Chapter 7 Gravitation

NCERT Solutions Class 11 Physics Chapter 7 – Gravitation

NCERT Solutions for Class 11 Physics Chapter 7 explain the universal force of gravitation and its role in the motion of objects, planets, and satellites. The chapter covers Newton’s Law of Gravitation, gravitational field, gravitational potential, acceleration due to gravity, escape velocity, and orbital motion.

NCERT Solutions for Class 11 Physics Chapter 7 by ALLEN covers all the important concepts like gravitational force, satellite motion, Kepler's laws, variation of acceleration due to gravity with height and depth with clear explanations and step-by-step solutions.

The chapter also connects Gravitation with Electrostatics through similar mathematical relationships. These solutions help students understand important concepts and solve numerical problems related to satellites, orbital velocity, escape velocity, and gravitational potential energy for CBSE, JEE, and NEET preparation.

1.0Class 11 Physics Chapter 7 : Key Concepts

This chapter examines the nature of gravitational interaction and its effects on celestial and terrestrial bodies. Key lessons include:

Kepler’s Laws of Planetary Motion:

Law of Orbits: Planets move in elliptical orbits with the Sun at one focus.
Law of Areas: A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
Law of Periods: The square of the time period of a planet is proportional to the cube of the semi-major axis of its orbit $(T^{2} \propto R^{3}) .$
Universal Law of Gravitation: Every particle in the universe attracts every other particle with a force: $F = G \frac{m _{1} m _{2}}{r ^{2}}$
The Gravitational Constant (G): Understanding the Cavendish experiment used to determine the value of $G \approx 6.67 \times 1 0^{- 11} N m^{2} / kg^{2} .$

Acceleration due to Gravity (g):

Relation with G: $g = \frac{GM}{R ^{2}}$
Variation of g with Altitude (h): g decreases as we go up.
Variation of g with Depth (d): g decreases as we go down toward the center of the Earth.
Gravitational Potential Energy: The work done in bringing a mass from infinity to a point $(U = - \frac{GM m}{r}) .$
Escape Speed ( $v_{e}$ ): The minimum speed required for an object to break free from Earth's gravitational pull is called escape speeed/velocity. $(v_{e} = \frac{2 GM}{R} \approx 11.2 km/s) .$

Earth Satellites:

Orbital Speed ( $v_{o}$ ): The speed required to keep a satellite in a circular orbit $(v_{o} = \frac{GM}{r}) .$
Geostationary Satellites: Satellites that appear fixed over a point on Earth (Period = 24 hours).
Polar Satellites: Satellites that orbit in a north-south direction for weather and environmental monitoring.
Weightlessness: Understanding why astronauts feel "weightless" in a state of free fall while orbiting the Earth.

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 7

Let the speed of the planet at the perihelion P in figure be $v_{P}$ and the sun-planet distance SP be $r_{p}$ . Relate ${r_{p}, v_{p}}$ . Will the planet take equal times to traverse BAC and CPB ?

Sol. The magnitude of the angular momentum at P is $L_{p} = m_{p} r_{p} v_{p}$ , since inspection tells us that $r_{p}$ and $v_{p}$ are mutually perpendicular. Similarly, $L_{A} = m_{p} r_{A} v_{A}$ . From angular momentum conservation

$m_{p} r_{p} v_{p} = m_{p} r_{A} v_{a}$

$\frac{v _{p}}{v _{A}} = \frac{r _{A}}{r _{p}}$

Since $r_{A} > r_{p}, v_{p} > v_{A}$ . The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in figure. From Kepler's second law, equal areas are swept in equal times. hence the planet will take a longer time traverse BAC than CPB.

Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass 2m placed at the centroid G of the triangle? (b) What is the force if the mass at the vertex A is doubled? Take $AG = BG = CG = 1 m$ (see figure)

Sol. (a) The angle between GC and the positive x-axis is 30° and so is the angle between GB and the negative x-axis. The individual forces in vector notation are

$F_{G A} = \frac{G m ( 2 m )}{1 ^{2}} \hat{j} = 2 G m^{2} \hat{j} F_{GB} = \frac{G m ( 2 m )}{1} (- cos 3 0^{\circ} \hat{i} - sin 3 0^{\circ} \hat{j}) = 2 G m^{2} (- cos 3 0^{\circ} \hat{i} - sin 3 0^{\circ} \hat{j}) F_{GC} = \frac{G m ( 2 m )}{1} (cos 3 0^{\circ} \hat{i} - sin 3 0^{\circ} \hat{j}) = 2 G^{2} (cos 3 0^{\circ} \hat{i} - sin 3 0^{\circ} \hat{j})$

From the principle of superposition and the law of vector addition, the resultant gravitational force $F_{R}$ on (2m) is

$F_{R} = F_{G A} + F_{GB} + F_{GC} F_{R} = 2 G^{2} \hat{j} + 2 Gm^{2} (- cos 3 0^{\circ} \hat{i} - sin 3 0^{\circ} \hat{j}) + 2 Gm^{2} (cos 3 0^{\circ} \hat{i} - sin 3 0^{\circ} \hat{j}) = 0$

Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero. (b) Now if the mass at vertex A is doubled then

$F_{GA}^{'} = \frac{G 2 m \cdot 2 m}{1} \hat{j} = 4 Gm^{2} \hat{j} F_{GB}^{'} = F_{GB} and F_{GC}^{'} = F_{GC} F_{R}^{'} = F_{GA}^{'} + F_{GB}^{'} + F_{GC}^{'} F_{R}^{'} = 2 Gm^{2} \hat{j}$

Find the potential energy of a system of four particles placed at the vertices of a square of side $l$ . Also obtain the potential at the centre of the square. Sol. Consider four masses each of mass m at the corners of a square of side $l$ ; See figure. We have four mass pairs at distance $l$ and two diagonal pairs at distance $2 l$ . Hence,

$U = - 4 \frac{Gm ^{2}}{l} - 2 \frac{Gm ^{2}}{2 l} = - \frac{2 Gm ^{2}}{l} (2 + \frac{1}{2}) = - 5.41 \frac{Gm ^{2}}{l}$

The gravitational potential at the centre of the square $(r = \frac{2 l}{2})$ is $V (r) = - 42 \frac{Gm}{l}$

Two uniform solid spheres of equal radii R, but mass M and 4M have a centre separation 6R, as shown in figure. The two spheres are held fixed. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Sol. The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see figure) is defined as the position where the two forces cancel each other exactly. If $ON = r$ , we have $\frac{GMm}{r ^{2}} = \frac{4 GMm}{( 6 R - r ) ^{2}}$

$(6 R - r)^{2} = 4 r^{2} 6 R - r = \pm 2 r r = 2 R or - 6 R$

The neutral point $r = - 6 R$ does not concern us in this example. Thus $ON = r = 2 R$ . It is sufficient to project the particle with a speed which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is

$E_{i} = \frac{1}{2} mv^{2} - \frac{GMm}{R} - \frac{4 GMm}{5 R}$

At the neutral point N , the speed approaches zero. The mechanical energy at N is purely potential.

$E_{N} = - \frac{GMm}{2 R} - \frac{4 GMm}{4 R}$

From the principle of conservation of mechanical energy.

$\frac{1}{2} v^{2} - \frac{GM}{R} - \frac{4 GM}{5 R} = - \frac{GM}{2 R} - \frac{GM}{R}$

$v^{2} = \frac{2 GM}{R} (\frac{4}{5} - \frac{1}{2}) v = (\frac{3 GM}{5 R})^{1/2}$

The planet Mars has two moons. phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of $9.4 \times 1 0^{3}$ km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days ?

Sol. (i) We employ Eq. $T^{2} = \frac{4 π ^{2} R ^{3}}{GM _{s}}$ with the Sun's mass replaced by the martian mass $M_{m}$ .

