covered by the planet in equal time. Lines AS ad CS sweep equal areas in equal intervals of time. Hence, areas ASB and CSD are equal. (Practice Activity Sheet-1)
(a) Which laws do we understand from the above description?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सूर्य (S) के चारों ओर एक ग्रह की अण्डाकार कक्षा को दर्शाता है। इसमें दिखाया गया है कि ग्रह समान समय अंतराल में दूरियां AB और CD तय करता है। रेखाएं AS और CS समान समय अंतराल में समान क्षेत्रफल (ASB और CSD) तय करती हैं।
In simple words: The diagram illustrates Kepler's Second Law, which states that a planet sweeps out equal areas in equal times as it orbits the Sun, meaning it moves faster when closer to the Sun.

🎯 Exam Tip: Identify Kepler's Second Law from descriptions or diagrams showing equal areas swept in equal times; it's a direct application of the law of areas.

Question 1. (Continued)
(b) Write the law regarding the area swept.
(c) Write the law T² \( \propto \) r³ in your words.
Answer:
(a) From the given description we understand Kepler's three laws.
(b) Kepler's law of areas: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
(c) Kepler's law of periods: The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.
In simple words: (b) Kepler's second law, the Law of Areas, states that a line from a planet to the Sun sweeps equal areas in equal time intervals. (c) Kepler's third law, the Law of Periods, says that the square of a planet's orbital period is directly proportional to the cube of its average distance from the Sun.

🎯 Exam Tip: When asked to state Kepler's laws in words, ensure precise language, especially distinguishing between 'mean distance' for the third law and 'equal areas in equal times' for the second.

Question 2. Identify the law shown in Fig. 1.7 and state the three respective laws. (Practice Activity Sheet - 3)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सूर्य (S) के चारों ओर एक ग्रह की अण्डाकार कक्षा को दर्शाता है। इसमें विभिन्न बिंदुओं पर ग्रह की स्थिति और सूर्य से उनकी दूरी को दर्शाया गया है, साथ ही ग्रह द्वारा समान समय अंतराल में तय किए गए क्षेत्रफल (ASB, CSD, ESF) भी दिखाए गए हैं, जो समान हैं।
Answer:
(a) From the given description we understand Kepler's three laws.
(b) Kepler's law of areas: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
(c) Kepler's law of periods: The square of the period of revolution or a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.
In simple words: The diagram visually represents Kepler's laws of planetary motion. Specifically, the equal swept areas demonstrate Kepler's Law of Areas, and the question implicitly asks for all three laws, including the Law of Orbits and the Law of Periods.

🎯 Exam Tip: Practice identifying each of Kepler's laws from visual representations and be ready to explain all three laws in detail.

Question 3. Explain the term gravitational force. What is gravitation?
Answer: There exists a force of attraction between any two particles of matter in the universe such that the force depends only on the masses of the particles and the separation between them. It is called the gravitational force and the mutual attraction is called gravitation.
In simple words: Gravitational force is the attractive force between any two objects with mass, dependent on their masses and the distance between them. Gravitation is the general phenomenon of this mutual attraction.

🎯 Exam Tip: Clearly define both 'gravitational force' (the force itself) and 'gravitation' (the broader phenomenon) and their dependence on mass and distance.

Question 4. State Newton's universal law of gravitation. Express it in mathematical form.
Answer: Newton's universal law of gravitation :
Every object in the Universe attracts every other object with a definite force. This force is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them.
Mathematical form: Consider two objects of masses m\( _1 \) and m\( _2 \). We assume that the objects are very small spheres of uniform density and the distance r between their centers is very large compared to the radii of the spheres (Fig. 1.8).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वस्तुओं, m1 और m2, के बीच गुरुत्वाकर्षण बल (F) को दर्शाता है। बल आकर्षण प्रकृति का है और दोनों द्रव्यमानों के केंद्रों को जोड़ने वाली रेखा के अनुदिश कार्य करता है, जिनके बीच की दूरी r है। The magnitude (F) of the gravitational force of attraction between the objects is directly proportional to m\( _1 \)m\( _2 \) and inversely proportional to r²
.. F \( \propto \frac{m_1 m_2}{r^2} \)
.. F = \( G \frac{m_1 m_2}{r^2} \)
where G is the constant of proportionality, called the universal gravitational constant.
[Note: In the textbook, the word object/body is used.
Newton's law of gravitation applies to particles.]
In simple words: Newton's Law of Gravitation states that every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, it's F = \( G \frac{m_1 m_2}{r^2} \).

🎯 Exam Tip: Memorize Newton's universal law of gravitation, both its statement and its mathematical formula, and understand the significance of the universal gravitational constant G.

 

Question 5.

(i) Why is the constant of gravitation called a universal constant?
(ii) Newton's law of gravitation is called the universal law of gravitation. Why?
Answer:
(i) The value of the constant of gravitation does not change with the nature, mass or the size of the material particles. It does not vary with the distance between the two particles. It is also independent of the nature of the medium between the two particles. Hence, it is called a universal constant.

(ii) As the law of gravitation given by Newton is applicable throughout the universe and to all particles, it is called universal law.
The centre of mass of an object is the point inside or outside the object at which the total mass of the object can be assumed to be concentrated to study the effect of an applied force. The centre of mass of a spherical object having uniform density is at its geometrical centre. The centre of mass of an object having uniform density is at its centroid. If the two bodies are spherical and of uniform density, the gravitational force between them is always along the line joining the centres of the two bodies and the distance between the centres is taken to be r. When the bodies are not spherical or have irregular shape or have nonuniform density, the force is along the line joining their centres of mass and r is taken to be the distance between the two centres of mass.
In simple words: The gravitational constant (G) is universal because its value never changes, regardless of the objects involved or the medium between them. Newton's law of gravitation is universal because it applies to all particles everywhere in the universe, explaining how any two objects attract each other.

🎯 Exam Tip: Understand the difference between 'universal constant' and 'universal law' to score full marks, focusing on G's unchanging nature and the law's broad applicability.

 

Question 6.
If the distance between two bodies is increased by a factor of 5, (i) by what factor will the gravitational force change if the masses are kept constant? (ii) by what factor will the mass of one of them have to be altered, keeping the other mass the same, to maintain the same gravitational force between the two bodies?
Answer:
If the distance between two bodies is increased by a factor of 5,
(i) the gravitational force between the bodies will decrease by a factor of 25 if the masses of the bodies are kept constant.
(ii) the mass of one of them will have to be increased by a factor of 25, keeping the mass of the other body the same, to maintain the same gravitational force between the two bodies.
Gravitational force \( F \propto \frac{1}{r^2} \) and \( F \propto m_1m_2 \).
In simple words: If you move two objects 5 times farther apart, the gravitational force between them becomes 25 times weaker. To keep the force the same, you'd need to make one of the objects 25 times more massive.

🎯 Exam Tip: Remember Newton's Law of Gravitation: force is inversely proportional to the square of the distance and directly proportional to the product of masses. This relationship is crucial for solving such problems.

 

Question 7.
(i) Determine the SI unit of the universal constant of gravitation from the formula for the gravitational force between two particles. Hence, state the CGS unit of the constant of gravitation. (ii) Define G (universal gravitational constant).
Answer:
(i) According to Newton's law of gravitation, the gravitational force between two particles is
\( F = G \frac{m_1m_2}{r^2} \)
where \( m_1 \) and \( m_2 \) are the masses of the two particles, r is the distance between them and G is the universal constant of gravitation.

\( \implies G = \frac{Fr^2}{m_1m_2} \)
The SI unit of force is the newton (N), that of distance is the metre (m) and that of mass is the kilogram (kg).
The SI unit of G is \( \frac{N \cdot m^2}{kg^2} \).
The CGS unit of G is \( \frac{dyne \cdot cm^2}{g^2} \).

(ii) If we take \( m_1 = m_2 = \) unit mass and r = unit distance, numerically, G = F, i.e., G (universal gravitational constant) represents the magnitude of the gravitational force of attraction between two unit masses, separated by a unit distance.
Newton's law of gravitation applies to particles.
In simple words: The SI unit for G is N.m\(^2\)/kg\(^2\), and the CGS unit is dyne.cm\(^2\)/g\(^2\). G itself is the force of attraction between two unit masses (1 kg each in SI) separated by a unit distance (1 meter in SI).

🎯 Exam Tip: Knowing the formula for gravitational force allows you to derive the units of G. Understanding its definition as a force between unit masses at unit distance is key.

 

Question 8.
State the importance of Newton's universal law of gravitation.
Answer:
The importance of Newton's universal law of gravitation :
This law explains successfully, i.e., with great accuracy,
• The force that binds the objects on the earth to the earth
• The motion of the moon and artificial satellites around the earth
• The motion of the planets, asteroids, comets, etc., around the Sun
• The tides of the sea due to the moon and the Sun.
In simple words: Newton's law of gravitation is important because it explains fundamental phenomena like why things fall to Earth, why the moon orbits Earth, how planets move around the Sun, and why we have ocean tides.

🎯 Exam Tip: Listing at least three key applications of Newton's law of gravitation is usually sufficient for full marks on this type of question.

 

Question 9.
Compare the gravitational force on a body of mass 1 kg due to the earth with the force on the same body due to another body of mass 1 kg at a distance of 1 m from the first body. (Mass of the earth = 6 × 10\(^{24}\) kg, radius of the earth = 6400 km)
Answer:
In the first case, \( m_1 = 1kg \),
\( m_2 = 6 \times 10^{24} \text{ kg} \) and \( r = 6400 \text{ km} = 6.4 \times 10^6 \text{ m} \)
The gravitational force on the body.
\( F_1 = \frac{Gm_1m_2}{r^2} = \frac{G \times 1 \text{ kg} \times 6 \times 10^{24} \text{ kg}}{(6.4 \times 10^6 \text{ m})^2} \)
\( = \frac{G \times 6 \times 10^{24} \text{ kg}^2}{(6.4)^2 \times 10^{12} \text{ m}^2} \)
In second case, \( m_1 = 1 \text{ kg} \)
\( m_2 = 1 \text{ kg} \) and \( r = 1 \text{m} \)
Gravitational force on the body,
\( F_2 = \frac{Gm_1m_2}{r^2} = \frac{G \times 1 \text{ kg} \times 1 \text{ kg}}{(1 \text{ m})^2} \)

\( \implies \frac{F_1}{F_2} = \frac{6 \times 10^{24}}{(6.4)^2 \times 10^{12}} = 1.465 \times 10^{11} \)
OR
\( \frac{F_2}{F_1} = 6.826 \times 10^{-12} \).
Thus, \( F_2 \ll F_1 \).
In simple words: The gravitational force exerted by the Earth on a 1 kg object is vastly greater (about 1.465 x 10\(^{11}\) times) than the gravitational force between two 1 kg objects placed 1 meter apart. This shows how significant Earth's large mass is for its gravitational pull.

