Chapter 08: Gravitation
CPhysics XI : Complete NCERT Exercise Solutions
Practice these examination-oriented questions to master the concepts of gravitation, Kepler's laws, orbital mechanics, and gravitational fields. Each question includes a detailed solution to enhance your understanding.
Question 8.1
Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the 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 the moon's pull. However, the tidal effect of the moon's pull is greater than the tidal effect of sun. Why?
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the 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 the moon's pull. However, the tidal effect of the moon's pull is greater than the tidal effect of sun. Why?
Answer & Explanation:
(a) No, gravitational shielding is not possible. Unlike electric forces where charges can be shielded by conductors, gravitational forces act on all masses and cannot be blocked by any material.
(b) Yes, in a large space station, tidal forces due to gravity could be detected because different parts of the station experience slightly different gravitational accelerations.
(c) Tidal effect depends on the difference in gravitational pull across Earth’s diameter, which varies as \( \propto M/d^3 \). Although Sun’s pull is stronger, Moon’s closeness makes its tidal effect about 2.2 times stronger.
(b) Yes, in a large space station, tidal forces due to gravity could be detected because different parts of the station experience slightly different gravitational accelerations.
(c) Tidal effect depends on the difference in gravitational pull across Earth’s diameter, which varies as \( \propto M/d^3 \). Although Sun’s pull is stronger, Moon’s closeness makes its tidal effect about 2.2 times stronger.
Question 8.2
Choose the correct alternative:
(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 mass of the earth/mass of the body.
(d) The formula \( -G M m \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \) is more/less accurate than the formula \( mg(r_2 - r_1) \) for the difference of potential energy between two points \( r_2 \) and \( r_1 \) distance away from the centre of the earth.
(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 mass of the earth/mass of the body.
(d) The formula \( -G M m \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \) is more/less accurate than the formula \( mg(r_2 - r_1) \) for the difference of potential energy between two points \( r_2 \) and \( r_1 \) distance away from the centre of the earth.
Answer & Explanation:
(a) decreases
(b) decreases
(c) mass of the body
(d) more accurate
Explanation: \( g \propto 1/(R+h)^2 \) with altitude; inside Earth \( g \propto (R-d) \); gravitational acceleration depends on Earth’s mass, not test mass; the exact formula uses Newton’s law of gravitation.
(b) decreases
(c) mass of the body
(d) more accurate
Explanation: \( g \propto 1/(R+h)^2 \) with altitude; inside Earth \( g \propto (R-d) \); gravitational acceleration depends on Earth’s mass, not test mass; the exact formula uses Newton’s law of gravitation.
Question 8.3
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 & Explanation:
Using Kepler’s Third Law: \( T^2 \propto R^3 \)
If \( T_p = T_e / 2 \), then \( (T_p/T_e)^2 = (R_p/R_e)^3 \)
\( (1/2)^2 = (R_p/R_e)^3 \) ⇒ \( 1/4 = (R_p/R_e)^3 \)
\( R_p/R_e = (1/4)^{1/3} \approx 0.63 \)
Answer: About 63% of Earth’s orbital radius.
If \( T_p = T_e / 2 \), then \( (T_p/T_e)^2 = (R_p/R_e)^3 \)
\( (1/2)^2 = (R_p/R_e)^3 \) ⇒ \( 1/4 = (R_p/R_e)^3 \)
\( R_p/R_e = (1/4)^{1/3} \approx 0.63 \)
Answer: About 63% of Earth’s orbital radius.
Question 8.4
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is \( 4.22 \times 10^8 \, \text{m} \). Show that the mass of Jupiter is about one-thousandth that of the sun.
Answer & Explanation:
Using \( T^2 = \frac{4\pi^2}{GM} R^3 \):
\( M_J = \frac{4\pi^2 R^3}{G T^2} \)
\( T = 1.769 \times 24 \times 3600 = 152841.6 \, \text{s} \)
\( M_J \approx 1.90 \times 10^{27} \, \text{kg} \)
\( M_{\text{Sun}} \approx 1.99 \times 10^{30} \, \text{kg} \)
Ratio \( \approx 0.000955 \approx 1/1000 \).
\( M_J = \frac{4\pi^2 R^3}{G T^2} \)
\( T = 1.769 \times 24 \times 3600 = 152841.6 \, \text{s} \)
\( M_J \approx 1.90 \times 10^{27} \, \text{kg} \)
\( M_{\text{Sun}} \approx 1.99 \times 10^{30} \, \text{kg} \)
Ratio \( \approx 0.000955 \approx 1/1000 \).
Question 8.5
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 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be \( 10^5 \) ly.