$T^{2} = M_{m} = \frac{4 π ^{2}}{GM _{m}} R^{3} = M_{m} = \frac{4 π ^{2}}{G} \frac{R ^{3}}{T ^{2}} = \frac{4 \times ( 3.14 ) ^{2} \times ( 9.4 ) ^{3} \times 1 0 ^{18}}{6.67 \times 1 0 ^{- 11} \times ( 459 \times 60 ) ^{2}} \frac{4 \times ( 3.14 ) ^{2} \times ( 9.4 ) ^{3} \times 1 0 ^{18}}{6.67 \times ( 4.59 \times 6 ) ^{2} \times 1 0 ^{- 5}} = 6.48 \times 1 0^{23} kg .$

(ii) Once again Kepler's third law comes to our aid,

$\frac{T _{M}^{2}}{T _{E}^{2}} = \frac{R _{MS}^{3}}{R _{ES}^{3}}$

where $R_{MS}$ is the Mars-Sun distance and $R_{ES}$ is the Earth-Sun distance.

$∴ T_{M} = (1.52)^{3/2} \times 365 = 684 days$

Weighing the Earth : You are given the following data: $g = 9.81 ms^{- 2}, R_{E} = 6.37 \times 1 0^{6} m$ , the distance to the moon $R = 3.84 \times 1 0^{8} m$ and the time period of the moon's revolution is 27.3 days. Obtain the mass of the Earth $M_{E}$ in two different ways. Sol. From Eq. $g = \frac{GM}{R ^{2}}$ , We have

$M_{E} = \frac{g R _{E}^{2}}{G} = \frac{9.81 \times ( 6.37 \times 1 0 ^{6} ) ^{2}}{6.67 \times 1 0 ^{- 11}} = 5.97 \times 1 0^{24} kg .$

The moon is a satellite of the Earth. From the derivation of Kepler's third law [See Eq.]

$T^{2} = \frac{4 π ^{2} R ^{3}}{G M _{E}} M_{E} = \frac{4 π ^{2} R ^{3}}{G T ^{2}} = \frac{4 \times 3.14 \times 3.14 \times ( 3.84 ) ^{3} \times 1 0 ^{24}}{6.67 \times 1 0 ^{- 11} \times ( 27.3 \times 24 \times 60 \times 60 ) ^{2}} = 6.02 \times 1 0^{24} kg$

Both methods yield almost the same answer, the difference between them being less than 1%.

Express the constant k of Eq. $T^{2} = kR^{3}$ in days and kilometres. Given $k = 1 0^{- 13} s^{2} m^{- 3}$ . The moon is at a distance of $3.84 \times 1 0^{5} km$ from the earth. Obtain its time-period of revolution in days. Sol. Given $k = 1 0^{- 13} s^{2} m^{- 3}$

$k = 1 0^{- 13} [\frac{1}{( 24 \times 60 \times 60 ) ^{2}} d^{2}] [\frac{1}{( 1/1000 ) ^{3} km ^{3}}] = 1.33 \times 1 0^{- 14} d^{2} km^{- 3} (d \to days)$

Using Eq. $T^{2} = kR^{2}$ , and the given value of k , the time period of the moon is

$T^{2} = (1.33 \times 1 0^{- 14}) (3.84 \times 1 0^{5})^{3} T = 27.3 d$

A 400 kg satellite is in a circular orbit of radius $2 R_{E}$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4 R_{E}$ ? What are the changes in the kinetic and potential energies? Sol. Initially, $E_{i} = - \frac{GM _{E} m}{4 R _{E}}$ While finally

$E_{f} = - \frac{GM _{E} m}{8 R _{E}}$

The change in the total energy is

$Δ E = E_{f} - E_{i} = \frac{GM _{E} m}{8 R _{E}} = (\frac{GM _{E}}{R _{E}^{2}}) \frac{mR _{E}}{8} Δ E = \frac{gmR _{E}}{8} = \frac{9.81 \times 400 \times 6.37 \times 1 0 ^{6}}{8} = 3.13 \times 1 0^{9} J$

