RBSE Solutions Class 9 Science Chapter 10 Gravitation

Objective Type Questions

 

Question 1. Newton's law of gravitation is called universal law of gravitation because:
(A) it is always valid for forces of attraction
(B) it is applicable to all particles and members of the solar family
(C) it is applicable for all masses at all distances, independent of the medium
(D) None of the options
Answer: (C) it is applicable for all masses at all distances, independent of the medium
In simple words: Newton's law of gravitation works for all types of masses, no matter how far apart they are or what material is between them. This makes it a universal rule.

🎯 Exam Tip: Remember that "universal" in scientific laws often means it applies everywhere, under all conditions, without exceptions.

 

Question 2. The necessary force required to circulate a body in circular motion is:
(A) gravitational force
(B) frictional force
(C) centripetal force
(D) None of the options
Answer: (C) centripetal force
In simple words: To make something move in a circle, a special force is needed that pulls it towards the center of the circle. This force is called centripetal force.

🎯 Exam Tip: Distinguish between centripetal (center-seeking) force, which keeps an object in circular motion, and centrifugal (center-fleeing) force, which is an apparent force felt outwards in a rotating frame.

 

Question 3. Universal gravitational constant, G depends:
(A) on the nature of the particle
(B) on the medium present between the particles
(C) on time
(D) does not depend on any of these factors.
Answer: (D) does not depend on any of these factors.
In simple words: The universal gravitational constant (G) is a fixed number that never changes. It stays the same no matter what objects are involved, what is between them, or when the measurement is taken.

🎯 Exam Tip: The capital G (universal gravitational constant) is a fundamental constant, while the small g (acceleration due to gravity) varies with location and mass of the planet.

 

Question 4. If the weight of a man on earth's surface is 60 N, then what will be the weight on the surface of the moon?
(A) 60 N
(B) 10 N
(C) 20 N
(D) 0 N
Answer: (D) 0 N
In simple words: The moon's gravity is much weaker than Earth's. Because of this, an object weighing 60 N on Earth would feel much lighter on the Moon. Specifically, it would be 1/6th of the weight. So 60 N / 6 = 10 N.

🎯 Exam Tip: Remember that weight changes with gravity, but mass remains constant everywhere. The question has a typo, the correct answer for 60N on Earth would be 10N on the moon (60/6). Option (D) 0 N usually implies weightlessness, which isn't the case on the moon's surface unless in orbit. However, if (D) 0N is the given answer, stick to it for the MCQ.

 

Question 6. we double the distance between two masses, the gravitational force between the masses:
(A) remains unchanged
(B) becomes one fourth
(C) becomes half
(D) becomes twice
Answer: (C) becomes half
In simple words: Gravitational force becomes one-fourth when the distance is doubled. The force decreases rapidly as objects move further apart. So if the original force was F, doubling the distance makes it F/4.

🎯 Exam Tip: Gravitational force is inversely proportional to the square of the distance. If distance doubles, the force becomes \( 1/2^2 = 1/4 \) of the original. There might be a typo in the provided answer (C). Based on the law, it should be 1/4 (option B).

Gravitation Very Short Answer Type Questions

 

Question 7. From where do satellites receive the necessary centripetal force to revolve around a planet?
Answer: Satellites get the necessary centripetal force from the planet's gravitational pull. This pull keeps them moving in their circular path. The planet's gravity acts like an invisible rope, constantly pulling the satellite towards its center.
In simple words: Satellites get the force needed to orbit from the planet's gravity.

🎯 Exam Tip: The gravitational force between a planet and its satellite provides the centripetal force for orbital motion.

 

Question 8. How does the gravitational mass of a body is determined in an artificial satellite?
Answer: The gravitational mass of a body cannot be easily determined inside an artificial satellite. This is because an artificial satellite is always in a state of free-falling around the Earth, which makes objects inside it appear weightless. In a weightless environment, typical methods for measuring mass based on weight don't work.
In simple words: You cannot find the gravitational mass of a body in a satellite because it's like being in free fall, so everything feels weightless.

🎯 Exam Tip: Remember that apparent weightlessness in orbit does not mean zero mass; the mass of an object remains constant, but its weight is effectively zero due to continuous free fall.

 

Question 9. What changes will arise in force of gravitation between the masses, if the distance between them is doubled?
Answer: If the distance (d) between two masses is doubled, the gravitational force between them will become one fourth. This is because gravitational force is inversely proportional to the square of the distance between the objects. So, if the distance doubles, the force becomes \( 1/(2^2) = 1/4 \).
In simple words: If you double the distance, the gravity force between objects becomes four times weaker.

🎯 Exam Tip: Clearly state the inverse square relationship in your answer: force is inversely proportional to the square of the distance.

 

Question 10. Find the weight of an object on the surface of the earth, if its mass is 10 kg.
Answer: To find the weight of an object on Earth, we use the formula: Weight \( = \) mass \( \times \) acceleration due to gravity (g). The mass of the object is 10 kg. The approximate value of acceleration due to gravity (g) on Earth is \( 9.8 \text{ m/s}^2 \).
So, Weight \( = 10 \text{ kg} \times 9.8 \text{ m/s}^2 = 98 \text{ N} \).
Thus, the weight of the object on the surface of the Earth is 98 Newtons.
In simple words: The weight of a 10 kg object on Earth is 98 N, found by multiplying its mass by Earth's gravity.

🎯 Exam Tip: Always specify the unit of weight as Newtons (N) and the unit of mass as kilograms (kg). Remember to use \( g = 9.8 \text{ m/s}^2 \) unless a different value is given.

 

Question 12. Name the principle on which ball pen works.
Answer: A ball pen works on the principle of the Universal Law of Gravitation, which is related to how ink is pulled down to the ball. More specifically, it relies on gravity pulling the ink towards the ball, and the ball rotating to spread the ink. It also uses surface tension which helps the ink stick to the ball and paper.
In simple words: A ball pen works using the principle of gravity, which helps ink flow to the ball.

🎯 Exam Tip: While gravity plays a role, for a more complete answer, also consider mentioning surface tension and the capillary action that helps the ink flow. However, if the source specifically mentions "Universal Law of Gravitation," adhere to that context.

 

Question 13. A man can jump higher on the moon. What is the cause of it?
Answer: A man can jump higher on the moon because the acceleration due to gravity on the moon is much less than on Earth. The moon's gravity is about one-sixth that of Earth's gravity. This means that a person experiences less downward pull, allowing them to jump higher and farther. Less gravity allows for greater vertical movement with the same effort.
In simple words: A man can jump higher on the moon because its gravity is much weaker than Earth's, so he feels less pull downwards.

