GSEB Class 11 Physics Solutions Chapter 8 Gravitation

Question 1. Answer the following:
(a) You can shield a body from electrical forces by putting it inside a hollow conductor. Can you shield a body from gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small space ship orbiting around the Earth cannot detect gravity. If the space station orbiting around the Earth has a large size can he hope to detect gravity?
(c) If you compare the gravitational force on the Earth due to the Sun to that due to the moon, you would find that the Sun's pull is greater than moon's pull. (You can check this yourself using the data available in the succeeding exercises.) However, the tidal effect of the moon's pull is greater than the tidal effect of the Sun, why?
Answer:
(a) No. Electrical forces rely on the kind of substance in between, but gravitational forces do not depend on the intervening material. Therefore, such protective actions are not possible for gravity, meaning gravity screens cannot be created.
(b) Yes, an astronaut can expect to notice gravity if the spacecraft is very large. In such a case, the strength of the gravity would become noticeable, and thus the gravitational impact of the spaceship might be measured.
(c) The distance between the Earth and the Moon is very small compared to the Earth-Sun distance. The tidal effect is inversely proportional to the cube of the distance, which means it is not governed by the inverse square law like gravitational force (which follows the inverse square law). Consequently, the tidal effect from the Moon is larger than that from the Sun.
In simple words: (a) No, you cannot block gravity with a hollow sphere because gravity is not affected by the material in between objects, unlike electrical forces. (b) Yes, if a spaceship is really big, you might feel gravity inside it. (c) The Moon's tidal pull is stronger than the Sun's because it is much closer, and tidal effects depend strongly on distance.

Exam Tip: Remember that gravitational force is a fundamental interaction and cannot be screened, unlike electromagnetic forces. Tidal effects are more sensitive to distance than direct gravitational attraction.

Question 2. Choose the correct alternatives:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.
(d) The formula \( -GMm \left( \frac{1}{r_{2}}-\frac{1}{r_{1}} \right) \) is more/less accurate than the formula \( mg (r_2 – r_1) \) for the difference of potential energy between two points \( r_1 \) and \( r_2 \) distance away from the centre of earth.
Answer:
(a) Acceleration due to gravity decreases with increasing altitude.
(b) Acceleration due to gravity decreases with increasing depth.
(c) Acceleration due to gravity is independent of the mass of the body.
(d) The formula \( -GMm \left( \frac{1}{r_{2}}-\frac{1}{r_{1}} \right) \) is more accurate than the formula \( mg (r_2 – r_1) \) for the difference of potential energy between two points \( r_1 \) and \( r_2 \) distance away from the centre of earth.
In simple words: (a) As you go higher, gravity gets weaker. (b) As you go deeper into Earth, gravity also gets weaker. (c) How strong gravity is does not depend on the object's mass, only the Earth's mass. (d) The first formula for potential energy is more precise than the second one.

Exam Tip: Understand how gravity changes with both height (altitude) and depth from the Earth's surface. Also, distinguish between the general formula for potential energy and approximations.

Question 3. Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Answer:
Let \( \omega \) = angular speed of the Earth around the Sun
Let \( \omega' \) = angular speed of the planet around the Sun
Let \( r' \) = radius of the orbit (orbital size) of the planet around the Sun.
The centripetal force needed for the Earth's revolution around the Sun is supplied by the gravitational force between the Sun and the Earth.
i.e. \( \frac{M_e v^2}{r} = \frac{GM_e M_s}{r^2} \)
or \( v^2 = \frac{GM_s}{r} \)
or \( (r\omega)^2 = \frac{GM_s}{r} \) \( ( \because v = r\omega ) \)
or \( \omega^2 = \frac{GM_s}{r^3} \) .... (1)
Similarly, for the planet
\( \omega'^2 = \frac{GM_s}{r'^3} \) .... (2)
Dividing (1) by (2), we get
\( \frac{\omega^2}{\omega'^2} = \frac{r'^3}{r^3} \)
or \( \left( \frac{\omega}{\omega'} \right)^2 = \left( \frac{r'}{r} \right)^3 \)
We are given that \( \omega' = 2\omega \)
or \( \frac{\omega}{\omega'} = \frac{1}{2} \) .... (4)
From (3) and (4), we obtain
\( \left( \frac{1}{2} \right)^2 = \left( \frac{r'}{r} \right)^3 \)
\( \frac{1}{4} = \left( \frac{r'}{r} \right)^3 \)
\( \frac{r'}{r} = \left( \frac{1}{4} \right)^{1/3} \)
\( \frac{r'}{r} = \frac{1}{1.587} \)
\( \frac{r'}{r} \approx 0.63 \)
i.e. the orbital size of the planet is about 0.63 times the orbital size of Earth. This means the planet's orbital size is smaller than Earth's orbital size by a factor of 0.63.
In simple words: If a planet circles the Sun twice as quickly as Earth, its orbit will be about 0.63 times smaller than Earth's orbit.

Exam Tip: This problem uses Kepler's Third Law (or a derivation of it). Remember that the square of the orbital period is proportional to the cube of the orbital radius (\( T^2 \propto R^3 \)). You can also express this in terms of angular speed, \( \omega \), where \( \omega = 2\pi/T \), leading to \( \omega^2 \propto 1/R^3 \).