The kinetic energy is reduced and it mimics $Δ E$ , namely, $Δ K = K_{f} - K_{i} = - 3.13 \times 1 0^{9} J$ . The change in potential energy is twice the change in the total energy, namely

$Δ U = U_{f} - U_{i} = - 6.25 \times 1 0^{9} J$

EXERCISE QUESTIONS WITH SOLUTIONS

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 we 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. (you can check this yourself the data available in the succeeding exercises). However, the tidal effect of the moon's pull is greater than the tidal effect of sun. Why ? Sol. (a) No, Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects. (b) Yes, If the size of the space station is large enough, then the astronaut will detect the change in Earth's gravity(g). (c) Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon's pull is greater than the tidal effect of the Sun's pull.
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 $- GMm (1/ r_{2} - 1/ r_{1})$ is more/less accurate than the formula $mg (r_{2} - r_{1})$ for the difference of difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the earth. Sol. (a) Decreases Explanation : Acceleration due to gravity at altitude h is given by the relation:

$g_{h} = (1 - \frac{2 h}{R _{E}}) g$

Where,

$R_{E} = Radius of the Earth$

$g =$ Acceleration due to gravity on the surface of the Earth. It is clear from the given relation that acceleration due to gravity decreases with an increase in height. (b) Decreases Explanation : Acceleration due to gravity at depth d is given by the relation :

$g_{d} = (1 - \frac{d}{R _{E}}) g$

It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

$g = \frac{G M _{E}}{R _{E}^{2}}$

Where, G = Universal gravitational constant $M_{E} =$ Mass of the Earth $R_{E} =$ Radius of the Earth Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body. (d) Gravitational potential energy of two points $r_{2}$ and $r_{1}$ distance away from the centre of the Earth is respectively given by:

$U (r_{1}) U (r_{2}) = \frac{- GmM}{r _{1}} = \frac{- GmM}{r _{2}}$

∴ Difference in potential energy,

$Δ U = U (r_{2}) - U (r_{1}) = - GmM (\frac{1}{r _{2}} - \frac{1}{r _{1}})$

Hence, this formula is more accurate than the formula $mg (r_{2} - r_{1})$ .

Suppose there existed a planet that went around the twice as fast as the earth. What would be it orbital size as compared to that of the earth? Sol. Time taken by the Earth to complete one revolution around the Sun, $T_{E} = 1$ year Orbital radius of the Earth in it orbit, $R_{E} = 1 AU$ Time taken by the planet to complete one revolution around the Sun, $T_{P} = \frac{1}{2} T_{E} = \frac{1}{2}$ year Orbital radius of the planet $= R_{p}$

From Kepler's third law of planetary motion, we can write :

$(\frac{R _{P}}{R _{E}})^{3} = (\frac{T _{P}}{T _{E}})^{2} \frac{R _{P}}{R _{E}} = (\frac{T _{P}}{T _{E}})^{2/3} = (\frac{1/2}{1})^{2/3} = (0.5)^{2/3} = 0.63$

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

$I_{0}$ , one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is $4.22 \times 1 0^{8} m$ . Show that the mass of Jupiter is about one-thousandth that of the sun. Sol. Orbital period of $I_{0}, T_{I_{0}} = 1.769$ days

$= 1.769 \times 24 \times 60 \times 60 s$

Orbital radius of $I_{0}, R_{I_{0}} = 4.22 \times 1 0^{8} m$ Satellite $I_{0}$ is revolving around the Jupiter Mass of the Jupiter is given by the relation :

$M_{J} = \frac{4 π ^{2} R _{I_{0}}^{3}}{G T _{I_{0}}^{2}}$

Where, $M_{J} =$ Mass of Jupiter G = Universal gravitational constant Orbital period of Earth, $T_{E} = 365.25$ days $= 365.25 \times 24 \times 60 \times 60 s$ Orbital radius of the Earth, $R_{E} = 1 AU = 1.496 \times 1 0^{11} m$ Mass of sun is given as :