🎯 Exam Tip: When explaining why things are different on the Moon, always refer to the difference in acceleration due to gravity (g) compared to Earth.

 

Question 14. Two bodies of masses 1 kg each are situated 1 metre apart. Find the gravitational force between them.
Answer: We are given the following values:
Mass of the first body (\( m_1 \)) \( = 1 \text{ kg} \)
Mass of the second body (\( m_2 \)) \( = 1 \text{ kg} \)
Distance between them (\( d \)) \( = 1 \text{ m} \)
The universal gravitational constant (\( G \)) \( = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \)
We use Newton's Law of Gravitation to find the force (\( F \)):
\( F = \frac{G m_1 m_2}{d^2} \)

\( \implies F = \frac{(6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2) \times (1 \text{ kg}) \times (1 \text{ kg})}{(1 \text{ m})^2} \)

\( \implies F = \frac{6.67 \times 10^{-11} \times 1 \times 1}{1^2} \)

\( \implies F = 6.67 \times 10^{-11} \text{ N} \)
The gravitational force between the two bodies is \( 6.67 \times 10^{-11} \text{ N} \). This force is very small because gravity is a weak force unless one of the masses is very large.
In simple words: With two 1 kg objects 1 meter apart, the gravitational force between them is \( 6.67 \times 10^{-11} \text{ N} \), which is a tiny amount.

🎯 Exam Tip: Clearly state the given values and the formula used. Ensure correct units for \( G \) and the final answer. Remember the magnitude of \( G \) means gravitational force is significant only for very massive objects.

 

Question 15. Earth is continuously pulling moon towards its centre by gravitational force, then why does not moon fall on the earth?
Answer: The Moon does not fall on Earth even though Earth pulls it with gravitational force because the Moon is also moving sideways at a very high speed. This sideways motion, combined with Earth's gravitational pull, provides the necessary centripetal force that keeps the Moon in its orbit around Earth. Instead of falling straight down, it keeps "falling" around the Earth in a continuous path, much like a ball swung on a string.
In simple words: The Moon doesn't fall to Earth because it's always moving sideways very fast while Earth pulls on it, making it orbit instead.

🎯 Exam Tip: Explain that the gravitational force acts as the centripetal force, continuously changing the Moon's direction but not its speed (ideally), keeping it in orbit.

 

Question 16. W Typesetting math: 18% cean tides?
Answer: The phenomenon of ocean tides occurs due to the gravitational force of attraction mainly exerted by the Moon and to some extent by the Sun on the Earth's surface. The Moon's gravitational pull causes the ocean water to bulge towards and away from it, creating high and low tides. This gravitational pull affects both the water and the solid Earth, but the water, being fluid, responds more noticeably.
In simple words: Ocean tides are caused by the Moon's gravity pulling on Earth's oceans.

🎯 Exam Tip: Mention both the Moon and the Sun's gravitational influence, but emphasize the Moon's greater effect due to its closer proximity to Earth.

Gravitation Short Answer Type Questions

 

Question 17. State Newton's law of gravitation.
Answer: Newton's Law of Gravitation states: "Every particle in the universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers." This means that heavier objects pull on each other with a stronger force, and objects further apart pull on each other with a weaker force.
In simple words: Every object in the universe pulls on every other object. This pull gets stronger if the objects are heavier and weaker if they are farther apart.

🎯 Exam Tip: Ensure you include both parts of the law: direct proportionality to masses and inverse proportionality to the square of the distance.

 

Question 18. At which place on the surface of the earth, the acceleration due to gravity is more and why
Answer: The acceleration due to gravity (g) is highest at the poles of the Earth. This happens because the Earth is not a perfect sphere; it is slightly flattened at the poles and bulges at the equator. This flattening means that the radius (r) of the Earth is smallest at the poles. Since the acceleration due to gravity is inversely proportional to the square of the radius (\( g \propto \frac{1}{r^2} \)), a smaller radius results in a larger value of g. Therefore, the gravitational pull is stronger at the poles than at the equator.
In simple words: Gravity is strongest at Earth's poles because the poles are slightly closer to the Earth's center than the equator.

🎯 Exam Tip: To score full marks, mention the Earth's oblate spheroid shape and the inverse square relationship between 'g' and radius.

 

Question 19. What is the value of universal gravitational constant? Write also its S.I. unit.
Answer: The value of the universal gravitational constant (\( G \)) is \( 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \). Its S.I. unit is Newton metre squared per kilogram squared (\( \text{Nm}^2/\text{kg}^2 \)). This constant is used in the equation for gravitational force to calculate the strength of the gravitational pull between any two objects.
In simple words: The universal gravitational constant (G) is \( 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \).

🎯 Exam Tip: Memorize both the numerical value and the correct S.I. unit for \( G \), as both are frequently asked in exams.

 

Question 20. What is acceleration due to gravity? Write its formula.
Answer: Acceleration due to gravity (denoted by 'g') is the acceleration that a body experiences due to the Earth's gravitational pull when near its surface. This acceleration makes objects fall faster and faster towards the ground. The formula for acceleration due to gravity is:
\[ g = \frac{G M_e}{R_e^2} \]
Here,
\( G \) = Universal gravitational constant
\( M_e \) = Mass of Earth
\( R_e \) = Radius of Earth
In simple words: Acceleration due to gravity is how fast an object speeds up when falling towards Earth, and its formula uses the gravitational constant, Earth's mass, and Earth's radius.

🎯 Exam Tip: Clearly define 'g' as an acceleration and remember the formula, understanding that it's derived from Newton's law of gravitation and the second law of motion.

 

Question 21. What are the differences between the mass of an object and its weight? Al Typesetting math: 18%
Answer: The main differences between mass and weight are:

Mass is an intrinsic property of an object, while weight is a force that depends on the gravitational field it is in.
In simple words: Mass is how much 'stuff' an object has and stays the same, but weight is how strongly gravity pulls on that object and changes depending on where you are.

🎯 Exam Tip: Present the differences in a clear table format, highlighting the key distinctions such as definition, constancy, measurement tool, and SI unit for both mass and weight.

 

Question 22. What do you mean by free fall? Give examples.
Answer: Free fall is when an object falls towards the Earth purely under the influence of Earth's gravitational force, with no other forces like air resistance acting on it. In this state, objects accelerate downwards at 'g' (approximately \( 9.8 \text{ m/s}^2 \)). An example of free fall is when you drop a stone from a height and it falls to the ground. Another example is an apple falling from a tree when air resistance is ignored.
In simple words: Free fall means an object is falling only because of gravity, like a stone dropped from a tall building.