Question 4. If one of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is \( 4.22 \times 10^8 \) m. Show the mass of Jupiter is about one thousandth that of Sun?
Answer:
Let \( M_J \) = Mass of Jupiter = ?
\( M_S \) = Mass of Sun = \( 2 \times 10^{30} \) kg
\( T \) = Time period of Io, a satellite of Jupiter = 1.769 days
\( = 1.769 \times 24 \times 3600 \) s
\( = 15.2841 \times 10^4 \) s
\( r \) = radius of its orbit around Jupiter = \( 4.22 \times 10^8 \) m.
\( G = 6.67 \times 10^{-11} \) Nm\(^2\)kg\(^{-2}\)
To prove \( M_J = \frac{1}{1000} M_S \)
Using the relation, \( \frac{GM}{r^2} = \omega^2 \), we get
\( \frac{GM_J}{r^2} = \left( \frac{2\pi}{T} \right)^2 \)
\( M_J = \frac{4\pi^2 r^3}{T^2G} \)
\( M_J = \frac{4 \times (3.14)^2 \times (4.22 \times 10^8)^3}{(15.2841 \times 10^4)^2 \times 6.67 \times 10^{-11}} \)
\( M_J = \frac{4 \times 9.87 \times (75.29 \times 10^{24})}{(233.5 \times 10^8) \times 6.67 \times 10^{-11}} \)
\( M_J \approx 1.9 \times 10^{27} \) kg
So, \( M_J \approx 2 \times 10^{27} \) kg
Now compare \( M_J \) with \( M_S \):
\( \frac{M_J}{M_S} = \frac{2 \times 10^{27}}{2 \times 10^{30}} = \frac{1}{1000} \)
Thus, the mass of Jupiter is approximately one thousandth of the Sun's mass. This proves the statement.
In simple words: We calculate Jupiter's mass using the orbital period and radius of one of its moons. After performing the calculations, we find that Jupiter's mass is about one-thousandth of the Sun's mass.

Exam Tip: This problem applies Kepler's third law and Newton's law of gravitation to determine the mass of a central body using the orbital parameters of its satellite. Make sure to convert units consistently (days to seconds, km to m).

Question 5. Let us assume that our galaxy consists of \( 2.5 \times 10^{11} \) stars each of one solar mass. How long will a star at a distance of 50,000 light years from the galactic centre take to complete one revolution? Take the diameter of the milky way to be \( 10^5 \) light years?
Answer:
Let \( M \) = Mass of stars in the galaxy
\( = 2.5 \times 10^{11} \times (2 \times 10^{30}) \) kg (since one solar mass = \( 2 \times 10^{30} \) kg)
\( = 5 \times 10^{41} \) kg
\( r \) = radius of orbit of a star.
\( = \) Distance of a star from galactic centre
\( = 50,000 \) light years.
\( = 50,000 \times (9.46 \times 10^{15}) \) m (since 1 light year = \( 9.46 \times 10^{15} \) m)
\( G = 6.67 \times 10^{-11} \) Nm\(^{-2}\)kg\(^{-2}\)
\( T \) = Time of one revolution = ?
Diameter of Milky Way = \( 10^5 \) light years (this information is extra and not used for this specific calculation)
Using the relation, \( \frac{GM}{r^2} = \omega^2 = \left( \frac{2\pi}{T} \right)^2 \), we get
\( T^2 = \frac{4\pi^2 r^3}{GM} \)
\( T^2 = \frac{4 \times (3.14)^2 \times (50,000 \times 9.46 \times 10^{15})^3}{6.67 \times 10^{-11} \times 5 \times 10^{41}} \)
\( T^2 = \frac{4 \times 9.87 \times (4.73 \times 10^{20})^3}{33.35 \times 10^{30}} \)
\( T^2 = \frac{4 \times 9.87 \times 105.7 \times 10^{60}}{33.35 \times 10^{30}} \)
\( T^2 \approx 12527.5 \times 10^{28} \) s\(^2\)
\( T = \sqrt{12527.5 \times 10^{28}} \) s
\( T = 111.93 \times 10^{14} \) s
To convert seconds to years, divide by (365 days/year * 24 hours/day * 3600 seconds/hour):
\( T = \frac{111.93 \times 10^{14}}{365 \times 24 \times 3600} \) years
\( T = \frac{111.93 \times 10^{14}}{31536000} \) years
\( T \approx 3.55 \times 10^8 \) yrs.
In simple words: We calculated the total mass of the galaxy and the star's distance from the center. Then, using a specific physics formula for orbits, we found that it would take approximately 355 million years for this star to complete one full trip around the galactic center.

Exam Tip: For large-scale astronomical problems, remember to convert all units (especially light years to meters and solar masses to kg) before performing calculations. Kepler's third law in its generalized form is key here.

Question 6. Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of the kinetic/ potential energy.
(b) The energy required to rocket an orbiting satellite out of Earth's gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth's influence.
Answer:
(a) Kinetic energy
(b) less.
In simple words: (a) When potential energy is zero at infinity, an orbiting satellite's total energy is the negative of its kinetic energy. (b) It takes less energy to move an orbiting satellite out of Earth's gravity compared to moving a still object at the same height.