$M_{S} = \frac{4 π ^{2} R _{E}^{3}}{G T _{E}^{2}}$

$∴ \frac{M _{x}}{M _{j}} = \frac{4 π ^{2} R _{E}^{3}}{G T _{E}^{2}} \times \frac{G T _{I_{0}}^{2}}{4 π ^{2} R _{I_{0}}^{3}} = \frac{R _{E}^{3}}{R _{I_{0}}^{3}} \times \frac{T _{I_{0}}^{2}}{T _{E}^{2}}$

$∴ \frac{M _{S}}{M _{J}} = 1000 \Rightarrow M_{J} = \frac{M _{s}}{1000}$

Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.

Let us assume that our galaxy consists of $2.5 \times 1 0^{11}$ stars 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 $1 0^{5} ly$ . Sol. Mass of our galaxy Milky Way,

$M = 2.5 \times 1 0^{11} solar mass Solar mass = Mass of Sun = 2.0 \times 1 0^{30} kg$

Mass of our galaxy,

$M = 2.5 \times 1 0^{11} \times 2 \times 1 0^{30} = 5 \times 1 0^{41} kg$

Diameter of Milky Way, $d = 1 0^{5} ly$ Radius of Milky Way, $r = 5 \times 1 0^{4} ly$

$1 ly = 9.46 \times 1 0^{15} m ∴ r = 5 \times 1 0^{4} \times 9.46 \times 1 0^{15} m = 4.73 \times 1 0^{20} m$

Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation :

$T = (\frac{4 π ^{2} r ^{3}}{GM})^{1/2} = (\frac{4 \times ( 3.14 ) ^{2} \times ( 4.73 ) ^{3} \times 1 0 ^{60}}{6.67 \times 1 0 ^{- 11} \times 5 \times 1 0 ^{41}})^{1/2} = (\frac{39.44 \times 105.82 \times 1 0 ^{30}}{33.35})^{1/2} = (125.14 \times 1 0^{30})^{1/2} = 1.12 \times 1 0^{16} s 1 year = 365 \times 324 \times 60 \times 60 s 1 s = \frac{1}{365 \times 324 \times 60 \times 60} years ∴ 1.12 \times 1 0^{16} s = \frac{1.12 \times 1 0 ^{16}}{365 \times 24 \times 60 \times 60} = 3.55 \times 1 0^{8} years$

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. Sol. (a) Kinetic Energy Explanation : Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (may be negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative. Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy. (b) Less Explanation : An orbiting satellite acquires a certain amount of energy that enables it to revolve around the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out of the influence of the Earth's gravitational field than a stationary object on the Earth's surface that initially contains no energy.
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?

Sol. (a) No

(b) No (c) No (d) Yes

Explanation : Escape velocity of a body from the Earth is given by the relation :

$v_{esc} = 2 gR$

$g =$ Acceleration due to gravity R = Radius of the Earth It is clear from equation (i) that escape velocity $v_{ese}$ is independent of the mass of the body and the direction of its projection. However, it depends on gravitational potential at the point from where the body is launched. Since this potential marginally depends on the height of the point, escape velocity also marginally depends on these factors.

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? Neglect any mass loss of the comet when it comes very close to the Sun.

Sol. (a) No

(b) No (c) Yes (d) No (e) No (f) Yes

Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Its linear speed, angular speed, kinetic and potential energy varies from point to point in the orbit.

Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem? Sol. (b), (c) and (d) Explanation :

(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not afflict him/her in space. (b) A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears nose, and mouth constitute a person's face. This symptom can affect an astronaut in space. (c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space. (d) Space has different orientations. Therefore, orientational problem can afflict an astronaut in space.

Choose the correct answer from among the given ones : The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see figure) (i) a, (ii) b, (iii) c, (iv) O

class-11-chap-7-physics-exercise-ques-10

Sol. (iii) c Explanation : Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient $(\frac{dV}{dr})$ is zero everywhere inside the spherical shell.