🎯 Exam Tip: Emphasize that free fall happens only under gravity, meaning air resistance is negligible or absent. Give clear, simple examples.

 

Question 23. A man moves from earth poles to equatorial line, then what changes will be there in man's weight and why?
Answer: When a man moves from the Earth's poles to the equatorial line, his weight will decrease. This happens because the Earth is not perfectly round; it bulges at the equator and is flattened at the poles. Due to this shape, the acceleration due to gravity (g) is slightly higher at the poles and lower at the equator because the distance to the Earth's center is greater at the equator. Since weight is calculated as mass multiplied by gravity (Weight \( = m \times g \)), a smaller 'g' at the equator means a smaller weight.
In simple words: A man's weight will become less when he moves from the poles to the equator because gravity is slightly weaker at the equator.

🎯 Exam Tip: Explain both the 'what' (weight decreases) and the 'why' (Earth's shape, varying 'g') to provide a complete answer.

 

Question 24. What is weightlessness? Give two examples.
Answer: Weightlessness is a situation where the effective weight of a body becomes zero. This doesn't mean the object loses its mass, but rather that it doesn't feel the force of gravity or is in a constant state of free fall. This can occur when there's no normal force supporting the body.
Example 01: When a body is inside a freely falling lift, its effective acceleration (a) is equal to the acceleration due to gravity (g). So, the effective acceleration due to gravity becomes \( g' = g - a = g - g = 0 \). This makes the weight of the body zero inside the falling lift.
Example 02: When a body is inside a satellite orbiting around the Earth, the Earth's gravitational pull on the body is balanced by the centrifugal force. This balance causes the effective weight of the body to become zero, leading to the sensation of weightlessness. Astronauts floating in space are a common example.
In simple words: Weightlessness is when an object feels like it has no weight. This happens inside a falling lift or in a satellite orbiting Earth.

🎯 Exam Tip: Clarify that weightlessness is an apparent state, not a loss of mass, and provide diverse examples like a freely falling lift and an orbiting satellite.

 

Question 25. Quoction 25 Typesetting math: 18% Wie me quaισης motion for a freely falling body. Also, write the meaning of symbols used in the
Answer: The equations of motion for a freely falling body, where 'g' is the acceleration due to gravity, are:
1. \( v = u + gt \)
2. \( s = ut + \frac{1}{2}gt^2 \)
3. \( v^2 = u^2 + 2gs \)
Here, if the object falls from rest, its initial velocity (u) is 0. Also, 'g' is taken as positive for downward motion. These equations help calculate velocity, distance, or time for objects under gravity.
The meaning of the symbols used is:
\( u \) = initial velocity (the speed at the start)
\( v \) = final velocity (the speed at the end)
\( t \) = time (how long it takes)
\( g \) = acceleration due to gravity (the rate at which speed changes due to gravity)
\( s \) = displacement/height (the distance covered)
In simple words: The equations of motion tell us how objects move when gravity pulls them down. The letters stand for starting speed (u), ending speed (v), time (t), gravity's pull (g), and distance (s).

🎯 Exam Tip: Be precise with your equations and accurately define each symbol. Also, specify the sign convention for 'g' (positive for downward, negative for upward motion).

 

Question 26. A body is thrown upwards with velocity 'u' and it attains a height h. Write the equation of motion for the body.
Answer: When an object is thrown upwards, the acceleration due to gravity (g) acts downwards, so its value is considered negative in the equations of motion. The body eventually stops at its highest point, meaning its final velocity (v) is zero there. The equations of motion for a body thrown upwards are:
1. \( v = u - gt \)
2. \( h = ut - \frac{1}{2}gt^2 \)
3. \( v^2 = u^2 - 2gh \)
These equations help describe the object's motion as it fights against gravity to reach a certain height.
In simple words: When an object is thrown upwards, gravity slows it down, so the equations of motion use a negative 'g' and help calculate its speed or height.

🎯 Exam Tip: Remember to use negative 'g' for upward motion and understand that final velocity (v) is zero at the maximum height for objects thrown vertically upwards.

 

Question 27. What problems occur for astronaut during weightlessness?
Answer: Astronauts face several challenges during weightlessness in space:
(a) Eating and drinking become difficult. Food and water do not stay in place and float around, making it hard to consume them without specialized tools. Liquids don't pour and must be squeezed from containers.
(b) Long periods in space adversely affect astronauts' bodies. They can experience bone density loss and muscle weakening because their muscles and bones are not working against gravity as they do on Earth. This leads to the need for regular exercise programs.
(c) Inside the spacecraft, all objects, including the astronauts themselves, are in a floating position. This means tools can drift away, and daily tasks require careful management to prevent items from floating off. This can make simple activities more complex.
In simple words: Astronauts face problems like food floating away, weaker bones and muscles, and everything floating inside the spacecraft due to weightlessness.

🎯 Exam Tip: Focus on the physiological effects (bone/muscle loss) and practical difficulties (eating/drinking, floating objects) as key problems of weightlessness.

 

Question 28. Before Kepler and Newton, name some ancient astrologers. Also, write their achievement in the gravitational field.
Answer: Before Kepler and Newton, several ancient scholars made significant observations and theories related to planetary motion and gravity:
(a) In the fifth century, the Indian astrologer Aryabhatta developed the geocentric model. This model, where Earth is at the center of the universe, helped explain the observed motions of planets and stars at that time.
(b) After Aryabhatta, Bhaskaracharya, another Indian scholar, presented his theories on gravity and planetary motion in his book 'Siddhanta Shiromani'. He even estimated the radius of the Earth and its circumference quite accurately. Bhaskaracharya's contributions laid some groundwork for later understanding of celestial mechanics.
In simple words: Before Kepler and Newton, Aryabhatta proposed the Earth-centered model, and Bhaskaracharya wrote about gravity and estimated Earth's size.

🎯 Exam Tip: When discussing historical figures, always associate their names with specific contributions or theories they developed, such as Aryabhatta's geocentric model and Bhaskaracharya's calculations.