Exam Tip: Remember that for an orbiting satellite, total energy \( E = -K.E. \), and the escape energy is \( -E \). Understand the difference between escaping from orbit versus escaping from a stationary position.

Question 7. Does the escape velocity of a body from the Earth depend on (a) the mass of the body, (b) the location from which it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched? Explain your answer?
Answer:
(a) No. We understand that the escape velocity of a body is given by \( V_e = \frac{\sqrt{2GM}}{R} \), where \( M \) and \( R \) are the mass and radius of Earth. Clearly, it does not depend on the mass of the body since \( V_e \) is independent of it.
(b) Yes. We know that \( V_e \) depends upon the gravitational potential at the point from where the body is launched. Since the gravitational potential depends on the latitude and height of the point, the escape velocity depends on the location from which it is projected. This can also be seen from \( V_e = \sqrt{2gr} \). As \( g \) has different values at different heights, \( V_e \) depends upon the height of the location.
(c) No, it does not depend on the direction of projection because \( V_e \) is independent of the direction of projection.
(d) Yes, it depends on the height of location from where the body is launched, as explained in part (b).
In simple words: (a) Escape velocity doesn't depend on how heavy the object is. (b) It does depend on where on Earth you launch it from, because gravity changes with location and height. (c) The direction you launch it in doesn't matter for escape velocity. (d) Yes, it depends on the starting height, as explained earlier.

Exam Tip: The formula for escape velocity \( V_e = \sqrt{2GM/R} \) shows its fundamental dependencies. However, remember that 'g' can vary with height and latitude, indirectly affecting \( V_e \).

Question 8. A comet orbits the sun in highly elliptical orbit. Does the comet has a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy and (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to sun?
Answer:
All quantities change over an orbit, except for angular momentum and the total energy, which can be explained as follows:
(a) According to the law of conservation of angular momentum, the comet moves faster when it is near the Sun and moves slower when it is farther away. Therefore, the linear speed of the comet does not stay constant.
(b) The angular speed of the comet also changes slightly.
(c) According to the law of conservation of angular momentum, the angular momentum of the comet remains constant.
(d) Since the linear speed of the comet around the Sun changes continuously, its kinetic energy also changes continuously.
(e) Potential energy depends on the distance between the Sun and the comet. Thus, potential energy changes in the elliptical orbit as the distance between the Sun and the comet continuously varies.
(f) As total energy is the sum of kinetic energy and potential energy, the total energy always stays constant, according to the law of conservation of energy.
In simple words: Most things about a comet's motion change during its elliptical orbit, except its total angular momentum and its total energy, which both stay the same because of conservation laws. Speed, kinetic energy, and potential energy all vary as its distance from the Sun changes.

Exam Tip: For objects in orbit under a central force, the total mechanical energy and angular momentum are conserved. Other quantities like linear speed, kinetic energy, and potential energy will vary as the object moves closer to or further from the central body.

Question 9. Which the following symptoms is likely to afflict an astronaut in space, (a) swollen feet, (b) swollen face, (c) headache and (d) orientational problem?
Answer:
(b), (c), (d) i.e. swollen face, headache, and orientational problems are the symptoms that an astronaut is likely to experience in space.
In simple words: Astronauts in space often get a puffy face, headaches, and trouble telling up from down. Swollen feet are less common because blood tends to shift upwards in microgravity.

Exam Tip: In microgravity, fluids in the body shift upwards towards the head, causing symptoms like a swollen face and headaches. Disorientation is also common due to the lack of a consistent 'down' direction.

Question 10. The gravitational intensity at the centre of a spherical shell of uniform mass density has the direction indicated by the arrow (see in the fig.) (i) a, (ii) b, (iii) c, (iv) zero?
Answer:
The gravitational potential inside a uniform spherical shell is constant. Because the potential is constant throughout the interior of the shell, the gravitational intensity (which is the negative gradient of the potential) at any point inside the shell, including its center, must be zero. If the intensity is zero, it has no direction.
Therefore, the correct option is (iv) zero.
In simple words: Inside a perfectly round, empty shell of uniform mass, gravity at the very center is zero. This means there is no force pulling in any direction, so the answer is zero.

Exam Tip: Remember Gauss's law for gravitation: the gravitational field inside a uniform spherical shell is always zero. This is a crucial concept for understanding gravity in such systems.

Question 11. For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g?
Answer:
As stated in the previous problem, for any point P *inside* a uniform spherical shell, the gravitational potential is constant. Because of this, the gravitational intensity (or force) at point P is zero. If the intensity is zero, it cannot have a direction. Therefore, none of the directional options (d, e, f, g) can correctly indicate the gravitational intensity at point P inside the shell.
In simple words: Inside a uniform hollow sphere, gravity is zero everywhere. So, at any point P inside, there's no force and no direction of gravity to point to.

Exam Tip: It is a fundamental property of uniform spherical shells that the gravitational field (intensity) is zero everywhere inside the shell, not just at its center. This means no net force acts on an object placed anywhere within the shell.