The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric. If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at centre O will be in the downward direction.

exercise-ques-10-(ii)-class-11-physics-chap-7

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at centre O of the given hemispherical shell has the direction as indicated by arrow c.

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.

Sol. (ii) e Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient $(\frac{dV}{dr})$ is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.

If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.

class-11-exercise-ques-11-chap-7-physics

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

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 $= 2 \times 1 0^{30} kg$ , mass of the Earth $= 6 \times 1 0^{24} kg$ . Neglect the effect of other planets etc. (Orbital radius $= 1.5 \times 1 0^{11} m$ ).

Sol. Mass of the Sun, $M_{s} = 2 \times 1 0^{30} kg$ Mass of the Earth, $M_{e} = 6 \times 1 0^{24} kg$ Orbital radius, $r = 1.5 \times 1 0^{11} m$ Mass of the rocket $= m$

physics-class-11-exercise-ques-12-chap-7

Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.

From Newton's law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as :

$\frac{GmM _{S}}{( r - x ) ^{2}} = Gm \frac{M _{e}}{x ^{2}} (\frac{r - x}{x})^{2} = \frac{M _{s}}{M _{e}} \frac{r - x}{x} = (\frac{2 \times 1 0 ^{30}}{6 \times 1 0 ^{24}})^{1/2} = 577.35 1.5 \times 1 0^{11} - x = 577.35 x 578.35 x = 1.5 \times 1 0^{11} x = \frac{1.5 \times 1 0 ^{11}}{578.35} = 2.59 \times 1 0^{8} m$

How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the Sun is $1.5 \times 1 0^{8} km$ .

Sol. Orbital radius of the Earth around the Sun,

$r = 1.5 \times 1 0^{11} m$

Time taken by the Earth to complete one revolution around the Sun,

$T = 1 year = 365.25 days = 365.25 \times 24 \times 60 \times 60 s$

Universal gravitational constant,

$G = 6.67 \times 1 0^{- 11} Nm^{2} kg^{- 2}$

Thus, mass of the Sun can be calculate using the relation,

$M_{s} M_{s} = \frac{4 π ^{2} r ^{3}}{G T ^{2}} = \frac{4 \times ( 3.14 ) ^{2} \times ( 1.5 \times 1 0 ^{11} ) ^{3}}{6.67 \times 1 0 ^{- 11} \times ( 365.25 \times 24 \times 60 \times 60 ) ^{2}} = \frac{133.24 \times 10}{6.64 \times 1 0 ^{4}} = 2.0 \times 1 0^{30} kg$

Hence, the mass of the Sun is $2 \times 1 0^{30} kg$ .

A Saturn year is 29.5 times the earth year. How far is the Saturn from the Sun if the Earth is $1.50 \times 1 0^{8} km$ away from the Sun?

Sol. Distance of the Earth from the Sun,

$r_{E} = 1.5 \times 1 0^{8} km = 1.5 \times 1 0^{11} m$

Time period of the Earth $= T_{E}$ Time period of Saturn, $T_{s} = 29.5 T_{E}$ Distance of Saturn from the Sun $= r_{s}$ From Kepler's third law of planetary motion, we have

$T = (\frac{4 π ^{2} r ^{3}}{GM})^{1/2}$

For Saturn and Sun, we can write

$\frac{r _{S}^{3}}{r _{E}^{3}} = \frac{T _{S}^{2}}{T _{E}^{2}} r_{S} = r_{E} (\frac{T _{S}}{T _{E}})^{2/3} = 1.5 \times 1 0^{11} (\frac{29.5 T _{E}}{T _{E}})^{2/3} = 1.5 \times 1 0^{11} (29.5)^{2/3} = 1.5 \times 1 0^{11} \times 9.55 = 14.32 \times 1 0^{11} m$

Hence, the distance between Saturn and the Sun is $1.43 \times 1 0^{12} m$ .

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 ?