Answer & Explanation:
Total galactic mass \( M = 2.5 \times 10^{11} \times 2 \times 10^{30} = 5 \times 10^{41} \, \text{kg} \)
Orbital radius \( r = 50000 \, \text{ly} = 4.73 \times 10^{20} \, \text{m} \)
Using \( T^2 = \frac{4\pi^2}{GM} r^3 \):
\( T \approx 1.88 \times 10^{15} \, \text{s} \approx 5.97 \times 10^7 \, \text{years} \approx 60 \, \text{million years} \).
Orbital radius \( r = 50000 \, \text{ly} = 4.73 \times 10^{20} \, \text{m} \)
Using \( T^2 = \frac{4\pi^2}{GM} r^3 \):
\( T \approx 1.88 \times 10^{15} \, \text{s} \approx 5.97 \times 10^7 \, \text{years} \approx 60 \, \text{million years} \).
Question 8.6
Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch 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.
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch 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 & Explanation:
(a) kinetic energy (since \( E = -K \) for circular orbit)
(b) less (orbiting satellite already has kinetic energy)
In circular orbit: \( E = -\frac{GMm}{2r} \), \( K = \frac{GMm}{2r} \), so \( E = -K \).
(b) less (orbiting satellite already has kinetic energy)
In circular orbit: \( E = -\frac{GMm}{2r} \), \( K = \frac{GMm}{2r} \), so \( E = -K \).
Question 8.7
Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?
Answer & Explanation:
Escape speed \( v_e = \sqrt{\frac{2GM}{R}} \)
(a) No
(b) Yes (depends on distance from Earth’s center)
(c) No (direction doesn’t matter in vacuum)
(d) Yes \( \left( v_e = \sqrt{\frac{2GM}{R+h}} \right) \)
(a) No
(b) Yes (depends on distance from Earth’s center)
(c) No (direction doesn’t matter in vacuum)
(d) Yes \( \left( v_e = \sqrt{\frac{2GM}{R+h}} \right) \)
Question 8.8
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit?
Answer & Explanation:
(a) Not constant (varies by Kepler’s second law)
(b) Not constant
(c) Constant (conserved in central force)
(d) Not constant
(e) Not constant (\( U \propto -1/r \))
(f) Constant (mechanical energy conserved)
(b) Not constant
(c) Constant (conserved in central force)
(d) Not constant
(e) Not constant (\( U \propto -1/r \))
(f) Constant (mechanical energy conserved)
Question 8.9
Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.
Answer & Explanation:
(b) swollen face, (c) headache, (d) orientational problem
In microgravity, fluids shift upward (facial puffiness), spatial orientation is disrupted, and headaches may occur due to fluid shifts.
In microgravity, fluids shift upward (facial puffiness), spatial orientation is disrupted, and headaches may occur due to fluid shifts.
Question 8.10
The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 8.12) (i) a, (ii) b, (iii) c, (iv) 0.
Answer & Explanation:
(iv) 0
Gravitational field at the centre of a symmetric shell is zero due to cancellation of forces.
Gravitational field at the centre of a symmetric shell is zero due to cancellation of forces.
Question 8.11
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 & Explanation:
(i) d (toward the centre of the hemisphere)
Gravitational force points toward the centre of mass of the shell.
Gravitational force points toward the centre of mass of the shell.
Question 8.12
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 \( = 2 \times 10^{30} \, \text{kg} \), mass of the earth \( = 6 \times 10^{24} \, \text{kg} \). Neglect the effect of other planets etc. (orbital radius \( = 1.5 \times 10^{11} \, \text{m} \)).
Answer & Explanation:
Let \( x \) = distance from Earth’s centre where forces balance:
\( \frac{GM_E}{x^2} = \frac{GM_S}{(R-x)^2} \)
\( \frac{R-x}{x} = \sqrt{\frac{M_S}{M_E}} \approx 577.35 \)
\( x \approx \frac{R}{578.35} \approx 2.59 \times 10^8 \, \text{m} \)
\( \frac{GM_E}{x^2} = \frac{GM_S}{(R-x)^2} \)
\( \frac{R-x}{x} = \sqrt{\frac{M_S}{M_E}} \approx 577.35 \)
\( x \approx \frac{R}{578.35} \approx 2.59 \times 10^8 \, \text{m} \)
Question 8.13
How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is \( 1.5 \times 10^8 \, \text{km} \).
Answer & Explanation:
Use Kepler’s Third Law: \( T^2 = \frac{4\pi^2}{GM_S} R^3 \)
\( M_S = \frac{4\pi^2 R^3}{G T^2} \)
With \( T = 1 \, \text{year} \approx 3.15 \times 10^7 \, \text{s} \), \( R = 1.5 \times 10^{11} \, \text{m} \)
\( M_S \approx 2.0 \times 10^{30} \, \text{kg} \)
\( M_S = \frac{4\pi^2 R^3}{G T^2} \)
With \( T = 1 \, \text{year} \approx 3.15 \times 10^7 \, \text{s} \), \( R = 1.5 \times 10^{11} \, \text{m} \)
\( M_S \approx 2.0 \times 10^{30} \, \text{kg} \)
Question 8.14
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is \( 1.50 \times 10^8 \, \text{km} \) away from the sun?