Gravitation Long Answer Type Questions

 

Question 30. Two spheres each of mass 10 kg are 50 cm apart. What is the gravitational force between them?
Answer: We are given the following values:
Mass of the first sphere (\( m_1 \)) \( = 10 \text{ kg} \)
Mass of the second sphere (\( m_2 \)) \( = 10 \text{ kg} \)
Distance between them (\( d \)) \( = 50 \text{ cm} \). We need to convert this to meters: \( 50 \text{ cm} = 0.50 \text{ m} = \frac{1}{2} \text{ m} = 50 \times 10^{-2} \text{ m} \).
The universal gravitational constant (\( G \)) \( = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \).
According to Newton's Law of Gravitation, the force (\( F \)) is given by:
\[ F = \frac{G m_1 m_2}{d^2} \]
Substitute the given values into the formula:
\[ F = \frac{(6.67 \times 10^{-11}) \times 10 \times 10}{(50 \times 10^{-2})^2} \]
\[ F = \frac{6.67 \times 10^{-11} \times 100}{2500 \times 10^{-4}} \]
\[ F = \frac{6.67 \times 10^{-11} \times 10^2}{(50)^2 \times (10^{-2})^2} \]
\[ F = \frac{6.67 \times 10^{-11} \times 10^2}{2500 \times 10^{-4}} \]
\[ F = \frac{6.67 \times 10^{-9}}{25 \times 10^{-2}} \]
\[ F = \frac{6.67 \times 10^{-9}}{0.25} \]
\[ F = 26.68 \times 10^{-9} \text{ N} \]
\[ F = 2.668 \times 10^{-8} \text{ N} \]
The gravitational force between the two spheres is approximately \( 2.67 \times 10^{-8} \text{ N} \). This calculation shows that the gravitational force between everyday objects is extremely small.
In simple words: For two 10 kg spheres 50 cm apart, the gravity pulling them together is a very tiny force, about \( 2.67 \times 10^{-8} \text{ N} \).

🎯 Exam Tip: Always convert all units to SI units (meters, kilograms) before starting calculations. Be careful with exponents, especially when squaring a value with a power of 10.

 

Question 31. Find the gravitational force on a body of man 40 kg situated on the surface of the earth; if radius of earth (R) = 6400 km and mass of earth (M) = 6 x 1024 kg.
Answer: We are given the following values:
Mass of the body (\( m \)) \( = 40 \text{ kg} \)
Mass of Earth (\( M \)) \( = 6 \times 10^{24} \text{ kg} \)
Radius of Earth (\( R \)) \( = 6400 \text{ km} \). We convert this to meters: \( 6400 \text{ km} = 6400 \times 10^3 \text{ m} = 6.4 \times 10^6 \text{ m} \).
Universal gravitational constant (\( G \)) \( = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \).
According to Newton's gravitational law, the force (\( F \)) is given by:
\[ F = \frac{G m M}{R^2} \]
Substitute the values into the formula:
\[ F = \frac{(6.67 \times 10^{-11}) \times 40 \times (6 \times 10^{24})}{(6.4 \times 10^6)^2} \]
\[ F = \frac{6.67 \times 10^{-11} \times 240 \times 10^{24}}{(6.4)^2 \times (10^6)^2} \]
\[ F = \frac{6.67 \times 240 \times 10^{13}}{40.96 \times 10^{12}} \]
\[ F = \frac{1600.8 \times 10^{13}}{40.96 \times 10^{12}} \]
\[ F = \frac{1600.8}{40.96} \times 10^{13-12} \]
\[ F = 39.082 \times 10^1 \text{ N} \]
\[ F = 390.82 \text{ N} \]
The gravitational force on the man of 40 kg on the Earth's surface is approximately 390.82 N. This force is what we perceive as the man's weight.
In simple words: The gravitational pull on a 40 kg person on Earth is 390.82 N.

🎯 Exam Tip: Pay careful attention to unit conversions (km to m) and handling large exponents correctly in your calculations.

 

Question 33. If the radius and mass of moon are 1738 km and 0.073 x 1024 kg, respectively. Find the acceleration due to gravity on the surface of the moon.
Answer: We are given the following values for the Moon:
Radius of Moon (\( R_m \)) \( = 1738 \text{ km} \). Convert to meters: \( 1738 \times 10^3 \text{ m} = 1.738 \times 10^6 \text{ m} \).
Mass of Moon (\( M_m \)) \( = 0.073 \times 10^{24} \text{ kg} = 7.3 \times 10^{22} \text{ kg} \).
Universal gravitational constant (\( G \)) \( = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \).
The acceleration due to gravity on the surface of the moon (\( g_m \)) is given by the formula:
\[ g_m = \frac{G M_m}{R_m^2} \]
Substitute the values:
\[ g_m = \frac{(6.67 \times 10^{-11}) \times (7.3 \times 10^{22})}{(1.738 \times 10^6)^2} \]
\[ g_m = \frac{48.691 \times 10^{11}}{(1.738)^2 \times (10^6)^2} \]
\[ g_m = \frac{48.691 \times 10^{11}}{3.020644 \times 10^{12}} \]
\[ g_m = \frac{48.691}{3.020644} \times 10^{11-12} \]
\[ g_m = 16.118 \times 10^{-1} \text{ m/s}^2 \]
\[ g_m = 1.6118 \text{ m/s}^2 \]
So, the acceleration due to gravity on the surface of the Moon is approximately \( 1.61 \text{ m/s}^2 \). This is much less than Earth's gravity of \( 9.8 \text{ m/s}^2 \), which explains why objects are lighter on the Moon.
In simple words: By using the Moon's mass and radius, we calculate that its gravity is about \( 1.61 \text{ m/s}^2 \), making objects much lighter there.

🎯 Exam Tip: Be careful with scientific notation and power calculations, especially when squaring numbers with exponents. Ensure all units are consistent (SI units) before calculation.

 

Question 34. A stone is released from the top of a tower 125 m high. Then find:
(1) the time with which the stone strikes the ground.
(2) the final velocity of the stone. (Take g = 10 m/s²)
Answer: We are given the following information:
Height of tower (\( h \)) \( = 125 \text{ m} \)
Since the stone is released, its initial velocity (\( u \)) \( = 0 \text{ m/s} \)
Acceleration due to gravity (\( g \)) \( = 10 \text{ m/s}^2 \)

(1) To find the time (\( t \)) with which the stone strikes the ground, we use the second equation of motion under gravity:
\( s = ut + \frac{1}{2}gt^2 \)
Here, \( s = h = 125 \text{ m} \).
\( 125 = (0)t + \frac{1}{2}(10)t^2 \)
\( 125 = 5t^2 \)
\( t^2 = \frac{125}{5} \)
\( t^2 = 25 \)
\( t = \sqrt{25} \)
\( t = 5 \text{ seconds} \)

(2) To find the final velocity (\( v \)) of the stone, we can use the first equation of motion:
\( v = u + gt \)
\( v = 0 + (10)(5) \)
\( v = 50 \text{ m/s} \)
Alternatively, using the third equation of motion:
\( v^2 = u^2 + 2gs \)
\( v^2 = (0)^2 + 2(10)(125) \)
\( v^2 = 0 + 2500 \)
\( v = \sqrt{2500} \)
\( v = 50 \text{ m/s} \)
The stone will strike the ground after 5 seconds with a final velocity of 50 m/s. These calculations show how quickly an object can accelerate under constant gravity.
In simple words: A stone dropped from 125 m will hit the ground in 5 seconds, reaching a speed of 50 m/s.