Question 12. A rocket is fired from the Earth towards the Sun. At what distance from the Earth's centre is the gravitational force on the rocket zero? Mass of the Sun = \( 2 \times 10^{30} \) kg., Mass of Earth = \( 6.0 \times 10^{24} \) kg. Neglect the effect of the other planets etc. (Orbital radius = \( 1.5 \times 10^{11} \) m)
Answer:
Let P be a point at a distance \( r \) from Earth's center where the gravitational force due to the Sun and Earth is equal, meaning the net gravitational force on the rocket is zero.
Let \( x \) = distance between Sun and Earth = orbital radius of Earth = \( 1.5 \times 10^{11} \) m
Mass of Sun, \( M_S = 2 \times 10^{30} \) kg
Mass of Earth, \( M_e = 6 \times 10^{24} \) kg
Also let \( m \) = mass of the rocket
At point P, the gravitational force between the Sun and the rocket equals the gravitational force between Earth and the rocket:
\( \frac{GM_S m}{(x-r)^2} = \frac{GM_e m}{r^2} \)
\( \frac{M_S}{(x-r)^2} = \frac{M_e}{r^2} \)
\( \frac{(x-r)^2}{r^2} = \frac{M_S}{M_e} \)
\( \left( \frac{x-r}{r} \right)^2 = \frac{2 \times 10^{30}}{6 \times 10^{24}} = \frac{1}{3} \times 10^6 \)
\( \frac{x-r}{r} = \sqrt{\frac{1}{3} \times 10^6} = \frac{10^3}{\sqrt{3}} \)
\( \frac{x}{r} - 1 = \frac{1000}{1.732} = 577.35 \)
\( \frac{x}{r} = 1 + 577.35 = 578.35 \)
\( r = \frac{x}{578.35} = \frac{1.5 \times 10^{11}}{578.35} \)
\( r \approx 2.594 \times 10^8 \) m
\( r \approx 2.6 \times 10^8 \) m from Earth.

In simple words: We need to find a point where the Sun's pull on the rocket is equal to Earth's pull. By setting the gravitational forces equal and solving the equation, we found that this point is approximately 2.6 × 10^8 meters away from the Earth's center.

Exam Tip: This is a classic "Lagrangian Point" type problem. The key is to equate the gravitational forces from both large bodies (Earth and Sun) on the smaller body (rocket) and then solve for the distance. Be careful with large numbers and exponents.

Question 13. How will you weigh the Sun i.e. estimate its mass? You will need to know the period of one of its planets and the radius of the planetary orbit. The mean orbital radius of the Earth around the Sun is \( 1.5 \times 10^8 \) km. Estimate the mass of the Sun?
Answer:
We know that Earth revolves around the Sun in an orbit with a radius of \( 1.5 \times 10^{11} \) m and completes one revolution around the Sun in 365 days.
So, \( R \) = radius of orbit of Earth = \( 1.5 \times 10^{11} \) m
\( T \) = Time period of Earth around the Sun = 365 days
\( = 365 \times 24 \times 60 \times 60 \) s.
\( G = 6.67 \times 10^{-11} \) Nm\(^2\)kg\(^{-2}\)
We know from Kepler's third law and Newton's law of gravitation that for an orbiting planet,
\( \frac{R^3}{T^2} = \frac{GM_S}{4\pi^2} \)
where \( M_S \) = mass of Sun = ?
Rearranging the equation to find \( M_S \):
\( M_S = \frac{4\pi^2 R^3}{T^2 G} \)
Now, substitute the values:
\( M_S = \frac{4 \times (3.14)^2 \times (1.5 \times 10^{11})^3}{(365 \times 24 \times 60 \times 60)^2 \times 6.67 \times 10^{-11}} \)
\( M_S = \frac{4 \times 9.87 \times (3.375 \times 10^{33})}{(3.1536 \times 10^7)^2 \times 6.67 \times 10^{-11}} \)
\( M_S = \frac{133.05 \times 10^{33}}{9.945 \times 10^{14} \times 6.67 \times 10^{-11}} \)
\( M_S = \frac{133.05 \times 10^{33}}{66.33 \times 10^3} \)
\( M_S \approx 2.0 \times 10^{30} \) kg
Thus, the mass of the Sun is approximately \( 2.0 \times 10^{30} \) kg.
In simple words: To find the Sun's mass, we use Earth's orbital period and distance from the Sun along with the gravitational constant. Applying a specific physics formula, we calculate the Sun's mass to be about 2.0 × 10^30 kilograms.

Exam Tip: This problem is a direct application of Kepler's Third Law (or its Newtonian derivation). Make sure to correctly convert the time period from days to seconds and the orbital radius from km to m for consistency in units.