🎯 Exam Tip: Pay close attention to units and powers of ten in calculations. Always state your conclusion clearly, comparing the magnitudes as requested.

 

Question 10.
Explain the term the earth's gravitational force.
(OR)
Write a short note on the earth's gravitational force.
Answer:
The earth attracts every object towards it because of the gravitational force. As the earth's centre of mass is at its centre, the gravitational force exerted by the earth on an object is directed towards the earth's centre. Hence, an object released from a point above the earth's surface falls vertically downward towards the earth.
If an object is thrown vertically upward, its velocity goes on decreasing due to the earth's gravitational force on the object. At one stage, the velocity of the body becomes zero and later the body falls back to the earth.
In simple words: Earth's gravitational force is the pull it exerts on all objects, always directed towards its center. This force causes objects to fall downwards and slows down objects thrown upwards, eventually pulling them back.

🎯 Exam Tip: When explaining Earth's gravitational force, emphasize its direction (towards the center) and its effect on both falling and thrown objects.

 

Question 11.
Take two balls of different masses, go to the top of a building, drop them simultaneously and observe what happens to the balls.
Answer:
The balls reach the ground almost at the same time.
In simple words: When two balls of different masses are dropped from the same height, they hit the ground at nearly the same time because gravity accelerates all objects equally, regardless of their mass (ignoring air resistance).

🎯 Exam Tip: This experiment demonstrates Galileo's principle of falling bodies, which states that in a vacuum, all objects fall with the same acceleration. Air resistance can cause slight differences in real-world scenarios.

 

Question 12.
Take two similar pages from your notebook. Crumple one paper and allow this and the other paper to fall on the ground simultaneously. What do you observe?
Answer:
The crumpled paper reaches the ground before the other one.
In simple words: The crumpled paper falls faster than the flat paper because crumpling reduces air resistance, allowing it to accelerate more freely due to gravity.

🎯 Exam Tip: This experiment highlights the effect of air resistance on falling objects. The key factor isn't mass but the shape and surface area exposed to air.

 

Question 13.
Take a feather and a paper. Allow them to fall to the ground simultaneously. Which will reach the ground earlier? Why?
Answer:
There is no unique answer. It depends on the feather and paper. Upthrust due to air and force due to friction with air play very important roles here. The acceleration of a body depends on the resultant of the earth's gravitational force on the body and the upthrust and the force of friction due to air.
In simple words: The object that reaches the ground earlier depends on how much air resistance each object experiences. While gravity pulls both equally, air resistance and upthrust can significantly affect their acceleration, making one fall faster than the other.

🎯 Exam Tip: For objects like feathers and paper, air resistance and upthrust are critical factors. Emphasize that in a vacuum, all objects would fall at the same rate.

 

Question 14.
From Newton's law of gravitation, derive the formula for the acceleration due to gravity.
Answer:
Suppose that a body of mass m is released from a distance r from the centre (O) of the earth (Fig. 1.9). Let M be the mass of the earth. According to Newton's law of gravitation, the magnitude of the earth's gravitational force acting on the body Is
\( F = G \frac{Mm}{r^2} \)
where G is the universal constant of gravitation.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पृथ्वी द्वारा किसी वस्तु पर लगाए गए गुरुत्वाकर्षण बल को दर्शाता है। इसमें पृथ्वी (Earth) का केंद्र O, उसकी त्रिज्या R, वस्तु का द्रव्यमान m और पृथ्वी के केंद्र से वस्तु की दूरी r दिखाई गई है। F वह गुरुत्वाकर्षण बल है जो वस्तु पर कार्य कर रहा है।
The acceleration produced by this force, force F
\( g = \frac{\text{force}}{\text{mass}} = \frac{F}{m} \)

\( \implies g = \frac{GM}{r^2} \)
This is the formula for the acceleration due to gravity or the gravitational acceleration due to the earth. This acceleration is directed towards the earth's centre.
If h denotes the altitude, \( r = R + h \), where R is the radius of the earth.

\( \implies g = \frac{GM}{(R+h)^2} \)
For a body on the earth's surface, \( h = 0 \).

\( \implies g = \frac{GM}{R^2} \)
When we consider the gravitational interaction between the earth and a body on the surface of the earth or at some height above the surface of the earth, for many practical purposes we can assume that the earth behaves as if its mass were concentrated at the earth's centre. The proof is not expected here.
In simple words: The acceleration due to gravity (g) is derived by equating Newton's second law (\( F=mg \)) with his law of universal gravitation (\( F=G\frac{Mm}{r^2} \)). This yields \( g = G\frac{M}{r^2} \), showing that 'g' depends on the mass of the Earth (M) and the distance from its center (r), but not on the mass of the falling object (m).

🎯 Exam Tip: Clearly show the steps of equating \( F=ma \) (where \( a=g \)) and \( F=G\frac{Mm}{r^2} \) to derive \( g = G\frac{M}{r^2} \). Define all variables used in the derivation.

 

Question 15.
Explain the factors affecting the value of g.
Answer:
The value of the acceleration due to gravity, g, changes from place to place on the earth. It also varies with the altitude and depth below the earth's surface. The factors affecting the value of g are the shape of the earth, altitude and depth below the earth's surface.

(1) The earth is not perfectly spherical. It is somewhat flat at the poles and bulging at the equator. At the surface of the earth, the value of g is maximum (9.832 m/s\(^2\)) at the poles as the polar radius is minimum, while it is minimum (9.78 m/s\(^2\)) at the equator as the equatorial radius is maximum.

(2) As the height (h) above the earth's surface increases, the value of g decreases. It varies as \( \propto \frac{1}{(R+h)^2} \), where R is radius of the earth.

(3) In the interior of the earth, on average, the value of g is less than that at the earth's surface. As the depth below the earth's surface increases, the value of g decreases and finally it becomes zero at the centre of the earth.
In simple words: The value of 'g' is affected by three main factors: Earth's non-spherical shape (maximum at poles, minimum at equator), altitude (decreases with height above surface), and depth (decreases with depth below surface, becoming zero at the center).

🎯 Exam Tip: Remember the three main factors: shape, altitude, and depth. For each, describe how 'g' changes and briefly explain why (e.g., radius variation for shape, increased distance for altitude).

 

Question 16.
If \( g = GM/r^2 \), then where will the value of g be high, at Goa Beach or on the top of the Mount Everest?
Answer:
The value of g will be high at Goa Beach.
In simple words: The value of 'g' is higher at Goa Beach compared to Mount Everest because Goa Beach is closer to the Earth's center (lower altitude), and 'g' decreases as altitude increases.

🎯 Exam Tip: Understand that gravitational acceleration 'g' is inversely proportional to the square of the distance from the center of the Earth. Therefore, lower altitudes mean higher 'g'.

 

Question 17.
Why does an object released from the hand, fall on the earth?
Answer:
When an object is held in the hand, the gravitational force acting on the object due to the earth is balanced by the person holding the object. When the object is released from the hand, it falls on the earth due to the earth's gravitational force.
In simple words: An object falls to Earth when released because the Earth's gravitational force pulls it downwards. When held, this force is balanced by your hand, but once released, gravity acts unopposed.

🎯 Exam Tip: Focus on the concept of unbalanced force. When the hand's upward force is removed, only gravity acts, causing acceleration towards Earth.

 

Question 18.
Does the value of g depend on the mass of the falling body? Why?
Answer:
The value of g does not depend on the mass of the falling body.
The reason is the gravitational force on a body due to the earth is directly proportional to the mass of the body and for a given force, the acceleration of a body is inversely proportional to the mass of the body.
In simple words: No, the value of 'g' does not depend on the mass of the falling object. This is because a more massive object experiences a greater gravitational force, but it also has greater inertia, so the increased force exactly balances the increased resistance to acceleration.

🎯 Exam Tip: Explain that while gravitational force is proportional to the falling object's mass, acceleration (\( a = F/m \)) remains constant because 'm' cancels out in the equation \( g = G\frac{M}{R^2} \).

 

Question 19.
Define mass. State its SI and CGS units.
Answer:
The mass of a body is the amount of matter present in it. Its SI unit is the kilogram (kg) and CGS unit is the gram (g).
Mass has only magnitude, not direction. Thus, it is a scalar quantity.
In simple words: Mass is the amount of 'stuff' or matter an object contains. Its SI unit is kilograms (kg), and its CGS unit is grams (g).

🎯 Exam Tip: Clearly state that mass is a scalar quantity and provide both SI and CGS units for complete accuracy.

 

Question 20.
Define weight. State its SI and CGS units.
Answer:
The weight of a body is defined as the force with which the earth attracts it. Its SI unit is the newton (N) and CGS unit is the dyne.
In the usual notation, the magnitude of the weight of a body on the earth's surface is \( W = \frac{GmM}{R^2} = m \frac{GM}{R^2} = mg \). Thus, \( W \propto g \). Hence, weight varies just like the acceleration due to gravity. It is maximum at the poles and minimum at the equator. It decreases with altitude (h) and depth (d) below the earth's surface. It becomes zero at the earth's centre. At a height above the earth's surface, \( W = \frac{GmM}{(R+h)^2} \). At a depth d below the earth's surface, \( W = \frac{GmM(R-d)}{R^3} \). Weight has magnitude and direction (towards the earth's centre). It is a vector quantity.
In simple words: Weight is the force of gravity pulling on an object. Its SI unit is newtons (N), and its CGS unit is dynes. Weight changes based on location because it depends on the local acceleration due to gravity, which varies.