🎯 Exam Tip: Clearly state your knowns and unknowns. Use the appropriate equation of motion. Remember that for objects released from rest, the initial velocity (u) is zero.

 

Question 35. If the diameter of the earth becomes half of its original value, then mass becomes 1/8 of its original value. Find the acceleration due to gravity of this part.
Answer: Let the original radius of the Earth be \( r \) and the original mass be \( m \). The original acceleration due to gravity (\( g \)) is given by:
\[ g = \frac{G m}{r^2} \]
Now, if the diameter becomes half, the new radius (\( R \)) will also be half of the original radius:
\[ R = \frac{r}{2} \]
If the mass becomes 1/8 of its original value, the new mass (\( M \)) will be:
\[ M = \frac{m}{8} \]
The effective (new) acceleration due to gravity (\( g' \)) will be:
\[ g' = \frac{G M}{R^2} \]
Substitute the new values for \( M \) and \( R \):
\[ g' = \frac{G \left(\frac{m}{8}\right)}{\left(\frac{r}{2}\right)^2} \]
\[ g' = \frac{G \frac{m}{8}}{\frac{r^2}{4}} \]
\[ g' = \frac{G m}{8} \times \frac{4}{r^2} \]
\[ g' = \frac{4}{8} \times \frac{G m}{r^2} \]
\[ g' = \frac{1}{2} \times g \]
Thus, the acceleration due to gravity becomes half of its original value. This illustrates how both mass and radius significantly influence gravity.
In simple words: If Earth's diameter becomes half and its mass becomes 1/8, then the new gravity will be half of the old gravity.

🎯 Exam Tip: Clearly define original and new variables. Show each step of algebraic substitution and simplification carefully to avoid errors, especially with fractions and squares.

Gravitation Addition Questions Solved

Multiple Choice Questions (MCQs)

 

Question 1. A ball is thrown vertically upwards. The acceleration due to gravity:
(A) is in the upward direction
(B) is in the downward direction
(C) is in the horizontal direction
(D) always in the direction of motion
Answer: (B) is in the downward direction
In simple words: When a ball is thrown up, the force of gravity always pulls it down, no matter which way the ball is moving.

🎯 Exam Tip: Remember that acceleration due to gravity always acts towards the center of the Earth (downwards), regardless of the object's motion (up, down, or sideways).

 

Question 2 Typesetting math: 18% The ruin vuon due to gravity:
Answer: The question is incomplete and cannot be fully answered. Based on the options and context, it likely asks about the nature or value of acceleration due to gravity.
In simple words: The question is broken, so a full answer cannot be given.

🎯 Exam Tip: If you encounter a mangled or incomplete question, state that it cannot be fully answered due to missing information, but try to infer the likely topic.

 

Question 3. The gravitational force between two objects is F. If the masses of both the objects are halved, keeping the distance between them the same, then the gravitational force would become:
(A) 1/2 F
(B) 1/4 F
(C) 2 F
(D) 4 F
Answer: (B) 1/4 F
In simple words: If you cut the mass of both objects in half, the gravitational force between them becomes one-fourth of what it was before.

🎯 Exam Tip: Gravitational force is directly proportional to the product of the masses. If both masses are halved, the product becomes \( (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{4} \) of the original.

 

Question 4. Two bodies of different masses are falling freely near the surface of the earth. They:
(A) have the same velocity at any instant
(B) move with different accelerations.
(C) experience forces of the same magnitude
(D) undergo a change in their inertia.
Answer: (A) have the same velocity at any instant
In simple words: When objects fall freely near Earth, no matter how heavy or light they are, they speed up at the same rate and have the same speed at any moment.

🎯 Exam Tip: In free fall (ignoring air resistance), all objects experience the same acceleration (g), which means they will have the same velocity at any given time if dropped from the same height.

 

Question 5. The law of gravitation is true for:
(A) the earth and any object only
(B) earth and the sun only
(C) any two massive bodies
(D) two charged bodies only
Answer: (C) any two massive bodies
In simple words: The law of gravity applies to any two objects that have mass, not just Earth or charged items.

🎯 Exam Tip: Understand that gravity is a universal force acting between *any* two objects with mass, making it applicable throughout the universe.

 

Question 6. At the centre of the earth:
(A) the mass of a body is zero but the weight is not zero
(B) the mass of a body is not zero but the weight is zero
(C) both mass and weight of a body are zero
(D) both mass and weight of a body are not zero
Answer: (B) the mass of a body is not zero but the weight is zero
In simple words: At the very center of the Earth, an object still has its mass, but its weight becomes zero because the gravitational forces pull it equally in all directions, canceling each other out.

🎯 Exam Tip: Recall that mass is an intrinsic property and never zero. Weight, however, depends on gravity, and at the Earth's center, the net gravitational force is zero, hence zero weight.

 

Question 8. If the distance between two objects is doubled, the gravitational force between them:
(A) remains unchanged
(B) becomes half
(C) becomes one-fourth
(D) becomes double
Answer: (C) becomes one-fourth
In simple words: If you make the distance between two objects twice as long, the gravity pulling them together becomes four times weaker.

🎯 Exam Tip: Always apply the inverse square law: force is proportional to \( 1/d^2 \). So if \( d \) becomes \( 2d \), the force becomes \( 1/(2d)^2 = 1/(4d^2) \).

 

Question 9. The S.I. unit of weight is:
(A) kg
(B)kg-¹ nr¹
(C) N-m
(D) N
Answer: (D) N
In simple words: The standard unit for measuring weight is Newtons (N) because weight is a force.

🎯 Exam Tip: Remember that weight is a force, so its SI unit is the Newton (N), not kilograms (kg), which is the unit for mass.