Question 14. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is \( 1.50 \times 10^8 \) km away from the Sun?
Answer:
According to Kepler's third law, we know that the square of the time period of a planet's revolution around the Sun is proportional to the cube of its semi-major axis (orbital radius):
\( T^2 \propto R^3 \)
i.e. \( \frac{T^2}{R^3} = \text{constant} \)
For Saturn, \( T_S^2 \propto R_S^3 \) ....(1)
And for Earth, \( T_e^2 \propto R_e^3 \) ....(2)
Dividing (1) by (2), we get:
\( \frac{T_S^2}{T_e^2} = \frac{R_S^3}{R_e^3} \)
or \( \left( \frac{T_S}{T_e} \right)^2 = \left( \frac{R_S}{R_e} \right)^3 \)
We are given that \( T_S = 29.5 \, T_e \), so \( \frac{T_S}{T_e} = 29.5 \)
Substitute this value into the equation:
\( (29.5)^2 = \left( \frac{R_S}{R_e} \right)^3 \)
\( (29.5)^2 = 870.25 \)
\( 870.25 = \left( \frac{R_S}{R_e} \right)^3 \)
\( \frac{R_S}{R_e} = (870.25)^{1/3} \)
\( \frac{R_S}{R_e} \approx 9.547 \)
So, \( R_S = 9.547 \times R_e \)
Given that the Earth's mean orbital radius is \( R_e = 1.50 \times 10^8 \) km.
\( R_S = 9.547 \times (1.50 \times 10^8) \) km
\( R_S = 14.32 \times 10^8 \) km
\( R_S \approx 1.43 \times 10^9 \) km.
In meters: \( R_S = 1.43 \times 10^{12} \) m.
In simple words: Saturn's year is much longer than Earth's. Using Kepler's law that connects orbital period and distance, we can figure out Saturn's distance from the Sun. Based on Earth's distance and the ratio of their years, Saturn is about 1.43 × 10^9 kilometers away from the Sun.

Exam Tip: This problem is a direct application of Kepler's Third Law. Ensure you correctly set up the ratio of periods squared to radii cubed. Units (km vs m) should be handled carefully for the final answer.

Question 15. A body weighs 63 N on the surface of Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of Earth?
Answer:
Here, \( h \) = height above Earth's surface = \( \frac{R}{2} \)
where \( R \) = radius of Earth.
We know that the acceleration due to gravity at a height 'h' above Earth's surface (\( g_h \)) is given by:
\( g_h = g \left( \frac{R}{R+h} \right)^2 \)
Substitute \( h = \frac{R}{2} \):
\( g_h = g \left( \frac{R}{R+\frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 \)
\( g_h = g \left( \frac{2}{3} \right)^2 = g \times \frac{4}{9} \)
So, \( g_h = \frac{4}{9}g \)
Let \( m \) = mass of the body.
If \( w \) and \( w_h \) are its weight on Earth's surface and at height \( h \) above Earth's surface, respectively.
Then, \( w = mg = 63 \) N (given)
And \( w_h = mg_h \)
Substitute \( g_h = \frac{4}{9}g \):
\( w_h = m \left( \frac{4}{9}g \right) = \frac{4}{9} (mg) \)
\( w_h = \frac{4}{9} \times 63 \)
\( w_h = 4 \times 7 \)
\( w_h = 28 \) N.
In simple words: The object weighs 63 N on Earth. At a height equal to half of Earth's radius, gravity becomes weaker. Using the formula for gravity at height, we find its new weight is 28 N.

Exam Tip: Remember the formula for acceleration due to gravity at a height \( h \) above the Earth's surface. It's inversely proportional to the square of the distance from the center of the Earth. Make sure to use \( R+h \) as the total distance.

Question 16. Assuming the Earth to be a sphere of uniform mass density, how much would body weigh halfway down to the centre of Earth if it weighed 250 N of the surface?
Answer:
Let \( g \) be the acceleration due to gravity on Earth's surface, and \( g_d \) be the acceleration due to gravity at a depth 'd' from Earth's surface.
Also, let \( W \) and \( W_d \) be the weight of a body on Earth's surface and at depth 'd', respectively.
We are given:
\( W = mg = 250 \) N ....(i)
And \( W_d = mg_d \) ....(ii)
Now we know that \( g_d = g \left(1 - \frac{d}{R}\right) \) ....(iii)
Here, \( d = \frac{R}{2} \), where \( R \) = radius of Earth ....(iv)
From (iii) and (iv), we get:
\( g_d = g \left(1 - \frac{R/2}{R}\right) = g \left(1 - \frac{1}{2}\right) \)
\( g_d = g \times \frac{1}{2} = \frac{g}{2} \) ....(v)
Now substitute \( g_d \) into the equation for \( W_d \):
\( W_d = m g_d = m \times \frac{g}{2} \) [using (v)]
\( W_d = \frac{1}{2} (mg) \)
Since \( W = mg \), we can write:
\( W_d = \frac{1}{2} W = \frac{1}{2} \times 250 = 125 \) N.
Thus, the weight of the body halfway down to the center of Earth is 125 N.
In simple words: An object weighing 250 N on Earth's surface would weigh less if moved halfway to the Earth's center. Because gravity decreases as you go deeper inside the Earth, its weight would be exactly half, or 125 N.

Exam Tip: Remember the specific formula for acceleration due to gravity at a depth \( d \) inside the Earth (assuming uniform density). It's a linear decrease, unlike the inverse square relationship for height above the surface.