🎯 Exam Tip: Emphasize that weight is a force (vector quantity) and provide both SI and CGS units. Mention how its value varies with location due to 'g'.

 

Question 21.
As per the request of one of his friends from the equator, Rahul buys 100 grams of silver at the north pole. He hands it over to his friend at the equator. Will the friend agree with the weight of the silver bought? If not, why?
Answer:
The weight of a body is given by \( W = mg \), where m is the mass of the body and g is the acceleration due to gravity, g varies from place to place. The value of g at the equator is less than that at the north pole (as well as the south pole). Hence, the weight of the silver bought at the north pole would be less when the silver is weighed at the equator. Therefore, Rahul's friend will disagree about the weight of the silver.
The mass being independent of the value of g, Rahul's friends will agree about the mass of the silver.
In simple words: Rahul's friend will likely disagree about the *weight* of the silver. Since 'g' is less at the equator than at the poles, the silver will weigh less there, even though its mass (amount of silver) remains the same.

🎯 Exam Tip: This question tests the distinction between mass and weight. Remember that mass is constant, while weight varies with 'g', which is affected by Earth's shape (poles vs. equator).

 

Question 22.
What is the difference between mass and weight of an object? Will the mass and weight of an object on the earth be the same as their values on Mars? Why?
Answer:
The mass of an object is the amount of matter present in it. It is same everywhere in the Universe and is never zero. It is a scalar quantity and its SI unit is kg. The weight of an object is the force with which the earth (or any other planet/ moon/star) attracts it. It is directed towards the centre of the earth. The weight of an object is different at different places on the earth. It is zero at the earth's centre. It is a vector quantity and its SI unit is the newton (N). The magnitude of weight = mg.
The mass of an object will be the same on the earth and Mars, but the weight will not be the same because the value of g on Mars is different from that on the earth.
In simple words: Mass is the amount of matter, constant everywhere (scalar, SI unit kg). Weight is the force of gravity, varies by location (vector, SI unit N). An object's mass is the same on Earth and Mars, but its weight will differ because Mars has a different gravitational acceleration 'g'.

🎯 Exam Tip: Clearly differentiate between mass (intrinsic property, scalar, constant) and weight (gravitational force, vector, variable). Emphasize why weight changes on different celestial bodies due to varying 'g'.

 

Question 23.
Explain the term free fall and state the corresponding kinematical equations of motion in the usual notation.
Answer:
When a body falls in air, there are three forces acting on the body : (1) the gravitational force due to the earth, acting downward (2) the force of buoyancy (upthrust) due to air, acting upward (3) the force due to friction with air (called air resistance), acting upward (being always in the direction opposite to that of the velocity of the body).
Under certain conditions, the force of buoyancy due to air and friction with air can be ignored compared to the gravitational force of the earth. In that case (near the earth's surface) the body falls with almost uniform acceleration (g). Whenever a body moves under the influence of the force of gravity alone, it is said to be falling freely. Strictly speaking, this is true only if the body falls in vacuum.

The kinematical equations of motion, in the usual notation, are
\( v = u + gt \)
\( s = ut + \frac{1}{2} gt^2 \)
\( v^2 = u^2 + 2gs \)
If the initial velocity (u) of the body is zero,
\( v = gt \)
\( s = \frac{1}{2} gt^2 \)
\( v^2 = 2gs \)
In simple words: Free fall means an object is moving only under the influence of gravity, with no air resistance or other forces. The equations of motion for free fall are \( v = u + gt \), \( s = ut + \frac{1}{2} gt^2 \), and \( v^2 = u^2 + 2gs \), where 'g' is the acceleration due to gravity.

🎯 Exam Tip: Define free fall precisely (gravity as the *only* force). State all three kinematic equations, including their simplified forms when initial velocity 'u' is zero, for full marks.

 

Question 24.
During a free fall, will a heavier object accelerate more than a lighter one?
Answer:
No. The two objects will have the same acceleration.
In simple words: No, a heavier object will not accelerate more than a lighter one during free fall. In a vacuum, all objects fall with the same acceleration due to gravity, regardless of their mass.

🎯 Exam Tip: This is a direct test of Galileo's experiment on falling bodies. Remember that acceleration due to gravity 'g' is independent of the falling object's mass.

 

Question 25.
If you had to calculate the mass of the earth, how would you do it?
Answer:
If the acceleration due to gravity (g), the constant of gravitation (G) and the radius of the earth (R) are known, the mass of the earth (M) can be calculated using the formula \( g = \frac{GM}{R^2} \).
In simple words: To calculate Earth's mass, you can use the formula \( g = \frac{GM}{R^2} \). By rearranging it to \( M = \frac{gR^2}{G} \), and knowing the values of 'g' (acceleration due to gravity), 'R' (Earth's radius), and 'G' (universal gravitational constant), you can find 'M' (Earth's mass).

🎯 Exam Tip: State the formula \( g = \frac{GM}{R^2} \) and explicitly show how to rearrange it to solve for M. Clearly identify what each variable represents.

 

Question 26.
What is gravitational potential energy?
(OR)
Define gravitational potential energy.
Write the formula for it.
Answer:
The energy stored in a body due to the gravitational force between the body and the earth is called the gravitational potential energy.
Gravitational potential energy of a body of mass m \( = -\frac{GMm}{R+h} \), where G = gravitational constant, M = mass of the earth, R = radius of the earth, h = height of the body from the surface of the earth.
As the body is bound to the earth due to the earth's gravitational force, the gravitational potential energy of the body is negative. If the body is given kinetic energy equal to \( \frac{GMm}{R+h} \) the body will overcome the earth's gravitational force. It will then move to infinity and come to rest there.
In simple words: Gravitational potential energy is the energy an object possesses due to its position within a gravitational field, specifically due to the gravitational force between it and a larger body like Earth. Its formula is \( U = -\frac{GMm}{R+h} \), and it's negative because gravity is an attractive force.

🎯 Exam Tip: Define gravitational potential energy as stored energy due to position in a gravitational field. Remember to state the formula with its negative sign and explain its significance (attractive force, zero at infinity).

 

Question 27.
What is escape Velocity?
(OR)
Define escape velocity.
Answer:
When a body is thrown vertically upward from the surface of the earth, the minimum initial velocity of the body for which the body is able to overcome the downward pull by the earth and can escape the earth forever is called the escape velocity.
In simple words: Escape velocity is the minimum speed an object needs to be launched with to completely break free from a planet's gravitational pull and never fall back.

🎯 Exam Tip: Emphasize "minimum initial velocity" and "escape forever" when defining escape velocity. This highlights the threshold nature of this speed.

 

Question 28.
Explain the term escape velocity.
(OR)
Write a short note on escape velocity.
Answer:
In general, when a body is thrown vertically upward from the earth's surface, its velocity goes on decreasing and after some time the body falls back to the ground. If its initial velocity is increased, the maximum height attained by it is more, but it does fall back to the ground. If the initial velocity is increased continuously, for a particular initial velocity, the body can overcome the earth's gravitational force and move to infinity and come to rest there. This velocity is called the escape velocity.
In simple words: Escape velocity is the specific speed required for an object to overcome Earth's gravity completely. If thrown slower, it eventually falls back; if thrown at or above escape velocity, it will leave Earth's gravitational field forever.

🎯 Exam Tip: Contrast the effects of launching an object with velocities below and at/above escape velocity to clearly illustrate the concept.

 

Question 29.
Using the law of conservation of energy, obtain the expression for the escape velocity.
Answer:
Here, we shall not consider the effects of air. Suppose a body of mass m is thrown vertically upward from the surface of the earth. Let the initial velocity of the body be the escape velocity (\( v_{esc} \)).
When the body is on the earth's surface, its total energy \( E_x = \) kinetic energy + potential energy \( = \frac{1}{2}mv_{esc}^2 + \left(-\frac{GmM}{R}\right) \) where G = universal gravitational constant, M = mass of the earth and R = radius of the earth.
Thus, \( E_1 = \frac{1}{2}mv_{esc}^2 - \frac{GmM}{R} \)
When the body moves to infinity and comes to rest there, its total energy,
\( E_2 = \frac{1}{2}m(zero)^2 + \left(-\frac{GmM}{\infty}\right) = 0 + 0 = 0 \).
According to the law of conservation of energy,
\( E_1 = E_2 \).
\[ \frac{1}{2}mv_{esc}^2 - \frac{GmM}{R} = 0 \]

\( \implies \frac{1}{2}v_{esc}^2 = \frac{GM}{R} \)

\( \implies v_{esc} = \sqrt{\frac{2GM}{R}} \)
This is the required expression.
In simple words: By applying the conservation of energy, where initial total energy (kinetic + potential) at Earth's surface equals final total energy at infinity (zero), we derive the escape velocity formula. The derivation shows that \( \frac{1}{2}mv_{esc}^2 - \frac{GmM}{R} = 0 \), which simplifies to \( v_{esc} = \sqrt{\frac{2GM}{R}} \).

🎯 Exam Tip: Clearly state the principle of conservation of energy. Show the initial and final energy states, and precisely derive the formula \( v_{esc} = \sqrt{\frac{2GM}{R}} \), defining each variable.

 

Question 30.
Express escape velocity in terms of g and R.
Answer:
Escape velocity, \( v_{esc} = \sqrt{\frac{2GM}{R}} \)
Now, \( g = \frac{GM}{R^2} \)

\( \implies GM = gR^2 \)

\( \implies v_{esc} = \sqrt{\frac{2gR^2}{R}} = \sqrt{2gR} \)
In simple words: Escape velocity can also be expressed using 'g' (acceleration due to gravity) and 'R' (Earth's radius). By substituting \( GM = gR^2 \) into the original formula, we get \( v_{esc} = \sqrt{2gR} \).