 

Question 10. A stone is dropped from the top of a tower 40 metre high. Its velocity after it has fallen for 2s is:
(Take g = 10m/s²).
(A) - 10 m/s
(B) 10 m/s
(C) - 20 m/s
(D) 20 m/s
Answer: (C) - 20 m/s
In simple words: A stone dropped from a 40m tower will have a speed of 20 m/s downwards after 2 seconds. The negative sign shows it's moving downwards.

🎯 Exam Tip: Use the equation \( v = u + gt \). Since it's dropped, \( u=0 \). Take downward direction as positive or negative consistently. If upward is positive, downward is negative. Here, the options show \( -20 \text{ m/s} \) meaning downward if upward is taken positive. If downward is positive, the answer would be \( +20 \text{ m/s} \).

Gravitation Very Short Answer Type Questions

 

Question 1. What will be the weight of an object on the earth whose mass is 10 kg? [g = 10 m/s²]
Answer: To find the weight of an object, we use the formula: Weight \( = \) mass \( \times \) acceleration due to gravity (g).
Given: Mass (\( m \)) \( = 10 \text{ kg} \)
Given: Acceleration due to gravity (\( g \)) \( = 10 \text{ m/s}^2 \)
Weight \( = 10 \text{ kg} \times 10 \text{ m/s}^2 \)
Weight \( = 100 \text{ N} \)
The weight of the object on Earth is 100 Newtons. This calculation shows the direct relationship between mass, gravity, and weight.
In simple words: A 10 kg object on Earth will weigh 100 N if we use \( g = 10 \text{ m/s}^2 \).

🎯 Exam Tip: Always use the given value of 'g' for calculations. Clearly state the formula and units in your answer.

 

Question 2. How Newton's second law of motion, related to the universal law of gravitation?
Answer: Newton's second law of motion (\( F = ma \)) and the universal law of gravitation (\( F = \frac{G M m}{r^2} \)) are closely related. The force of gravity acting on an object is its weight. From Newton's second law, this force (\( F \)) can also be written as \( mg \) (where \( a \) becomes \( g \)).
So, by Newton's second law of motion: \( F = mg \)
And by universal law of gravitation: \( F = \frac{G M m}{r^2} \)
Comparing these two expressions for force, we get:
\( mg = \frac{G M m}{r^2} \)

\( \implies g = \frac{G M}{r^2} \)
This equation shows that the acceleration due to gravity (\( g \)) depends on the mass of the planet (\( M \)) and its radius (\( r \)), not on the mass of the object (\( m \)). This connection explains why all objects fall with the same acceleration near the Earth's surface, regardless of their mass. This is a fundamental concept in understanding falling objects.
In simple words: Newton's second law (F=ma) and his law of gravity (F=GMm/r²) are linked because the force of gravity is what causes the acceleration (a=g). By combining them, we find that the acceleration due to gravity depends on the planet, not the falling object.

🎯 Exam Tip: Derive \( g = \frac{G M}{r^2} \) by equating \( F=mg \) and \( F=\frac{GMm}{r^2} \). This derivation shows the fundamental relationship between the two laws.

 

Question 3. What will be the weight of the book whose mass is 500 gm, placed at c height equal to the radius of the earth?
Answer: We need to find the weight of a book with a mass of 500 gm at a height equal to the Earth's radius. First, convert the mass to kg: \( 500 \text{ gm} = 0.5 \text{ kg} \).
The acceleration due to gravity on the Earth's surface is approximately \( g = 9.8 \text{ m/s}^2 \).
When an object is at a height \( h \) equal to the Earth's radius \( R \), the effective radius from the center of the Earth becomes \( R + h = R + R = 2R \).
The acceleration due to gravity at this height (\( g' \)) is given by:
\[ g' = g \left( \frac{R}{R+h} \right)^2 \]
Since \( h = R \):
\[ g' = g \left( \frac{R}{R+R} \right)^2 \]
\[ g' = g \left( \frac{R}{2R} \right)^2 \]
\[ g' = g \left( \frac{1}{2} \right)^2 \]
\[ g' = g \times \frac{1}{4} = \frac{g}{4} \]
So, the acceleration due to gravity at this height is one-fourth of the surface gravity: \( g' = \frac{9.8}{4} = 2.45 \text{ m/s}^2 \).
Now, calculate the weight of the book at this height:
Weight \( = \text{mass} \times g' \)
Weight \( = 0.5 \text{ kg} \times 2.45 \text{ m/s}^2 \)
Weight \( = 1.225 \text{ N} \)
Therefore, the weight of the book at a height equal to the Earth's radius is 1.225 N. This shows how gravity weakens with distance.
In simple words: A 500 gm book placed at a height equal to Earth's radius will weigh 1.225 N, because gravity becomes four times weaker there.

🎯 Exam Tip: Remember to convert all units to SI (grams to kilograms) and correctly calculate the effective radius. The formula for 'g' at height 'h' is crucial for such problems.

 

Question 4. The gravitational force acting on all objects is proportional to their masses. Why, then, a heavy object does not fall faster than a light object?
Answer: Although the gravitational force (weight) acting on an object is proportional to its mass (a heavy object experiences a greater gravitational force than a light one), a heavy object does not fall faster than a light object in a vacuum. This is because the acceleration produced by gravity (\( g \)) is independent of the mass of the falling body. When we combine Newton's second law (\( F=ma \)) with the law of universal gravitation (\( F=\frac{GMm}{r^2} \)), we find that the mass of the falling object (\( m \)) cancels out, leaving \( a = g = \frac{GM}{r^2} \). So, all objects fall with the same acceleration (\( 9.8 \text{ m/s}^2 \)) regardless of their individual masses. This means they will fall at the same rate and reach the ground at the same time if air resistance is ignored.
In simple words: Heavy objects don't fall faster than light ones because the extra gravitational pull on a heavy object is exactly balanced by its greater resistance to changing speed, so all objects accelerate the same amount due to gravity.

🎯 Exam Tip: The key is to explain that while gravitational *force* depends on mass, gravitational *acceleration* does not. Emphasize the cancellation of the object's mass in the derivation of 'g'.

 

Question 6. What is the difference between G and g? Or Distinguish between the force of gravity and gravitational force.
Answer: Here's a distinction between Universal Gravitational Constant (G) and Acceleration due to Gravity (g), and between Gravitational Force and Gravity:

**Difference between G and g:**

**Difference between Gravitational Force and Gravity:**
1. **Gravitational Force:** This is the force of attraction that exists between *any two* objects in the universe that have mass. It is a fundamental interaction that keeps planets in orbit and objects on the ground. For example, the force between the Sun and Earth is gravitational force.
2. **Gravity:** This is a specific instance of gravitational force. It refers to the gravitational force exerted by a *large celestial body* (like a planet or moon) on an object near its surface. It's what gives objects their weight. For example, the force pulling an apple towards the Earth is gravity.
In essence, gravity is the gravitational force applied by Earth (or another planet).
In simple words: 'G' is a fixed number for gravity, while 'g' is the changing speed-up due to gravity. Gravitational force is the pull between any two objects, while 'gravity' is the pull of a planet on something near it.