Question 17. A rocket is fired vertically with a speed of \( 5 \) km s\(^{-1} \) from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth? Mass of the Earth = \( 6.0 \times 10^{24} \) kg, mean radius of Earth = \( 6.4 \times 10^6 \) m, G = \( 6.67 \times 10^{-11} \) Nm\(^2\)kg\(^{-2} \).
Answer:
Let \( v \) be the initial speed of the rocket, and \( h \) be the maximum height above the Earth's surface where its velocity becomes zero before it returns.
If \( m \) is the mass of the rocket, then its total energy at Earth's surface is:
K.E. + P.E. \( = \frac{1}{2}mv^2 - \frac{GMm}{R} \) ....(i)
where \( M \) = mass of Earth, \( R \) = radius of Earth, \( G \) = Universal gravitational constant.
At the highest point, K.E. = 0 (since velocity = 0).
And P.E. \( = - \frac{GMm}{R+h} \) ....(ii)
According to the law of conservation of energy, Initial Total Energy = Final Total Energy:
\( \frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h} \)
\( \frac{1}{2}v^2 - \frac{GM}{R} = - \frac{GM}{R+h} \)
Multiply by \( 2/G \):
\( \frac{v^2}{G} - \frac{2M}{R} = - \frac{2M}{R+h} \)
From \( g = \frac{GM}{R^2} \), we have \( GM = gR^2 \). Substitute this into the equation:
\( \frac{1}{2}v^2 - \frac{gR^2}{R} = - \frac{gR^2}{R+h} \)
\( \frac{1}{2}v^2 - gR = - \frac{gR^2}{R+h} \)
Rearranging the terms to solve for \( h \):
\( gR - \frac{1}{2}v^2 = \frac{gR^2}{R+h} \)
\( R+h = \frac{gR^2}{gR - \frac{1}{2}v^2} \)
\( h = \frac{gR^2}{gR - \frac{1}{2}v^2} - R \)
\( h = \frac{2gR^2}{2gR - v^2} - R \)
\( h = \frac{2gR^2 - R(2gR - v^2)}{2gR - v^2} \)
\( h = \frac{2gR^2 - 2gR^2 + Rv^2}{2gR - v^2} \)
\( h = \frac{Rv^2}{2gR - v^2} \)
Here, \( v = 5 \) km s\(^{-1} = 5000 \) m s\(^{-1} \)
\( R = 6.4 \times 10^6 \) m.
Let's use \( g = 9.8 \) m s\(^{-2} \).
\( h = \frac{(6.4 \times 10^6) \times (5000)^2}{2 \times 9.8 \times (6.4 \times 10^6) - (5000)^2} \)
\( h = \frac{(6.4 \times 10^6) \times (25 \times 10^6)}{125.44 \times 10^6 - 25 \times 10^6} \)
\( h = \frac{160 \times 10^{12}}{100.44 \times 10^6} \)
\( h \approx 1.593 \times 10^6 \) m
\( h \approx 1.6 \times 10^6 \) m = 1600 km.
The distance from the center of Earth is \( R+h = 6.4 \times 10^6 + 1.6 \times 10^6 = 8.0 \times 10^6 \) m.
In simple words: We used the idea of energy conservation to find how high the rocket goes. Starting with its initial speed and the Earth's gravity, we calculated that the rocket will reach a maximum height of about 1600 kilometers above the Earth's surface before falling back. From the Earth's center, this is 8.0 × 10^6 meters.

Exam Tip: Conservation of mechanical energy is the key principle for this type of problem. At the maximum height, the kinetic energy becomes zero. Ensure all units are consistent (meters, seconds, kilograms) and perform calculations carefully with large exponents.

Question 18. The escape speed of a projectile on Earth's surface is \( 11.2 \) km s\(^{-1} \). A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of Sun and other planets?
Answer:
Note: In this type of problem, using the law of conservation of energy is the most suitable method.
Let \( v_e \) be the escape speed of the body from Earth's surface, and \( v \) be the initial speed of projection.
Given \( v_e = 11.2 \) km s\(^{-1} \).
The body is projected with thrice this speed, so \( v = 3v_e \).
Let \( v_f \) be the final speed of the body far away from Earth (at infinity).
According to the principle of conservation of energy:
Initial K.E. + Initial P.E. = Final K.E. + Final P.E.
\( \frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2 + 0 \)
We know that the escape velocity is defined by \( \frac{1}{2}mv_e^2 = \frac{GMm}{R} \).
Substitute \( \frac{GMm}{R} \) with \( \frac{1}{2}mv_e^2 \):
\( \frac{1}{2}m(3v_e)^2 - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_f^2 \)
\( \frac{1}{2}m(9v_e^2) - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_f^2 \)
Divide by \( \frac{1}{2}m \):
\( 9v_e^2 - v_e^2 = v_f^2 \)
\( 8v_e^2 = v_f^2 \)
Take the square root of both sides:
\( v_f = \sqrt{8}v_e \)
\( v_f = 2\sqrt{2}v_e \)
Given \( v_e = 11.2 \) km s\(^{-1} \)
\( v_f = 2 \times 1.414 \times 11.2 \)
\( v_f = 31.6736 \) km s\(^{-1} \)
So, the speed of the body far away from the Earth is approximately \( 31.7 \) km s\(^{-1} \).

In simple words: If an object is launched from Earth at three times the escape speed, it will still have a lot of speed when it gets very far away. We use energy conservation to calculate that its speed far from Earth will be about 31.7 kilometers per second.

Exam Tip: Remember the relationship between launch velocity, escape velocity, and the velocity at infinity. For velocities greater than escape velocity, the body retains kinetic energy even at an infinite distance, which can be calculated using energy conservation.