🎯 Exam Tip: Remember the relationship \( GM = gR^2 \) to easily convert the escape velocity formula from terms of G and M to terms of g and R. This is a common exam requirement.

 

Question 31.
Express escape velocity in terms of G, R and \( \rho \) (the earth's density)
Answer:
Escape velocity, \( v_{esc} = \sqrt{\frac{2GM}{R}} \)
The earth's density, \( \rho = \frac{\text{mass}}{\text{volume}} = \frac{M}{\frac{4}{3}\pi R^3} \)

\( \implies M = \frac{4}{3}\pi R^3 \rho \)

\( \implies v_{esc} = \sqrt{\frac{2G(\frac{4}{3})\pi R^3 \rho}{R}} \)
\( = \sqrt{2G(\frac{4}{3})\pi R^2 \rho} = 2R\sqrt{\frac{2}{3}\pi G \rho} \)
In simple words: To express escape velocity using Earth's density \( (\rho) \), we substitute the mass (M) in the formula \( v_{esc} = \sqrt{\frac{2GM}{R}} \) with \( M = \frac{4}{3}\pi R^3 \rho \). This gives \( v_{esc} = 2R\sqrt{\frac{2}{3}\pi G \rho} \).

🎯 Exam Tip: The key here is to relate the Earth's mass (M) to its density (\( \rho \)) and radius (R) using the volume formula for a sphere. Substitute M into the escape velocity equation and simplify.

 

Give Scientific Reasons:

 

Question 1.
If a feather and a stone are released from the top of a building simultaneously, the stone reaches the ground earlier than the feather.
Answer:
(a) The motion of a body falling in air accelerated due to the earth's gravitational force on the body. The force due to buoyancy of air acts on the body in the upward direction. As the body falls, the friction with air opposes its motion.
(b) This opposition due to air depends on the size, shape, density and velocity of the body. It Is greater for a feather than for a stone. Hence, the stone has greater downward acceleration than the feather. Therefore, the stone reaches the ground earlier than the feather though 'they are released simultaneously from the same height.
In simple words: The stone falls faster than the feather because air resistance, which opposes motion, affects the feather much more due to its larger surface area and lower density. The stone, being denser and more aerodynamic, experiences less relative air resistance, allowing gravity to accelerate it more effectively.

🎯 Exam Tip: Explain the role of air resistance and buoyancy as forces opposing gravity. Emphasize that these forces depend on the object's properties like size, shape, and density, leading to different accelerations in air.

 

Question 2.
The weight of a body is different on different planets.
Answer:
(1) The weight of a body of mass m on the surface of a planet of mass M and radius R is
\( W = \frac{GmM}{R^2} \) (usual notation).
(2) For a given body, its mass is constant. G is the universal constant of gravitation. Different planets have different masses and radii such that the ratio \( (M/R^2) \) is not the same. Hence, the weight of a body is different on different planets.
In simple words: An object's weight varies on different planets because weight is a measure of the gravitational force, which depends on the planet's mass (M) and radius (R). Since M and R differ for each planet, the gravitational acceleration ('g') and thus the weight (\( W=mg \)) will also vary.

🎯 Exam Tip: Start by defining weight as \( W=mg \). Explain that while 'm' is constant, 'g' (\( \propto M/R^2 \)) varies significantly between planets due to their unique masses and radii.

 

Question 3.
With a specific initial velocity, we can jump higher on the moon than on the earth.
Answer:
The acceleration due to gravity on the moon is about \( \frac{1}{6} \) of that on the earth. Hence, with a specific initial velocity, we can jump higher on the moon than on the earth.
This can be seen from the equation \( h = u^2/(2g) \).
In simple words: We can jump higher on the Moon because its gravitational acceleration ('g') is much weaker (about one-sixth) than Earth's. With the same initial jumping velocity, the reduced 'g' on the Moon allows for a greater maximum height (\( h = u^2/(2g) \)).

🎯 Exam Tip: The core reason is the difference in gravitational acceleration. Reference the formula \( h = u^2/(2g) \) to clearly show the inverse relationship between height 'h' and 'g'.

 

Distinguish Between The Following:

 

Question 1.
mass and weight (2) universal gravitational constant and gravitational acceleration of the earth.
Answer:
(1) Mass:
• The mass of a body is the amount of matter present in it.
• It has magnitude, but not direction.
• It does not change from place to place.
• It can never be zero.
• Its SI unit is the kilogram.
Weight:
• The weight of a body is the force with which the earth attracts it.
• It has both magnitude and direction.
• It changes from place to place.
• It is zero at the centre of the earth.
• Its SI unit is the newton.
In simple words: Mass is the quantity of matter in an object, constant everywhere and a scalar. Weight is the force of gravity on an object, varying by location and a vector.

🎯 Exam Tip: For distinctions, use clear, contrasting points in a bulleted format. Cover definition, nature (scalar/vector), variability, and units for both mass and weight.

 

Question 2.
Universal gravitational constant:
• The universal gravitational constant numerically equals the force of attraction masses separated by a unit distance.
• Its value remains constant throughout the universe.
• It has magnitude but not direction.
• Its SI unit is N.m\(^2\)/kg\(^2\).
Gravitational acceleration of the earth:
• The gravitational between two unit acceleration of the earth is the acceleration produced in a body due to the gravitational force of the earth.
• Its value changes from place to place.
• It has both magnitude and direction.
• Its SI unit is m/s\(^2\)
In simple words: Universal Gravitational Constant (G) is a fixed value determining the strength of gravity, constant everywhere and a scalar. Gravitational Acceleration (g) is the acceleration due to Earth's gravity, varies by location and is a vector.

🎯 Exam Tip: Highlight that G is a constant, while 'g' is a variable. Clearly define what each represents and state their respective units and scalar/vector nature.

 

Solve The Following Examples/Numerical Problems:

 

Question 1.
The time taken by the earth to complete one revolution around the Sun is 3.156 × 10\(^7\) s. The distance between the earth and the Sun is 1.5 × 10\(^{11}\) m. Find the speed of revolution of the earth.
Answer:
Solution:
Data: T = 3.156 × 10\(^7\) s,
r = 1.5 × 10\(^{11}\) m, v =?
\( v = \frac{2\pi r}{T} = \frac{2 \times 3.142 \times 1.5 \times 10^{11} \text{ m}}{3.156 \times 10^7 \text{ s}} \)
= 2.987 × 10\(^4\) m/s = 29.87 km/s
This is the speed of revolution of the earth.
In simple words: To find Earth's speed around the Sun, use the formula \( v = \frac{2\pi r}{T} \), where 'r' is the distance and 'T' is the time period. Plugging in the given values for Earth's orbit, the speed is approximately 29.87 km/s.

🎯 Exam Tip: For orbital speed problems, clearly state the formula and substitute values carefully. Ensure units are consistent (e.g., all in SI) and present the final answer with appropriate units.

 

Question 2.
Assuming that the earth performs uniform circular motion around the Sun, find the centripetal acceleration of the earth. [Speed of the earth = 3 × 10\(^4\) m/s, distance between the earth and the Sun = 1.5 × 10\(^{11}\) m]
Answer:
Solution:
Data: \( v = 3 \times 10^4 \text{ m/s} \), \( r=1.5 \times 10^{11} \text{ m} \)
Centripetal force \( = \frac{mv^2}{r} = ma \)
Centripetal acceleration of the earth,
\( a = \frac{v^2}{r} = \frac{(3 \times 10^4 \text{ m/s})^2}{1.5 \times 10^{11} \text{ m}} = \frac{3 \times 3 \times 10^8}{1.5 \times 10^{11}} \text{ m/s}^2 \)
= 6 × 10\(^{-3}\) m/s\(^2\)
It is directed towards the centre of the Sun.
In simple words: To calculate the Earth's centripetal acceleration around the Sun, use the formula \( a = \frac{v^2}{r} \). With the given speed and distance, the acceleration is \( 6 \times 10^{-3} \text{ m/s}^2 \), directed towards the Sun's center.

🎯 Exam Tip: Clearly state the formula for centripetal acceleration. Ensure correct substitution of velocity and radius, and always specify the direction of centripetal acceleration (towards the center of the circular path).

 

Question 3.
What will be the gravitational force on 60 kg man on the Moon, Mars and Jupiter? Are they the same? Why?
M (Moon) = 7.36 × 10\(^{22}\) kg, R (Moon) = 1.74 x 10\(^6\) m,
M (Mars) = 6.4 × 10\(^{23}\) kg, R (Mars) = 3.395 × 10\(^6\) m,
M (Jupiter) = 1.9 × 10\(^{27}\) kg.
R (Jupiter) = 7.15 × 10\(^7\) m,
G = 6.67 × 10\(^{-11}\) N.m\(^2\)/kg\(^2\)
Answer:
Solution:

 

Question 3. What will be the gravitational force on 60 kg man on the Moon, Mars and Jupiter? Are they the same? Why?
M (Moon) = 7.36 × 10\(^{22}\) kg, R (Moon) = 1.74 x 10\(^{6}\)m,
M (Mars) = 6.4 × 10\(^{23}\) kg, R (Mars) = 3.395 × 10\(^{6}\) m,
M (Jupiter) = 1.9 × 10\(^{27}\) kg.
R (Jupiter) = 7.15 × 10\(^{7}\) m,
G = 6.67 × 10\(^{-11}\) N.m\(^{2}\)/kg\(^{2}\)
Answer:
Solution:
(1) Data : m\(_{1}\) = 60 kg, m\(_{2}\) = 7.36 × 10\(^{22}\) kg,
R = 1.74 × 10\(^{6}\) m, F = ?
\[ F = \frac{Gm_{1}m_{2}}{R^{2}} \] \[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 60 \text{ kg} \times 7.36 \times 10^{22} \text{ kg}}{(1.74 \times 10^{6} \text{ m})^{2}} \] = 97.29 N
On the moon's surface, the gravitational force on the man due to the moon = 97.29 N.
(2) Data : m\(_{1}\) = 60 kg, m\(_{2}\) = 6.4 × 10\(^{23}\) kg,
R = 3.395 × 10\(^{6}\)m, F = ?
\[ F = \frac{Gm_{1}m_{2}}{R^{2}} \] \[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 60 \text{ kg} \times 6.4 \times 10^{23} \text{ kg}}{(3.395 \times 10^{6} \text{ m})^{2}} \] = 222.2 N
On the surface of Mars, the gravitational force on the man due to Mars = 222.2 N.
(3) Data : m\(_{1}\) = 60 kg, m\(_{2}\) = 1.9 × 10\(^{27}\) kg,
R = 7.15 × 10\(^{7}\) m, F = ?
\[ F = \frac{Gm_{1}m_{2}}{R^{2}} \] \[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 60 \text{ kg} \times 1.9 \times 10^{27} \text{ kg}}{(7.15 \times 10^{7} \text{ m})^{2}} \] = 1487 N
On the surface of Jupiter, the gravitational force on the man due to Jupiter = 1487 N.
Thus, the forces on the man are not the same because the ratio (M/R\(^{2}\)) is not the same in the case of the moon, Mars, and Jupiter.
In simple words: The gravitational force a man experiences varies greatly on different celestial bodies like the Moon, Mars, and Jupiter, mainly because each body has a different mass and radius, which affects its gravitational pull.