Sol. Weight of the body, $W = 63 N$ Acceleration due to gravity at height h from the Earth's surface is given by the relation :

$g^{'} = \frac{g}{( \frac{1 + h}{R _{E}} ) ^{2}}$

Where, $g =$ Acceleration due to gravity on the Earth's surface $R_{E} =$ Radius of the Earth

$For h = \frac{R _{E}}{2} g^{'} = \frac{g}{( 1 + \frac{R _{E}}{2 \times R _{E}} ) ^{2}} = \frac{g}{( 1 + \frac{1}{2} ) ^{2}} = \frac{4}{9} g$

Weight of a body of mass m at height h is given as :

$W^{'} = mg^{'} = m \times \frac{4}{9} g = \frac{4}{9} \times mg = \frac{4}{9} W = \frac{4}{9} \times 63 = 28 N$

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?

Sol. Weight of a body of mass m at the Earth's surface, $W = mg = 250 N$ Body of mass m is located at depth,

$d = \frac{1}{2} R_{E}$

Where, $R_{E} =$ Radius of the Earth Acceleration due to gravity at depth g(d) is given by the relation :

$g^{'} = (1 - \frac{d}{R _{E}}) g = (1 - \frac{R _{E}}{2 \times R _{E}}) g = \frac{1}{2} g$

Weight of the body at depth d,

$W^{'} = mg^{'} = m \times \frac{1}{2} g = \frac{1}{2} mg = \frac{1}{2} W = \frac{1}{2} \times 250 = 125 N$

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 $= 6.0 \times 1 0^{24} kg$ ; mean radius of the earth $= 6.4 \times 1 0^{6} m$ ; $G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$

Sol. Velocity of the rocket,

$v = 5 km / s = 5 \times 1 0^{3} m / s$

Mass of the Earth, $M_{E} = 6.0 \times 1 0^{24} kg$ Radius of the Earth, $R_{E} = 6.4 \times 1 0^{6} m$ Mass of rocket $= m$ Height reached by rocket $= h$ At the surface of the Earth, Total energy of the rocket

$= Kinetic energy + Potential energy = \frac{1}{2} m v^{2} + (\frac{- G M _{E} m}{R _{E}})$

At highest point h ,

$v = 0$

And, potential energy $= - \frac{GM _{E} m}{R _{E} + h}$ Total energy of the rocket

$= 0 + (\frac{GM _{E} m}{R _{E} + h}) = - \frac{GM _{E} m}{R _{E} + h}$

From the law of conservation of energy, we have Total energy of the rocket at the Earth's surface = Total energy at height h

$\frac{1}{2} m v^{2} + (- \frac{G M _{E} m}{R _{E}}) = - \frac{G M _{E} m}{R _{E} + h} \frac{1}{2} v^{2} = G M_{E} (\frac{1}{R _{E}} - \frac{1}{R _{E} + h}) = G M_{E} (\frac{R _{E} + h - R}{R _{E} ( R _{E} + h )})$

$\frac{1}{2} v^{2} = \frac{G M _{E} h}{R _{E} ( R _{E} + h )} \times \frac{R _{E}}{R _{E}} \frac{1}{2} \times v^{2} = \frac{g R _{E} h}{R _{E} + h}$

Where $g = \frac{GM _{E}}{R _{E}^{2}} = 9.8 m / s^{2}$ (acceleration due to gravity on the Earth's surface)

$∴ v^{2} (R_{E} + h) = 2 g R_{E} h v^{2} R_{E} = h (2 g R_{E} - v^{2}) h = \frac{R _{E} v ^{2}}{2 g R _{E} - v ^{2}} = \frac{6.4 \times 1 0 ^{6} \times ( 5 \times 1 0 ^{3} ) ^{2}}{2 \times 9.8 \times 6.4 \times 1 0 ^{6} - ( 5 \times 1 0 ^{3} ) ^{2}} h = \frac{6.4 \times 25 \times 1 0 ^{12}}{100.44 \times 1 0 ^{6}} = 1.6 \times 1 0^{6} m$