Question 19. A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the Earth's gravitational influence? Mass of the satellite = 200 kg, mass of the Earth = \( 6.0 \times 10^{24} \) kg, radius of the Earth = \( 6.4 \times 10^6 \) m, G = \( 6.67 \times 10^{-11} \) Nm\(^2\)kg\(^{-2} \)?
Answer:
Let \( M \) and \( R \) be the mass and radius of Earth.
Let \( m \) be the mass of the satellite orbiting around Earth at a height \( h \) above Earth's surface.
The kinetic energy (K.E.) of the satellite in orbit is \( \frac{1}{2}mv^2 \).
The potential energy (P.E.) of the satellite at height \( h \) is \( - \frac{GMm}{R+h} \).
For a satellite in a circular orbit, the centripetal force is provided by gravity: \( \frac{mv^2}{R+h} = \frac{GMm}{(R+h)^2} \).
From this, \( v^2 = \frac{GM}{R+h} \).
So, K.E. \( = \frac{1}{2}m \left( \frac{GM}{R+h} \right) = \frac{GMm}{2(R+h)} \).
Thus, the total energy of the orbiting satellite is given by K.E. + P.E.:
\( E = \frac{GMm}{2(R+h)} - \frac{GMm}{R+h} \)
\( E = - \frac{GMm}{2(R+h)} \)
When the satellite is just removed from Earth's gravitational pull (at infinity), its gravitational P.E. will be zero, and its K.E. will also be zero. Therefore, the final energy of the satellite at infinity is 0.
The energy needed to put the satellite out of Earth's gravitational influence is the negative of its total orbital energy.
Energy to be expended \( = -E = - \left( - \frac{GMm}{2(R+h)} \right) = \frac{GMm}{2(R+h)} \)
Here, given values are:
\( h = 400 \) km \( = 400 \times 10^3 \) m
\( R = 6.4 \times 10^6 \) m
\( m = 200 \) kg
\( M = 6 \times 10^{24} \) kg
\( G = 6.67 \times 10^{-11} \) Nm\(^2\)kg\(^{-2} \)
Calculate \( R+h = 6.4 \times 10^6 + 400 \times 10^3 = 6.4 \times 10^6 + 0.4 \times 10^6 = 6.8 \times 10^6 \) m.
Now, substitute these values into the energy expression:
Energy to be expended \( = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times 200}{2 \times (6.8 \times 10^6)} \)
\( = \frac{6.67 \times 6 \times 200 \times 10^{13}}{13.6 \times 10^6} \)
\( = \frac{8004 \times 10^{13}}{13.6 \times 10^6} \)
\( = \frac{8004}{13.6} \times 10^7 \)
\( \approx 588.5 \times 10^7 \) J
\( \approx 5.885 \times 10^9 \) J.
The energy that must be expended is approximately \( 5.9 \times 10^9 \) J.
In simple words: To free a satellite from Earth's gravity, you need to add enough energy to overcome its current orbital energy. We calculate the satellite's total energy in orbit and find that about 5.9 × 10^9 Joules of energy must be supplied to make it escape Earth's pull.

Exam Tip: For an orbiting satellite, the total mechanical energy is negative and equal to half its potential energy, or the negative of its kinetic energy. The energy needed for escape from orbit is simply the positive value of its total mechanical energy.

Question 20. Two stars each of 1 solar mass (= \( 2 \times 10^{30} \) kg) are approaching each other for a head-on collision. When they are at a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G)
Answer: Let M be the mass of each star, which is \( 2 \times 10^{30} \) kg. The initial separation distance between the two stars is \( r = 10^9 \, \text{km} = 10^{12} \, \text{m} \). The radius of each star is \( R = 10^4 \, \text{km} = 10^7 \, \text{m} \). When the stars collide, they just touch each other, meaning the final distance between their centers is \( 2R = 2 \times 10^7 \, \text{m} \). The initial speeds are considered negligible, so the initial kinetic energy (K.E.) is zero.
The potential energy (P.E.) of one star with respect to the other, when they are separated by a distance \( r \), is calculated as:
\( \text{P.E.}_{initial} = -\frac{GMM}{r} \)
At the point of collision, when the stars are touching, the distance between their centers becomes \( 2R \), and their final potential energy is:
\( \text{P.E.}_{final} = -\frac{GMM}{2R} \)
If \( v \) is the speed of each star just before the collision, the total final kinetic energy of the two stars is \( Mv^2 \).