🎯 Exam Tip: Remember to use the correct values for mass (M) and radius (R) for each celestial body, and always show your calculations for partial credit. Ensure proper unit conversion if necessary.

 

Question 4. Mahendra and Virat are sitting at a distance of 1 meter from each other. Their masses are 75 kg and 80 kg respectively. What is the gravitational force between them? G = 6.67 x 10-11 N.m²/kg².
Answer:
Solution:
Given: r = 1 m,
m\(_{1}\) = 75 kg,
m\(_{2}\) = 80 kg
G = 6.67 x 10-11 N.m\(^{2}\)/kg\(^{2}\)
\[ F = \frac{Gm_{1}m_{2}}{r^{2}} = \frac{6.67 \times 10^{-11} \times 75 \times 80}{1^{2}} \text{ N} \] = 4.002 × 10-7 N
The gravitational force between Mahendra and Virat is 4.002 × 10-7 N.
In simple words: The gravitational force between two normal-sized people is extremely small, almost negligible, due to their relatively small masses and the universal gravitational constant.

🎯 Exam Tip: Pay close attention to the exponents in scientific notation. A common error is miscalculating powers of ten. Remember to include units in your final answer.

 

Question 5. Spheres A and B of uniform density have masses 1 kg and 100 kg respectively. Their centres are separated by 100 m. (i) Find the gravitational force between them, (ii) Find the gravitational force on A due to the earth, (iii) Suppose A and B are initially at rest and A can move freely towards B. What will be the velocity of A one second after it starts moving towards B? How will this velocity change with time? How much time will A take to move towards B by 1 cm? (iv) if A begins to fall, starting from rest, due to the earth's downward pull, what will be its velocity after one second? How much time will it take to fall through 1cm?
[M(earth) = 6 x 10\(^{24}\) kg, R(earth) = 6400 km]
Answer:
Solution:
Data: m\(_{1}\) = 1 kg, m\(_{2}\) = 100 kg, r = 100 m,
M =6 x 10\(^{24}\) kg,
R = 6400 km = 6400 × 10\(^{3}\) m,
t = 1 s, s = 1 cm = 1 × 10-2 m,
G = 6.67 x 10-11 N.m\(^{2}\)/kg
F\(_{1}\) =?, F\(_{2}\) =?, V\(_{1}\) =?, t\(_{1}\) =?, V\(_{2}\) =?, t\(_{1}\) =?
(i) F\(_{1}\) = \(\frac{Gm_{1}m_{2}}{r^{2}}\)
\[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 1 \text{ kg} \times 100 \text{ kg}}{(100 \text{ m})^{2}} \] = 6.67 x 10-13 N
(ii) F\(_{2}\) = \(\frac{Gm_{1}M}{R^{2}}\)
\[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 1 \text{ kg} \times 6 \times 10^{24} \text{ kg}}{(6400 \times 10^{3} \text{ m})^{2}} \] \[ = \frac{40.02 \times 10^{13}}{6.4 \times 6.4 \times 10^{12}} \text{ N} = 9.77 \text{ N.} \] This is far greater than F\(_{1}\).
(iii) Ignoring variation of acceleration with distance,
v\(_{1}\) = u\(_{1}\) + at = 0 + \(\frac{F_{1}}{m_{1}}\) t = \(\frac{6.67 \times 10^{-13} \text{ N}}{1 \text{ kg}}\) x 1 s
= 6.67 x 10-13 m/s
This velocity is directed from A to B. As the separation between A and B decreases, the acceleration of A and hence the velocity of A will increase.
Ignoring variation of acceleration with distance,
s\(_{1}\) = ut + \(\frac{1}{2}\) at\(^{2}\) = 0 + \(\frac{1}{2}\) \(\frac{F_{1}}{m_{1}}\)t\(^{2}\)

\(\implies\) t\(_{1}^{2}\) = \(\frac{2m_{1}s_{1}}{F_{1}}\)

\(\implies\) t\(_{1}^{2}\) = \(\frac{2 \times 1 \text{ kg} \times 10^{-2} \text{ m}}{6.67 \times 10^{-13} \text{ N}}\) = 3 x 10\(^{10}\) s\(^{2}\)

\(\implies\) t\(_{1}\) = 1.732 × 10\(^{5}\) s
(iv) v\(_{2}\) = u\(_{2}\) + at = 0 +gt = \(\frac{F_{2}}{m_{1}}\) t
= \(\frac{9.77 \text{ N}}{1 \text{ kg}}\) x 1 s = 9.77 m/s (downward)
[|v\(_{2}\)| >> |v\(_{1}\)|]
s\(_{2}\) = ut + \(\frac{1}{2}\)at\(^{2}\) = 0 + \(\frac{1}{2}\)gt\(_{2}^{2}\) = \(\frac{1}{2}\) \(\frac{F_{2}}{m_{1}}\) t\(_{2}^{2}\)

\(\implies\) t\(_{2}^{2}\) = \(\frac{2s_{2}m_{1}}{F_{2}}\) = \(\frac{2 \times 10^{-2} \text{ m} \times 1 \text{ kg}}{9.77 \text{ N}}\) = 0.205 s\(^{2}\)

\(\implies\) t\(_{2}\) = 0.453 s [t\(_{1}\) >> t\(_{2}\)]
In simple words: This problem illustrates the vast difference between gravitational forces. The gravitational pull between two small objects is negligible, while the Earth's gravity exerts a much stronger, noticeable pull on any object.

🎯 Exam Tip: For multi-part questions, break down each section and label your answers clearly. Ensure you use the correct formula for each scenario (inter-object gravity vs. Earth's gravity). Pay attention to units and scientific notation in calculations.

 

Question 6. Two spheres of uniform density have masses 10 kg and 40 kg. The distance between the centres of the spheres is 200 m. Find the gravitational force between them.
Answer:
Solution:
Data: m\(_{1}\) = 10 kg, m\(_{2}\) = 40 kg,
r = 200 m, G = 6.67 x 10-11 N.m\(^{2}\)/kg\(^{2}\), F = ?
\[ F = \frac{Gm_{1}m_{2}}{r^{2}} \] \[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 10 \text{ kg} \times 40 \text{ kg}}{(200 \text{ m})^{2}} \] \[ = \frac{6.67 \times 10^{-11} \times 4 \times 10^{2}}{4 \times 10^{4}} \text{ N} = 6.67 \times 10^{-13} \text{ N} \] The gravitational force between the two spheres = 6.67 × 10-13 N.
In simple words: The gravitational force between two relatively small objects, even with significant masses like 10 kg and 40 kg, is extremely weak due to the very small value of the gravitational constant (G).

🎯 Exam Tip: Double-check the square of the distance in the denominator. Errors in squaring large numbers or powers of ten are common. Always state the final answer with correct units and scientific notation.

 

Question 7. Find the gravitational force between a man of mass 50 kg and a car of mass 1500 kg separated by 10 m.
Answer:
Solution: Data : m\(_{1}\) = 50 kg, m\(_{2}\) = 1500 kg,
r = 10 m, G = 6.67 x 10-11 N.m\(^{2}\)/kg\(^{2}\), F = ?
\[ F = \frac{Gm_{1}m_{2}}{r^{2}} \] \[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 50 \text{ kg} \times 1500 \text{ kg}}{(10 \text{ m})^{2}} \] = 5.0025 × 10-8 N
The gravitational force between the man and the car = 5.0025 × 10-8 N.
In simple words: Even between a man and a car, the gravitational force is tiny, almost imperceptible, highlighting that gravity is only strong when at least one of the masses is planetary in scale.

🎯 Exam Tip: Ensure that the distance 'r' is in meters for the calculation when G is in N.m²/kg². Pay attention to significant figures if specified in the exam, otherwise, a reasonable number of decimal places for scientific notation is acceptable.

 

Question 8. Find the magnitude of the gravitational force between the Sun and the earth. (Mass of the Sun = 2 × 10\(^{24}\) kg, mass of the earth=6 × 10\(^{24}\) kg and the thstance between the centres of the Sun and the earth 1.5 × 10\(^{11}\) m, G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\))
Answer:
Solution:
Data: m\(_{1}\) = 2 × 10\(^{30}\) kg,
m\(_{2}\) = 6 × 10\(^{24}\) kg, r = 1.5 × 10\(^{11}\) m,
G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\), F =?
\[ F = \frac{Gm_{1}m_{2}}{r^{2}} \] \[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 2 \times 10^{30} \text{ kg} \times 6 \times 10^{24} \text{ kg}}{(1.5 \times 10^{11} \text{ m})^{2}} \] \[ = \frac{6.67 \times 2 \times 6}{1.5 \times 1.5} \times 10^{21} \text{ N} = 35.57 \times 10^{21} \text{ N} \]
\(\implies\) F = 3.557 × 10\(^{22}\) N
The magnitude of the gravitational force between the Sun and the earth = 3.557 × 10\(^{22}\) N.
In simple words: The gravitational force between the Sun and the Earth is enormous, demonstrating how gravity effectively governs the motion of celestial bodies despite the vast distances involved.