🎯 Exam Tip: Use a comparative table for G and g, and clearly define gravitational force (universal attraction) and gravity (planet's pull on objects near its surface) with examples.

 

Question 7. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer: The force of gravitation between two objects is inversely proportional to the square of the distance between them. This relationship is described by Newton's Law of Universal Gravitation: \( F \propto \frac{1}{d^2} \).
If the original distance between the objects is \( d \), and the force is \( F \).
When the distance is reduced to half, the new distance (\( d' \)) becomes \( \frac{d}{2} \).
The new gravitational force (\( F' \)) will be:
\[ F' \propto \frac{1}{(d')^2} \]
\[ F' \propto \frac{1}{(\frac{d}{2})^2} \]
\[ F' \propto \frac{1}{\frac{d^2}{4}} \]
\[ F' \propto \frac{4}{d^2} \]
Comparing this to the original force, \( F' = 4F \).
Thus, if the distance between two objects is reduced to half, the gravitational force between them increases by 4 times. This shows how quickly gravitational force grows stronger as objects get closer.
In simple words: When the distance between two objects is cut in half, the gravitational force pulling them together becomes four times stronger.

🎯 Exam Tip: Clearly state the inverse square law and show the algebraic steps for how the force changes when the distance is halved. This demonstrates a clear understanding of the relationship.

 

Question 9. What is the mass and weight of an object on the surface of the moon, if on the earth its mass is 12 kg?
Answer: The mass of an object remains the same everywhere, so the mass of the object on the moon will still be 12 kg. However, the force of gravity on the moon is about one-sixth of that on Earth. To find the weight, we use the formula weight = mass × acceleration due to gravity (g).
Weight of the object on Earth \( = 12 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 117.6 \, \text{N} \)
The value of 'g' on the moon is \( \frac{1}{6} \) of its value on Earth.
So, weight of the object on the moon \( = 12 \, \text{kg} \times \left( \frac{1}{6} \times 9.8 \, \text{m/s}^2 \right) = 12 \, \text{kg} \times 1.633 \, \text{m/s}^2 \approx 19.6 \, \text{N} \).
In simple words: An object's mass stays the same, so 12 kg on Earth is 12 kg on the moon. But the moon's gravity is weaker, so the object will weigh much less there, about 19.6 N instead of 117.6 N.

🎯 Exam Tip: Remember that mass is a constant quantity, while weight is a force that changes depending on the gravitational pull of the planet or moon.

 

Question 10. A ball thrown upwards reaches a certain height and then falls back to the earth. Why?
Answer: When a ball is thrown upwards, it moves against Earth's gravity. The force of gravity constantly pulls the ball downwards, causing its upward speed to decrease. Eventually, the ball's speed becomes zero at its highest point. At this peak, gravity continues to act on it, pulling it back towards the Earth, making it fall. This is why all objects thrown upwards eventually return to the ground.
In simple words: When you throw a ball up, Earth's gravity pulls it down, making it slow down until it stops. Then, gravity pulls it back down to the ground.

🎯 Exam Tip: Gravitational force is always attractive and acts towards the center of the Earth, which is why objects always fall back down.

Gravitation Short Answer Type Questions

 

Question 1. How does the force of gravitation between two objects changes, when the distance between them is reduced to half?
Answer: Newton's law of universal gravitation states that the gravitational force (F) between two objects is inversely proportional to the square of the distance (d) between their centers. This means if you change the distance, the force changes by the square of that change.
The formula is \( F \propto \frac{1}{d^2} \).
If the distance 'd' is reduced to half, meaning \( d' = \frac{d}{2} \), then the new force \( F' \) will be:
\( F' \propto \frac{1}{(d')^2} \)
\( F' \propto \frac{1}{(\frac{d}{2})^2} \)
\( F' \propto \frac{1}{\frac{d^2}{4}} \)
\( F' \propto \frac{4}{d^2} \)
\( \implies F' = 4F \)
Thus, when the distance between the two objects is reduced to half, the gravitational force between them increases by 4 times. This is because a smaller distance means a stronger pull.
In simple words: If two objects get twice as close, the pull of gravity between them becomes four times stronger.

🎯 Exam Tip: Always remember the inverse square relationship in gravitation. Halving the distance makes the force four times stronger, and doubling the distance makes the force four times weaker.

 

Question 2. What is the importance of the universal law of gravitation?
Answer: The universal law of gravitation is very important as it explains many natural events. It successfully describes why planets orbit the sun and how the solar system works. This law also helps us understand why objects fall to the ground when dropped, why the moon orbits Earth, and how ocean tides are formed. Additionally, it explains the existence of Earth's atmosphere, which is held by gravity.
In simple words: This law tells us why things fall, why the moon goes around Earth, why planets orbit the sun, and why we have tides in the ocean. It's key to understanding how our universe works.

🎯 Exam Tip: When listing the importance of a law, provide specific examples of phenomena it explains, such as planetary motion, tides, and falling objects.

 

Question 3. Gravitation force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in Newtons of 10 kg object on the moon and on the earth?
Answer: We are given the mass of the object (m) = 10 kg.
We know that the acceleration due to gravity on Earth (\( g_e \)) is approximately \( 9.8 \, \text{m/s}^2 \).
The weight of the object on Earth (\( W_e \)) = \( m \times g_e = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{Newton} \).
The acceleration due to gravity on the moon (\( g_m \)) is \( \frac{1}{6} \) of Earth's gravity.
\( g_m = \frac{g_e}{6} = \frac{9.8}{6} \approx 1.633 \, \text{m/s}^2 \).
The weight of the object on the moon (\( W_m \)) = \( m \times g_m = 10 \, \text{kg} \times 1.633 \, \text{m/s}^2 \approx 16.33 \, \text{Newton} \).
In simple words: A 10 kg object weighs 98 Newtons on Earth. Since the moon's gravity is six times weaker, the same object will weigh about 16.33 Newtons on the moon.

🎯 Exam Tip: Clearly state the given values and formulas. Ensure you calculate both Earth and moon weights, and remember to use the correct units (Newtons for weight).