Following the law of energy conservation:
\( \text{Total energy}_{initial} = \text{Total energy}_{final} \)
\( \text{K.E.}_{initial} + \text{P.E.}_{initial} = \text{K.E.}_{final} + \text{P.E.}_{final} \)
\( 0 + (-\frac{GMM}{r}) = Mv^2 + (-\frac{GMM}{2R}) \)
Rearranging the equation to solve for \( v^2 \):
\( Mv^2 = \frac{GMM}{2R} - \frac{GMM}{r} \)
\( v^2 = GM (\frac{1}{2R} - \frac{1}{r}) \)
Now, substitute the given values:
\( G = 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \)
\( M = 2 \times 10^{30} \, \text{kg} \)
\( R = 10^7 \, \text{m} \)
\( r = 10^{12} \, \text{m} \)
\( v^2 = (6.67 \times 10^{-11}) \times (2 \times 10^{30}) \left( \frac{1}{2 \times 10^7} - \frac{1}{10^{12}} \right) \)
\( v^2 = 13.34 \times 10^{19} \left( 0.5 \times 10^{-7} - 10^{-12} \right) \)
\( v^2 = 13.34 \times 10^{19} \left( 5 \times 10^{-8} - 10^{-12} \right) \)
Since \( 10^{-12} \) is very small compared to \( 5 \times 10^{-8} \), we can approximate:
\( v^2 \approx 13.34 \times 10^{19} \times 5 \times 10^{-8} \)
\( v^2 \approx 66.7 \times 10^{11} \)
\( v^2 \approx 6.67 \times 10^{12} \)
To find \( v \), take the square root:
\( v = \sqrt{6.67 \times 10^{12}} \)
\( v \approx 2.583 \times 10^6 \, \text{ms}^{-1} \)
The speed with which the stars collide is approximately \( 2.6 \times 10^6 \, \text{ms}^{-1} \).
In simple words: We used the law of energy conservation, where the initial potential energy transforms into kinetic energy just before the stars hit. By plugging in the given masses, distances, and gravitational constant, we could calculate their speed at impact.

Exam Tip: Remember to convert all units to SI (meters, kilograms, seconds) before starting calculations. When one term in an addition/subtraction is much smaller than another (like \( 10^{-12} \) vs \( 5 \times 10^{-8} \)), you can often neglect the smaller term to simplify calculations.

Question 21. Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid-point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Answer: Let the two spheres be positioned at points A and B respectively. The distance between their centers, AB, is 1.0 m. The mass of each sphere, M, is 100 kg, and their radius is 0.10 m.
Consider point O as the mid-point of the line segment AB.
Thus, the distance from each sphere to the midpoint O is AO = OB = \( \frac{1}{2} \times 1 \, \text{m} = 0.5 \, \text{m} \).
**Gravitational Force at the Mid-point:**
We understand that the gravitational force between two objects with masses \( M \) and \( m \), separated by a distance \( d \), is calculated by the formula \( F = \frac{GMm}{d^2} \).
If \( F_A \) and \( F_B \) denote the gravitational forces acting at point O from spheres A and B respectively, on a small test mass \( m \) placed at O, then:
Force from sphere A: \( F_A = \frac{G \times M \times m}{(AO)^2} = \frac{G \times 100 \times m}{(0.5)^2} \) along OA (towards A)
Force from sphere B: \( F_B = \frac{G \times M \times m}{(OB)^2} = \frac{G \times 100 \times m}{(0.5)^2} \) along OB (towards B)
Since the magnitudes of \( F_A \) and \( F_B \) are equal, and they pull in opposite directions (one towards A and the other towards B), the total force at point O becomes zero.
\( |F_A| = |F_B| \), and they act in opposite directions, so the net force at O = 0.
**Equilibrium at the Mid-point:**
A zero net force at point O suggests that an object placed at O feels no net force, so it is in equilibrium. However, this equilibrium is unstable because a small shift away from the center (towards A or B) will prevent the object from returning to its original position. For example, if the object moves slightly towards A, the force from A will become stronger (as it's closer), and the force from B will become weaker (as it's farther), creating a net force that pulls the object further towards A, away from the midpoint.
**Gravitational Potential at the Mid-point:**
We also know that gravitational potential is described by the formula \( V = -\frac{GM}{d} \).
If \( V_A \) and \( V_B \) represent the gravitational potentials at point O (the midpoint between the spheres' centers) due to sphere A and sphere B respectively, then:
Potential due to sphere A: \( V_A = -\frac{G \times M}{AO} = -\frac{G \times 100}{0.5} \)
Potential due to sphere B: \( V_B = -\frac{G \times M}{OB} = -\frac{G \times 100}{0.5} \)
The total gravitational potential \( V \) at point O is the sum of the potentials from both spheres:
\( V = V_A + V_B \)
\( V = -\frac{G \times 100}{0.5} + (-\frac{G \times 100}{0.5}) \)
\( V = -\frac{2 \times G \times 100}{0.5} \)
Substitute the value of G (\( 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \)):
\( V = -\frac{2 \times 6.67 \times 10^{-11} \times 100}{0.5} \)
\( V = -\frac{1334 \times 10^{-11}}{0.5} \)
\( V = -2668 \times 10^{-11} \, \text{J kg}^{-1} \)
\( V \approx -2.7 \times 10^{-8} \, \text{J kg}^{-1} \)
Therefore, an object positioned at the midpoint experiences an unstable equilibrium, and the gravitational potential at this point is approximately \( -2.7 \times 10^{-8} \, \text{J kg}^{-1} \).
In simple words: At the exact middle point between two identical heavy spheres, the pulls from both sides cancel out, so there's no net force on an object. This means it's balanced, but it's an "unstable" balance because if it moves even a tiny bit, it won't go back to the center. The total gravitational energy at that point is calculated by adding up the potential from each sphere, and it comes out to be a specific negative value.

Exam Tip: For equilibrium questions, always calculate both net force (zero for equilibrium) and check stability (a slight displacement should lead to a restoring force for stable equilibrium, or a further displacing force for unstable equilibrium).