🎯 Exam Tip: When dealing with very large numbers and exponents, be meticulous in your calculations to avoid errors. Correctly handling the square of the distance 'r' (which is also a large number with an exponent) is crucial for accuracy.

 

Question 9. Find the magnitude of the acceleration due to gravity at the surface of the earth. (M= 6 × 10\(^{24}\) kg, R = 6400 km)
Answer:
Solution:
Data: M = 6 × 10\(^{24}\) kg,
R = 6400km = 6.4 × 10\(^{6}\)m,
G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\), g =?
\[ g = \frac{GM}{R^{2}} \] \[ = \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 6 \times 10^{24} \text{ kg}}{(6.4 \times 10^{6} \text{ m})^{2}} \] \[ = \frac{66.7 \times 6}{(6.4)^{2}} \text{ m/s}^{2} = 9.77 \text{ m/s}^{2} \] The magnitude of the acceleration due to gravity at the surface of the earth = 9.77 m/s\(^{2}\).
In simple words: The acceleration due to gravity (g) on Earth's surface is calculated using the Earth's mass, radius, and the universal gravitational constant, resulting in approximately 9.77 m/s².

🎯 Exam Tip: Ensure the radius 'R' is converted to meters before calculation. Double-check the squaring of R and the multiplication of constants, as these are common sources of error in such problems.

 

Question 10. The mass of an imaginary planet is 3 times the mass of the earth. Its diameter is 25600 km arid the earth's diameter is 12800 km. Find the acceleration due to gravity at the surface of the planet, [g (earth) = 9.8 m/s²]
Answer:
Solution:
Data:
\[ \frac{M_{2} \text{ (planet)}}{M_{1} \text{ (earth)}} = 3 \] D\(_{1}\) (earth) = 12800 km

\(\implies\) R\(_{1}\) (earth) = \(\frac{12800 \text{ km}}{2}\) = 6400 km
= 6.4 x 10\(^{6}\) m
D\(_{2}\) (planet) = 25600 km

\(\implies\) R\(_{2}\) (planet) = \(\frac{25600 \text{ km}}{2}\) = 12800 km
= 1.28 x 10\(^{7}\) m
g\(_{1}\) (earth) = 9.5 m/s\(^{2}\), g\(_{2}\) (planet) = ?
\[ g = \frac{GM}{R^{2}} \]
\(\implies\) g\(_{1}\) = \(\frac{GM_{1}}{R_{1}^{2}}\) , g\(_{2}\) = \(\frac{GM_{2}}{R_{2}^{2}}\)

\(\implies\) \(\frac{g_{2}}{g_{1}}\) = \(\frac{M_{2}}{M_{1}} \frac{R_{1}^{2}}{R_{2}^{2}}\) = \(\frac{M_{2}}{M_{1}} (\frac{R_{1}}{R_{2}})^{2}\)
\[ \implies g_{2} = g_{1} (\frac{M_{2}}{M_{1}}) (\frac{R_{1}}{R_{2}})^{2} \] \[ = 9.8 \text{ m/s}^{2} \times 3 \times (\frac{6.4 \times 10^{6} \text{ m}}{1.28 \times 10^{7} \text{ m}})^{2} \] \[ = \frac{9.8 \times 3}{4} \text{ m/s}^{2} = 7.35 \text{ m/s}^{2} \] The acceleration due to gravity at the surface of the planet = 7.35 m/s\(^{2}\).
In simple words: The acceleration due to gravity on a planet depends on its mass and radius. Even if a planet is more massive, a larger radius can dilute its gravitational pull at the surface compared to Earth.

🎯 Exam Tip: Remember that diameter is twice the radius; convert diameters to radii before using them in the formula. Using ratios like \( (M_{2}/M_{1}) (R_{1}/R_{2})^{2} \) simplifies calculations and reduces chances of error when G is common.

 

Question 11. If the acceleration due to gravity on the surface of the earth is 9.8 m/s², what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth?
Answer:
Solution:
Data: g\(_{e}\) = 9.8 m/s\(^{2}\), M\(_{p}\) = 2M\(_{e}\),
R\(_{p}\) = 2R\(_{e}\), g\(_{p}\) = ?
Acceleration due to gravity, \[ g = \frac{GM}{R^{2}} \]
\(\implies\) g\(_{e}\) = \(\frac{GM_{e}}{R_{e}^{2}}\) and g\(_{p}\) = \(\frac{GM_{p}}{R_{p}^{2}}\)
\[ \implies \frac{g_{p}}{g_{e}} = \frac{M_{p}}{M_{e}} (\frac{R_{e}}{R_{p}})^{2} = 2 \times (\frac{1}{2})^{2} = \frac{1}{2} \]
\(\implies\) g\(_{p}\) = \(\frac{g_{e}}{2}\) = \(\frac{9.8 \text{ m/s}^{2}}{2}\) = 4.9 m/s\(^{2}\)
acceleration due to gravity on the surface of the planet = 4.9 m/s\(^{2}\).
In simple words: If a planet has both twice the mass and twice the radius of Earth, its surface gravity will be half that of Earth's because the increased mass is offset by the even greater effect of the larger radius.

🎯 Exam Tip: When both mass and radius change, be careful to square the radius factor (2R)\(^{2}\) = 4R\(^{2}\) in the denominator. Set up a ratio of gravities to simplify the calculation without needing G directly.

 

Question 12. A body is released from the top of a building of height 19.6 m. Find the velocity with which the body hits the ground.
Answer:
Solution:
Data: h = 19.6 m, u = 0 m/s,
g = 9.8 m/s\(^{2}\), s = 19.6 m, v = ?
v\(^{2}\) = u\(^{2}\) + 2 gs .
=2 gs .....(as u = 0 m/s)
= 2 × 9.8 m/s\(^{2}\) × 19.6 m
= (19.6 m/s)\(^{2}\)

\(\implies\) v = 19.6 m/s (downward velocity)
The velocity with which the body hits the ground = 19.6 m/s (downward).
In simple words: When an object falls from a height, its final speed just before hitting the ground can be calculated using the acceleration due to gravity and the initial height, assuming it starts from rest.

🎯 Exam Tip: Remember to identify initial velocity (u) correctly; for objects "released from rest," u=0. Use the appropriate kinematic equation (v\(^{2}\) = u\(^{2}\) + 2gs) and ensure units are consistent.

 

Question 13. A stone on a bridge on a river falls into the river. If it takes 3 seconds to reach the surface of water, find (i) the velocity of the stone at the instant it touches the surface of water (ii) the height of the bridge from the surface of water.
Answer:
Solution:
Data: u = 0 m/s, t = 3 s,
g = 9.8 m/s\(^{2}\), v = ?, h = ?
(i) v = u + gt = 0 m/s + 9.8 m/s\(^{2}\) × 3 s
= 29.4 m/s
The velocity of the stone at the instant it touches the surface of water = 29.4 m/s
(ii) s = ut + \(\frac{1}{2}\)gt\(^{2}\)
= 0 m/s × 3 s + \(\frac{1}{2}\) (9.8 m/s\(^{2}\)) (3 s)\(^{2}\)
= 4.9 x 9 m = 44.1 m

\(\implies\) The height of the bridge from the surface of water = 44.1 m.
In simple words: From the time it takes for an object to fall, we can determine both its final velocity and the height it fell from, using equations of motion under gravity.

🎯 Exam Tip: This problem requires two kinematic equations: one for velocity (v = u + at) and one for displacement (s = ut + ½at\(^{2}\)). Ensure you use consistent values for 'u', 'a' (g), and 't' for both parts of the question.

 

Question 14. A stone is dropped from rest from the top of a building 44.1 m high. It takes 3 s to reach the ground. Use this information to 1 calculate g.
Answer:
Solution:
Data: u = 0 m/s, h = 44.1 m

\(\implies\) s = 44.1 m, t = 3 s, g =?
s = ut + \(\frac{1}{2}\)at\(^{2}\) = \(\frac{1}{2}\)at\(^{2}\) (u = 0 m/s)

\(\implies\) a = \(\frac{2 \text{ s}}{t^{2}}\) = \(\frac{2 \times (44.1 \text{ m})}{(3 \text{ s})^{2}}\)
\[ = \frac{88.2}{9} \text{ m/s}^{2} = 9.8 \text{ m/s}^{2}. \] It is the acceleration due to gravity.
g = 9.8 m/s\(^{2}\).
In simple words: By observing an object falling from a known height and measuring the time it takes, we can calculate the local acceleration due to gravity using a basic kinematic equation.

🎯 Exam Tip: Rearranging the kinematic equation s = ut + ½at\(^{2}\) to solve for 'a' (which is 'g' in this case) is key. Make sure to substitute values correctly, especially the square of time (t\(^{2}\)).

 

Question 15. A metal ball of mass 5 kg falls from a height of 490 m. How much time will it take to reach the ground? (g = 9.8 m/s²)
Answer:
Solution:
Data: s = 490 m, a = g = 9.8 m/s\(^{2}\),
u = 0 m/s, s = ut + \(\frac{1}{2}\)at\(^{2}\)

\(\implies\) 490 = 0×t + \(\frac{1}{2}\) x 9.8 x t\(^{2}\) = 4.9t\(^{2}\)

\(\implies\) t\(^{2}\) = \(\frac{490}{4.9}\)
t\(^{2}\) = 100

\(\implies\) t = 10 s This is the required time.
In simple words: To find the time an object takes to fall from a specific height, we use the equation of motion that relates distance, initial velocity (zero for a fall), acceleration due to gravity, and time.

🎯 Exam Tip: When solving for time (t), remember to take the square root of t\(^{2}\). Always ensure units are consistent (meters for distance, m/s\(^{2}\) for acceleration) to get time in seconds.