Answer & Explanation:
\( T_S/T_E = 29.5 \)
\( (R_S/R_E)^3 = (T_S/T_E)^2 = 870.25 \)
\( R_S/R_E = 870.25^{1/3} \approx 9.54 \)
\( R_S \approx 1.43 \times 10^9 \, \text{km} \)
\( (R_S/R_E)^3 = (T_S/T_E)^2 = 870.25 \)
\( R_S/R_E = 870.25^{1/3} \approx 9.54 \)
\( R_S \approx 1.43 \times 10^9 \, \text{km} \)
Question 8.15
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer & Explanation:
\( F \propto 1/(R+h)^2 \), \( h = R/2 \)
\( \frac{F_h}{F_s} = \frac{R^2}{(1.5R)^2} = \frac{4}{9} \)
\( F_h = \frac{4}{9} \times 63 = 28 \, \text{N} \)
\( \frac{F_h}{F_s} = \frac{R^2}{(1.5R)^2} = \frac{4}{9} \)
\( F_h = \frac{4}{9} \times 63 = 28 \, \text{N} \)
Question 8.16
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth, if it weighed 250 N on the surface?
Answer & Explanation:
Inside Earth: \( g(d) = g(0) \left(1 - \frac{d}{R}\right) \)
At \( d = R/2 \): \( g = g(0)/2 \)
Weight = \( \frac{1}{2} \times 250 = 125 \, \text{N} \)
At \( d = R/2 \): \( g = g(0)/2 \)
Weight = \( \frac{1}{2} \times 250 = 125 \, \text{N} \)
Question 8.17
A rocket is fired vertically with a speed of \( 5 \, \text{km s}^{-1} \) from the earth's surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth \( = 6.0 \times 10^{24} \, \text{kg} \); mean radius of the earth \( = 6.4 \times 10^6 \, \text{m} \); \( G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \).
Answer & Explanation:
Energy conservation: \( \frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{r} \)
\( \frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{r} \right) \)
Solve for \( r \): \( r \approx 8.0 \times 10^6 \, \text{m} \)
Height \( = r - R \approx 1.6 \times 10^6 \, \text{m} = 1600 \, \text{km} \)
\( \frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{r} \right) \)
Solve for \( r \): \( r \approx 8.0 \times 10^6 \, \text{m} \)
Height \( = r - R \approx 1.6 \times 10^6 \, \text{m} = 1600 \, \text{km} \)
Question 8.18
The escape speed of a projectile on the earth's surface is \( 11.2 \, \text{km s}^{-1} \). 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 the sun and other planets.
Answer & Explanation:
Energy conservation: \( \frac{1}{2} m (3v_e)^2 - \frac{1}{2} m v_e^2 = \frac{1}{2} m v_f^2 \)
\( 9v_e^2 - v_e^2 = v_f^2 \) ⇒ \( v_f^2 = 8v_e^2 \)
\( v_f = \sqrt{8} \, v_e \approx 31.7 \, \text{km/s} \)
\( 9v_e^2 - v_e^2 = v_f^2 \) ⇒ \( v_f^2 = 8v_e^2 \)
\( v_f = \sqrt{8} \, v_e \approx 31.7 \, \text{km/s} \)
Question 8.19
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 \( = 200 \, \text{kg} \); mass of the earth \( = 6.0 \times 10^{24} \, \text{kg} \); radius of the earth \( = 6.4 \times 10^6 \, \text{m} \); \( G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \).
Answer & Explanation:
\( r = R + h = 6.8 \times 10^6 \, \text{m} \)
Energy needed \( = \frac{GMm}{2r} \)
\( \approx \frac{(6.67\times10^{-11})(6.0\times10^{24})(200)}{2 \times 6.8\times10^6} \)
\( \approx 5.9 \times 10^9 \, \text{J} \)
Energy needed \( = \frac{GMm}{2r} \)
\( \approx \frac{(6.67\times10^{-11})(6.0\times10^{24})(200)}{2 \times 6.8\times10^6} \)
\( \approx 5.9 \times 10^9 \, \text{J} \)
Question 8.20
Two stars each of one solar mass (\( = 2 \times 10^{30} \, \text{kg} \)) are approaching each other for a head on collision. When they are a distance \( 10^9 \, \text{km} \), their speeds are negligible. What is the speed with which they collide? The radius of each star is \( 10^4 \, \text{km} \).