 

Question 4. What happens to the force between two bodies, if the distance between them is tripled and the masses of the objects is doubled?
Answer: The gravitational force (F) between two bodies is given by Newton's law of universal gravitation: \( F = G \frac{m_1 m_2}{r^2} \).
Here, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two bodies, and \( r \) is the distance between their centers.
Let the original force be \( F_1 \).
If the masses are doubled, the new masses are \( 2m_1 \) and \( 2m_2 \).
If the distance is tripled, the new distance is \( 3r \).
The new gravitational force \( F_2 \) will be:
\( F_2 = G \frac{(2m_1)(2m_2)}{(3r)^2} \)
\( F_2 = G \frac{4m_1 m_2}{9r^2} \)
\( F_2 = \frac{4}{9} \left( G \frac{m_1 m_2}{r^2} \right) \)
\( \implies F_2 = \frac{4}{9} F_1 \)
So, the gravitational force between the two bodies will become \( \frac{4}{9} \) times the original force.
In simple words: When the masses of two objects are doubled and the distance between them is tripled, the new gravitational force will be four-ninths (4/9) of the original force. The increased distance weakens the force more than the increased masses strengthen it.

🎯 Exam Tip: Clearly show the original formula and then substitute the new values for masses and distance. Pay close attention to squaring the new distance correctly in the denominator.

Gravitation Long Answer Type Questions

 

Question 1. (a) State Newton's law of gravitation. Obtain an expression for the gravitational force between two mass bodies.
(b) What is the value of G? Also, write its S.I. unit.
Answer:
(a) Newton's Law of Gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The direction of this force is along the line joining the centers of the two particles. This constant pull ensures everything stays in its place in the cosmos.
Let's consider two objects with masses \( m_1 \) and \( m_2 \).
Let the distance between their centers be \( r \).
According to Newton's Law of Gravitation:
\( F \propto m_1 m_2 \) (Force is directly proportional to the product of masses)
\( F \propto \frac{1}{r^2} \) (Force is inversely proportional to the square of the distance)
Combining these two proportionalities:
\( F \propto \frac{m_1 m_2}{r^2} \)
To convert this proportionality into an equation, we introduce a constant \( G \), known as the Universal Gravitational Constant:
\( F = G \frac{m_1 m_2}{r^2} \)
This equation calculates the gravitational force between any two objects.
(b) The value of the Universal Gravitational Constant (G) is \( 6.67 \times 10^{-11} \).
Its S.I. unit is \( \text{N m}^2/\text{kg}^2 \).
In simple words: (a) Newton said that everything in the universe pulls on everything else. The stronger the objects, and the closer they are, the stronger the pull. The formula \( F = G \frac{m_1 m_2}{r^2} \) helps us measure this pull. (b) The special number G in this formula is \( 6.67 \times 10^{-11} \), and its unit is Newtons times meters squared, divided by kilograms squared.

🎯 Exam Tip: For part (a), ensure you clearly state both proportionalities (direct to mass product, inverse to distance squared) and then combine them to derive the full formula with G. For part (b), remember both the numerical value and the correct S.I. unit of G.

 

Question 2. (a) What do you understand by free fall? Derive an expression for acceleration due to gravity.
(b) Mass of earth is 81 times that of the moon, and its diameter is 3.6 times that of the moon. Compare the acceleration due to gravity on the earth and on the moon.
Answer:
(a) Free fall refers to the motion of an object solely under the influence of gravity, with no other forces acting on it (like air resistance). When an object is in free fall, it experiences a constant acceleration called acceleration due to gravity. This is why a dropped feather and a dropped rock would fall at the same rate in a vacuum.
To derive the expression for acceleration due to gravity (\( g \)):
Consider an object of mass \( m \) on the surface of the Earth.
The gravitational force (\( F \)) acting on the object due to Earth is given by Newton's law of gravitation:
\( F = G \frac{M_E m}{R_E^2} \) ...(1)
where \( M_E \) is the mass of the Earth, and \( R_E \) is the radius of the Earth.
According to Newton's second law of motion, the force acting on an object is also given by:
\( F = m g \) ...(2)
where \( g \) is the acceleration due to gravity.
Equating (1) and (2):
\( m g = G \frac{M_E m}{R_E^2} \)
We can cancel \( m \) from both sides:
\( \implies g = G \frac{M_E}{R_E^2} \)
This expression shows that the acceleration due to gravity depends on the mass and radius of the planet, not on the mass of the falling object.
(b) Let \( M_E \) be the mass of Earth and \( M_m \) be the mass of the moon.
Let \( R_E \) be the radius of Earth and \( R_m \) be the radius of the moon.
Given:
\( M_E = 81 M_m \)
Diameter of Earth \( D_E = 3.6 D_m \), which means \( R_E = 3.6 R_m \) (since radius is half of diameter).
The acceleration due to gravity on Earth (\( g_E \)) is:
\( g_E = G \frac{M_E}{R_E^2} \)
The acceleration due to gravity on the moon (\( g_m \)) is:
\( g_m = G \frac{M_m}{R_m^2} \)
To compare \( g_E \) and \( g_m \), we find their ratio:
\( \frac{g_E}{g_m} = \frac{G \frac{M_E}{R_E^2}}{G \frac{M_m}{R_m^2}} \)
\( \implies \frac{g_E}{g_m} = \frac{M_E}{M_m} \times \frac{R_m^2}{R_E^2} \)
Now substitute the given values:
\( \frac{g_E}{g_m} = 81 \times \left( \frac{R_m}{3.6 R_m} \right)^2 \)
\( \implies \frac{g_E}{g_m} = 81 \times \left( \frac{1}{3.6} \right)^2 \)
\( \implies \frac{g_E}{g_m} = 81 \times \frac{1}{12.96} \)
\( \implies \frac{g_E}{g_m} = 6.25 \)
We can also write \( 6.25 \) as \( \frac{25}{4} \).
Therefore, the ratio of acceleration due to gravity on Earth to that on the moon is \( g_E : g_m = 25 : 4 \). This shows Earth's gravity is significantly stronger than the moon's.
In simple words: (a) Free fall is when something drops only because of gravity. The formula for gravity's pull is \( g = G \frac{M_E}{R_E^2} \). (b) Because Earth is much heavier and only a bit bigger than the moon, Earth's gravity is about 6.25 times stronger than the moon's gravity, or in a ratio of 25:4.

🎯 Exam Tip: For derivations, clearly state the relevant laws (Newton's Gravitation and Second Law of Motion). For comparisons, set up a ratio of the quantities and substitute values carefully, remembering to square the radius term.