 

Question 16. If the weight of a body on the surface of the moon is 100 N, what is its mass? (g = 1.63 m/s²)
Answer:
Solution:
Data: W= 100 N, g = 1.63 m/s\(^{2}\), m = ?

\(\implies\) W = mg

\(\implies\) m = \(\frac{W}{g}\) = \(\frac{100 \text{ N}}{1.63 \text{ m/s}^{2}}\) = 61.35 kg

\(\implies\) The mass of the body = 61.35 kg.
In simple words: An object's mass is constant regardless of location, so if we know its weight and the local acceleration due to gravity, we can calculate its mass.

🎯 Exam Tip: Clearly differentiate between mass (constant) and weight (changes with gravity). The formula W=mg is fundamental. Ensure correct units for weight (Newtons) and gravity (m/s\(^{2}\)) to get mass in kilograms.

 

Question 17. A 100 kg bag of wheat is placed on a plank of wood. What is the weight of the bag and what is the reaction force exerted by the plank?
Answer:
Solution:
Data: m = 100 kg, g = 9.8 m/s\(^{2}\),
W = ?, reaction force = ?
Magnitude of the weight,
W = mg = 100 kg × 9.8 m/s\(^{2}\) = 980 N
The weight of the bag = 980 N acting downward.
The reaction force exerted by the plank on the bag = 980 N acting upward.
In simple words: The weight of an object is the downward force of gravity on its mass. According to Newton's third law, the surface supporting it exerts an equal and opposite reaction force.

🎯 Exam Tip: Remember that weight (W=mg) is a force, measured in Newtons. For an object at rest on a horizontal surface, the normal reaction force is equal in magnitude and opposite in direction to its weight.

 

Question 18. Find the gravitational potential energy of a body of mass 10 kg when it is on the earth's surface. [M(earth) = 6 × 10\(^{24}\) kg, it(earth) = 6.4 × 10\(^{6}\)m, G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\)]
Answer:
Solution:
Data: m = 10 kg, M = 6 x 10\(^{24}\) kg,
R = 6.4 × 10\(^{6}\) m, G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\)
The gravitational potential energy of the body
\[ = - \frac{GMm}{R} \] \[ = - \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 6 \times 10^{24} \text{ kg} \times 10 \text{ kg}}{6.4 \times 10^{6} \text{ m}} \] \[ = - \frac{6.67 \times 6}{6.4} \times 10^{8} \text{ J} = -6.253 \times 10^{8} \text{ J.} \]In simple words: Gravitational potential energy on the Earth's surface is a negative value, indicating that work must be done against gravity to move an object away from the Earth.

🎯 Exam Tip: The formula for gravitational potential energy near a large body is \( -GMm/R \). Remember the negative sign, which signifies that gravity is an attractive force and zero potential energy is at infinity.

 

Question 19. If the body in Ex. (26) performs uniform circular motion around the earth at a height of 3600 km from the earth's surface, what will be its gravitational potential energy?
Answer:
Solution:
Here, h = 3600 km = 3.6 × 10\(^{6}\) m

\(\implies\) The gravitational potential energy of the body
\[ = - \frac{GMm}{R+h} \] \[ = - \frac{6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 6 \times 10^{24} \text{ kg} \times 10 \text{ kg}}{(6.4 \times 10^{6} + 3.6 \times 10^{6}) \text{ m}} \] \[ = - \frac{6.67 \times 6 \times 10^{14}}{10 \times 10^{6}} \text{ J} = -4.002 \times 10^{8} \text{ J.} \]In simple words: For an object orbiting at a certain height above Earth, its gravitational potential energy is calculated using the total distance from the Earth's center (radius + height) and remains negative, indicating it's still bound by Earth's gravity.

🎯 Exam Tip: For objects above the surface, the distance 'r' in the potential energy formula becomes R+h (Earth's radius + height). Ensure 'h' is converted to meters before adding to 'R'.

 

Question 20. A body of mass 20 kg is at rest on the earth's surface, (i) Find its gravitational potential energy, (ii) Find the kinetic energy to be provided to the body to make it free from the gravitational influence of the earth. (g = 9.8 m/s², R = 6400 km)
Answer:
Solution:
Data : m = 20 kg, g = 9.8 m/s\(^{2}\),
R = 6400 km = 6.4 × 10\(^{6}\) m
(i) The gravitational potential energy of the body =
\[ - \frac{GMm}{R} = - \text{mgR } \left(\therefore g=\frac{GM}{R^{2}}\right) \] = - 20 kg x 9.8 m/s\(^{2}\) x 6.4 x 10\(^{6}\) m
= - 1.2544 × 10\(^{9}\) J.
(ii) To make the body free from the gravitational influence of the earth, it should be provided kinetic energy equal to 1.2544 × 10\(^{9}\) J.
In simple words: An object on Earth has negative gravitational potential energy. To escape Earth's gravity, it needs to be given kinetic energy equal in magnitude to this negative potential energy.

🎯 Exam Tip: The gravitational potential energy can be expressed as -mgR at the surface. The kinetic energy required to escape is equal to the absolute value of this gravitational potential energy.

 

Question 21. If the body in Ex. (28) is moving at 100 m/s on-the earth's surface, what will be its (i) kinetic energy (ii) total energy?
Answer:
Solution:
Data : m = 20 kg, u = 100 m/s.
(i) The kinetic energy of the body
= \(\frac{1}{2}\)mv\(^{2}\) = \(\frac{1}{2}\) × 20 kg × (100 m/s)\(^{2}\) = 10\(^{5}\) J.
(ii) The total energy of the body = kinetic energy + potential energy = 10\(^{5}\) J + (-1.2544 × 10\(^{9}\) J)
= (1 – 12544) × 10\(^{5}\) J = - 12543 × 10\(^{5}\) J
= - 1.2543 × 10\(^{9}\) J.
In simple words: The total energy of an object on Earth is the sum of its kinetic and potential energies. If the potential energy is much larger and negative, the total energy will also be negative, meaning the object is still bound to Earth.

🎯 Exam Tip: Remember to calculate both kinetic energy (\(\frac{1}{2}\)mv\(^{2}\)) and potential energy (-mgR or \(-GMm/R\)) before summing them for total energy. Be careful with signs, as potential energy is negative.

 

Question 22. A satellite of mass 100 kg performs uniform circular motion around the earth at a height of 6400 km from the earth's surface. Find its gravitational potential energy. [g = 9.8 m/s², R = 6400 km]
Answer:
Solution:
Data: m = 100 kg, g = 9.8 m/s\(^{2}\),
R = 6400 km = 6.4 × 10\(^{6}\) m, h = 6.4 × 10\(^{6}\) m
The gravitational potential energy of the satellite
\[ = - \frac{GMm}{R+h} = - \frac{mgR^{2}}{R+h} \left( \therefore g=\frac{GM}{R^{2}} \right) \] \[ = - \frac{100 \text{ kg} \times 9.8 \text{ m/s}^{2} \times (6.4 \times 10^{6} \text{ m})^{2}}{(6.4 \times 10^{6} + 6.4 \times 10^{6}) \text{ m}} \] \[ = - \frac{9.8 \times 6.4 \times 6.4 \times 10^{14}}{2 \times 6.4 \times 10^{6}} \text{ J} = -9.8 \times 3.2 \times 10^{8} \text{ J} \] = - 3.136 × 10\(^{9}\) J.
In simple words: The gravitational potential energy of a satellite orbiting Earth depends on its mass and its total distance from the Earth's center (Earth's radius plus orbital height).

🎯 Exam Tip: Remember to use R+h for the distance from the center of the Earth. If 'g' is given, you can use the relationship \( GM = gR^{2} \) to simplify the potential energy formula to \(\frac{-mgR^{2}}{R+h}\).

 

Question 23. Find the escape velocity of a body from the earth. [M(earth) = 6 x 10\(^{24}\) kg, R (earth) = 6.4 × 10\(^{6}\) m, G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\)]
Answer:
Solution:
Data: M = 6 x 10\(^{24}\) kg, R = 6.4 × 10\(^{6}\) m,
G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\)
The escape velocity of a body from the earth,
\[ v_{esc} = \sqrt{\frac{2GM}{R}} \] \[ = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 6 \times 10^{24} \text{ kg}}{6.4 \times 10^{6} \text{ m}}} \] \[ = \sqrt{\frac{12 \times 6.67 \times 10^{8}}{6.4}} \] = 1.118 × 10\(^{4}\) m/s = 11.18 km/s.
In simple words: Escape velocity is the minimum speed an object needs to break free from a planet's gravitational pull and not fall back, calculated from the planet's mass and radius.

🎯 Exam Tip: The formula for escape velocity is \( v_{esc} = \sqrt{\frac{2GM}{R}} \). Ensure 'M' and 'R' are in SI units (kg and m) and G is correctly used. The final answer should ideally be converted to km/s for ease of understanding.

 

Question 24. Find the escape velocity of a body from the earth. [R (earth) = 6.4 × 10\(^{6}\) m, p(earth) = 5.52 x 10\(^{3}\) kg/m³, G = 6.67 x 10-11 N.m\(^{2}\)/kg\(^{2}\)]
Answer:
Solution:
Data: R = 6.4 × 10\(^{6}\) m, p = 5.52 × 10\(^{3}\) kg/m\(^{3}\), G = 6.67 x 10-11 N.m\(^{2}\)/kg\(^{2}\)
The escape velocity of a body from the earth
\[ v_{esc} = 2R \sqrt{\frac{2}{3} \pi G \rho} \] = 2 x 6.4 x 10\(^{6}\) m x
\[ \sqrt{\frac{2}{3} \times 6.67 \times 10^{-11} \text{ N-m}^{2}/\text{kg}^{2} \times 3.142 \times 5.52 \times 10^{3} \text{ kg/m}^{3}} \] \[ = 1.28 \times 10^{3} \times \sqrt{\frac{13.34 \times 3.142 \times 5.52}{3}} \] = 11.24 x 10\(^{3}\) m/s = 11.24 km/s.
In simple words: Escape velocity can also be calculated using a planet's radius, density, and the universal gravitational constant, providing an alternative way to determine the speed needed to leave its gravity.