Answer & Explanation:
Energy conservation: \( -\frac{GM^2}{r_i} = Mv^2 - \frac{GM^2}{r_f} \)
\( v^2 = GM \left( \frac{1}{r_f} - \frac{1}{r_i} \right) \)
\( r_f = 2R = 2 \times 10^7 \, \text{m} \), \( r_i = 10^{12} \, \text{m} \)
\( v \approx \sqrt{ \frac{GM}{r_f} } \approx 2.58 \times 10^6 \, \text{m/s} \)
\( v^2 = GM \left( \frac{1}{r_f} - \frac{1}{r_i} \right) \)
\( r_f = 2R = 2 \times 10^7 \, \text{m} \), \( r_i = 10^{12} \, \text{m} \)
\( v \approx \sqrt{ \frac{GM}{r_f} } \approx 2.58 \times 10^6 \, \text{m/s} \)
Question 8.21
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 & Explanation:
Net force = 0 (equal and opposite).
Potential \( V = -2 \times \frac{Gm}{r} \approx -2.67 \times 10^{-8} \, \text{J/kg} \)
Equilibrium at midpoint: unstable (displacement increases net force toward nearer sphere).
Potential \( V = -2 \times \frac{Gm}{r} \approx -2.67 \times 10^{-8} \, \text{J/kg} \)
Equilibrium at midpoint: unstable (displacement increases net force toward nearer sphere).
Question 8.22
As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth \( = 6.0 \times 10^{24} \, \text{kg} \), radius = 6400 km.
Answer & Explanation:
\( r = R + h = 4.24 \times 10^7 \, \text{m} \)
\( V = -\frac{GM}{r} \approx -\frac{(6.67\times10^{-11})(6.0\times10^{24})}{4.24\times10^7} \)
\( \approx -9.44 \times 10^6 \, \text{J/kg} \)
\( V = -\frac{GM}{r} \approx -\frac{(6.67\times10^{-11})(6.0\times10^{24})}{4.24\times10^7} \)
\( \approx -9.44 \times 10^6 \, \text{J/kg} \)
Question 8.23
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun \( = 2 \times 10^{30} \, \text{kg} \)).
Answer & Explanation:
\( g = \frac{GM}{R^2} \approx 2.3 \times 10^{12} \, \text{m/s}^2 \)
Centripetal acceleration \( a_c = \omega^2 R \approx 6.8 \times 10^5 \, \text{m/s}^2 \)
Since \( g \gg a_c \), object remains stuck.
Centripetal acceleration \( a_c = \omega^2 R \approx 6.8 \times 10^5 \, \text{m/s}^2 \)
Since \( g \gg a_c \), object remains stuck.
Question 8.24
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the sun \( = 2 \times 10^{30} \, \text{kg} \); mass of mars \( = 6.4 \times 10^{23} \, \text{kg} \); radius of mars = 3395 km; radius of the orbit of mars \( = 2.28 \times 10^8 \, \text{km} \); \( G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \).
Answer & Explanation:
Escape Mars: \( E_1 = \frac{GM_M m}{R_M} \approx 1.26 \times 10^{10} \, \text{J} \)
Escape Sun from Mars orbit: \( E_2 = \frac{GM_S m}{r_{\text{orbit}}} \approx 5.85 \times 10^{11} \, \text{J} \)
Total \( \approx 5.85 \times 10^{11} \, \text{J} \)
Escape Sun from Mars orbit: \( E_2 = \frac{GM_S m}{r_{\text{orbit}}} \approx 5.85 \times 10^{11} \, \text{J} \)
Total \( \approx 5.85 \times 10^{11} \, \text{J} \)
Question 8.25
A rocket is fired vertically from the surface of mars with a speed of \( 2 \, \text{km s}^{-1} \). If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars \( = 6.4 \times 10^{23} \, \text{kg} \); radius of mars = 3395 km; \( G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \).
Answer & Explanation:
Effective initial energy = \( 0.8 \times \frac{1}{2}mv^2 \)
\( 0.4 v^2 = GM \left( \frac{1}{R} - \frac{1}{r} \right) \)
Solve for \( r \): \( r \approx 3.89 \times 10^6 \, \text{m} \)
Height \( = r - R \approx 496 \, \text{km} \)
\( 0.4 v^2 = GM \left( \frac{1}{R} - \frac{1}{r} \right) \)
Solve for \( r \): \( r \approx 3.89 \times 10^6 \, \text{m} \)
Height \( = r - R \approx 496 \, \text{km} \)
📘 Exam Preparation Tip:
These exercise questions will help you understand universal gravitation and celestial mechanics. You'll learn to apply Newton's law of gravitation, solve problems using Kepler's laws, calculate gravitational field and potential, determine satellite speeds and periods, compute escape velocity, and understand weightlessness concepts. Essential for astronomy and space science foundations.