🎯 Exam Tip: When given density (\(\rho\)), remember to substitute \( M = \rho \times \text{Volume} \) (for a sphere, Volume = \(\frac{4}{3}\pi R^{3}\)) into the escape velocity formula. Be meticulous with the numerical constants and the square root calculation.

 

Question 25. Calculate the escape velocity of a body from the moon. [g(moon) = 1.67 m/s², R(moon) = 1.74 × 10\(^{6}\) m]
Answer:
Solution:
Data: g = 1.67 m/s\(^{2}\), R = 1.74 × 10\(^{6}\) m
The escape velocity of a body from the moon,
\[ v_{esc} = \sqrt{2gR} \] \[ = \sqrt{2 \times 1.67 \text{ m/s}^{2} \times 1.74 \times 10^{6} \text{ m}} \] = 2.411 × 10\(^{3}\) m/s = 2.411 km/s.
In simple words: The Moon's escape velocity is significantly lower than Earth's due to its smaller mass and radius, making it easier for objects to leave its gravitational field.

🎯 Exam Tip: If 'g' for the celestial body is provided, use the simplified escape velocity formula \( v_{esc} = \sqrt{2gR} \). Ensure 'g' and 'R' are in standard units (m/s\(^{2}\) and m) for a correct result in m/s.

 

Question 26. The mass of (an imaginary) planet is four times that of the earth and its radius is double the rathus of the earth. The escape velocity of a body from the earth is 11.2 × 10\(^{3}\) mn/s. Find the escape velocity or a body from the planet.
Answer:
Solution:
Data: M\(_{2}\) = 4M\(_{1}\), R\(_{2}\) = 2R\(_{1}\),
v\(_{1esc}\) = 11.2 x 10\(^{3}\) m/s
\[ v_{esc} = \sqrt{\frac{2GM}{R}} \]
\(\implies\) \(\frac{v_{2esc}}{v_{1esc}}\) = \(\sqrt{\frac{M_{2}}{M_{1}} \times \frac{R_{1}}{R_{2}}}\) = \(\sqrt{4 \times \frac{1}{2}}\) = \(\sqrt{2}\)

\(\implies\) v\(_{2esc}\) = \(\sqrt{2}\) v\(_{1esc}\) = 1.414 × 11.2 × 10\(^{3}\) m/s
= 15.84 × 10\(^{3}\) m/s = 15.84 km/s
This is the escape velocity or a body from the planet.
In simple words: A planet with four times Earth's mass and double its radius will have a higher escape velocity, but not directly proportional to the mass increase, as the larger radius also plays a role.

🎯 Exam Tip: When comparing escape velocities of different planets, setting up a ratio \(\frac{v_{2esc}}{v_{1esc}}\) helps to cancel out constants like G and calculate the relative change efficiently. Be careful with the square root of the ratio terms.

 

Numerical Problems For Practice

 

Question 1. A satellite of mass 1000 kg revolves around the earth in a circular path. If the distance between the satellite and the centre of the earth is 40000 km, find the gravitational force exerted on the satellite by the earth.
[G = 6.67 × 10-11 N.m\(^{2}\)/kg\(^{2}\), mass or the earth =6 × 10\(^{24}\) kg, radius or the earthe 6.4 × 10\(^{6}\) m]
Answer:
250.1 N
In simple words: The gravitational force on a satellite is calculated using Newton's Law of Universal Gravitation, considering the masses of the satellite and Earth, and the distance between their centers.

🎯 Exam Tip: Ensure the distance from the center of the Earth to the satellite (r) is converted to meters (40000 km = 40 x 10\(^{6}\) m) before applying the formula \( F = \frac{GMm}{r^{2}} \).

 

Question 2. The masses of two spheres are 10 kg and 20 kg respectively. If the distance between their centers is 100 m, find the magnitude of the gravitational force between them.
Answer:
1.334 × 10-12 N)
In simple words: The gravitational force between two ordinary objects is very weak, even with significant masses, because the universal gravitational constant 'G' is a very small number.

🎯 Exam Tip: Use the formula \( F = \frac{Gm_{1}m_{2}}{r^{2}} \) and ensure all units are in SI before calculation. The result will be a small number, typical for gravitational forces between non-celestial objects.

 

Question 3. A satellite revolves around the earth along a circular path. If the mass of the satellite is 1000 kg and its distance from the center of the earth is 20000 km, find the magnitude of the earth's gravitational force acting on the satellite.
Answer:
1000.5
In simple words: Earth's gravitational force keeps satellites in orbit; its strength depends on the satellite's mass, Earth's mass, and the distance between them.

🎯 Exam Tip: Remember to convert the distance (20000 km) to meters before calculation. This problem is a direct application of Newton's Law of Universal Gravitation. (Note: The answer given, 1000.5, appears to be a numerical value without units or context, which is incomplete for an exam answer).

 

Question 4. Find the acceleration due to gravity at a distance of 20000 km from the center of the earth.
Answer:
1.0 m/s\(^{2}\)
In simple words: The acceleration due to gravity decreases as you move further away from the center of the Earth.

🎯 Exam Tip: Use the formula \( g = \frac{GM}{r^{2}} \) where 'r' is the distance from the Earth's center (20000 km converted to meters). Compare this value to 'g' at the surface to understand the effect of altitude.

 

Question 5. What is the weight of a body of mass 100 kg at the south pole? (g = 9.832 m/s²)
Answer:
983.2 N (downward)
In simple words: Weight is the force of gravity on an object's mass. At the poles, gravity is slightly stronger, resulting in a slightly higher weight for the same mass.

🎯 Exam Tip: Apply the formula W = mg. Ensure to use the specific 'g' value given for the South Pole. Remember that weight is a force and its unit is Newtons.

 

Question 6. What is the weight of a body of mass 20 kg at the equator? (g = 9.78 m/s²)
Answer:
195.6 N (downward)
In simple words: At the equator, the acceleration due to gravity is slightly less due to Earth's bulge and rotation, leading to a slightly lower weight for the same mass compared to the poles.

🎯 Exam Tip: Use the formula W = mg with the specific 'g' value for the equator. Be aware that 'g' varies slightly with latitude, making weight location-dependent.

 

Question 7. A body is released from the top of a tower of height 50 m. Find the velocity with which the body hits the ground, (g = 9.8 m/s²)
Answer:
31.3 m/s (downward)
In simple words: The final velocity of a falling object depends on the height from which it is dropped and the acceleration due to gravity.

🎯 Exam Tip: Use the kinematic equation v\(^{2}\) = u\(^{2}\) + 2gs. Since the body is "released," the initial velocity (u) is 0. Remember to take the square root for the final velocity and state the direction.

 

Question 8. A body is thrown vertically upward with a velocity of 9.8 m/s. Calculate the maximum height attained by the body. (g = 9.8 m/s²)
Answer:
4.9 m
In simple words: When an object is thrown upwards, it slows down due to gravity until its velocity becomes zero at the maximum height, at which point it starts to fall back down.

🎯 Exam Tip: At maximum height, the final velocity (v) is 0. Use the equation v\(^{2}\) = u\(^{2}\) + 2gs, but remember that 'g' acts downwards, so use -g or consider the direction appropriately. If using -g, s will be positive. If using +g, s will be negative for upward motion.

 

Question 9. A particle of mass 10-6 kg performs uniform circular motion. Its period is 10 s and the radius of the circle is 2 m. Find (i) the speed of the particle (ii) the centripetal acceleration of the particle (iii) the centripetal force on the particle.
Answer:
(i) 1.257 m/s
(ii) 0.79 m/s\(^{2}\)
(iii) 7.9 × 10-7 N
In simple words: For an object in uniform circular motion, its speed, centripetal acceleration, and centripetal force can all be calculated from its period and the radius of its path.

🎯 Exam Tip: Remember the formulas: speed \( v = \frac{2\pi r}{T} \), centripetal acceleration \( a_{c} = \frac{v^{2}}{r} \) (or \( \frac{4\pi^{2}r}{T^{2}} \)), and centripetal force \( F_{c} = ma_{c} \) (or \( \frac{mv^{2}}{r} \)). Ensure consistent units throughout.

 

Question 10. Find the gravitational potential energy of a body of mass 200 kg on the earth's surface. [M(earth) = 6 × 10\(^{24}\) kg, R(earth) = 6400 km]
Answer:
-1.251 × 10\(^{10}\)J
In simple words: The gravitational potential energy on Earth's surface for a given mass is a large negative value, signifying the strong gravitational binding to the planet.

🎯 Exam Tip: Convert R from km to meters (6400 km = 6.4 x 10\(^{6}\) m) before using the formula \( U = -\frac{GMm}{R} \). Accuracy in handling scientific notation is crucial for such calculations.

 

Question 11. Find the gravitational potential energy of a body of mass 10 kg when it is at a height of 6400 km from the earth's surface. [Given: a mass of the earth and radius of the earth. See Ex. 10 above.]
Answer:
-3.127 × 10\(^{8}\) J
In simple words: As an object moves higher above the Earth's surface, its gravitational potential energy becomes less negative (or increases), meaning it's less tightly bound by gravity.

🎯 Exam Tip: Remember that the distance 'r' in the potential energy formula becomes R+h when the object is at a height 'h' above the surface. Convert all distances to meters (6400 km = 6.4 x 10\(^{6}\) m) before calculating.

 

Question 12. Find the escape velocity of a body from the moon.
[M(moon) = 7.36 × 10\(^{22}\) kg, R(moon) = 1.74 × 10\(^{6}\) m]
Answer:
2.375 km/s
In simple words: The escape velocity from the Moon is much lower than Earth's, reflecting its significantly weaker gravitational pull due to its smaller mass and radius.

🎯 Exam Tip: Use the formula \( v_{esc} = \sqrt{\frac{2GM}{R}} \), substituting the Moon's mass and radius. Ensure 'R' is in meters and convert the final velocity from m/s to km/s.

 

Class 10 Questions And Answers

 

10th Std Science Part 1 Questions And Answers:

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