Height achieved by the rocket with respect to the centre of the Earth

$= R_{E} + h = 6.4 \times 1 0^{6} + 1.6 \times 1 0^{6} = 8.0 \times 1 0^{6} m$

The escape speed of a projectile on the 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 the sun and other planets. Sol. Escape velocity of a projectile from the Earth, $v_{ess} = 11.2 km / s$ Projection velocity of the projectile,

$v_{P} = 3 v_{esc}$

Mass of the projectile $= m$ Velocity of the projectile far away from the Earth $= v_{f}$ Total energy of the projectile on the Earth

$= \frac{1}{2} m v_{p}^{2} - \frac{1}{2} m v_{esc}^{2}$

Gravitational potential energy of the projectile far away from the Earth is zero. Total energy of the projectile far away from the Earth $= \frac{1}{2} mv_{f}^{2}$

From the law of conservation of energy, we have

$= \frac{1}{2} m v_{p}^{2} - \frac{1}{2} m v_{esc}^{2} = \frac{1}{2} m v_{f}^{2} v_{f} = v_{p}^{2} - v_{esc}^{2} = (3 v_{esc})^{2} - (v_{esc})^{2} = 8 v_{esc} = 8 \times 11.2 = 31.68 km / s$

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 $= 6.0 \times 1 0^{24} kg$ ; radius of the earth $= 6.4 \times 1 0^{6} m; G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ . Sol. Mass of the Earth, $M_{E} = 6.0 \times 1 0^{24} kg$ Mass of the satellite, $m = 200 kg$ Radius of the Earth, $R_{E} = 6.4 \times 1 0^{6} m$ Universal gravitational constant

$G = 6.67 \times 1 0^{- 11} Nm^{2} kg^{- 2}$

Height of the satellite, $h = 400 km$

$= 4 \times 1 0^{5} m = 0.4 \times 1 0^{6} m$

Total energy of the satellite at height h ,

$= \frac{1}{2} m v^{2} + (\frac{- GM _{E} m}{R _{E} + h})$

Orbital velocity of the satellite,

$v = \frac{GM _{E}}{R _{E} + h}$

Total energy of satellite at height h ,

$= \frac{1}{2} m (\frac{GM _{E}}{R _{E} + h}) - \frac{GM _{E} m}{R _{E} + h} = - \frac{1}{2} (\frac{GM _{E} m}{R _{E} + h})$

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite. Energy required to send the satellite out of its orbit $= -$ (Bound energy)

$= \frac{1}{2} \frac{GM _{E} m}{( R _{E} + h )} = \frac{1}{2} \times \frac{6.67 \times 1 0 ^{- 11} \times 6.0 \times 1 0 ^{24} \times 200}{( 6.4 \times 1 0 ^{6} + 0.4 \times 1 0 ^{6} )} = \frac{1}{2} \times \frac{6.67 \times 6 \times 2 \times 1 0 ^{15}}{6.8 \times 1 0 ^{6}} = 5.9 \times 1 0^{9} J$

Two stars each of one solar mass $(= 2 \times 1 0^{30} kg)$ are approaching each other for a head on collision. When they are a distance $1 0^{9} km$ , their speeds are negligible. What is the speed with which they collide? The radius of each star is $1 0^{4} km$ . Assume the stars to remain undistorted until they collide. (Use the known value of G). Sol. Mass of each star, $M = 2 \times 1 0^{30} kg$ Radius of each star, $R = 1 0^{4} km = 1 0^{7} m$ Distance between the stars,

$r = 1 0^{9} km = 1 0^{12} m$

For negligible speeds, $v = 0$ total energy of two stars separated at distance r,

$= \frac{- GMM}{r} + 2 \times (\frac{1}{2} Mv^{2}) = \frac{- GMM}{r} + 0$

Now, consider the case when the stars are about to collide: Velocity of the stars = v Distance between the centers of the stars

$= 2 R$

Total kinetic energy of both stars

$= \frac{1}{2} Mv^{2} + \frac{1}{2} Mv^{2} = Mv^{2}$

Total potential energy of both stars

$= \frac{- GMM}{2 R}$

Total energy of the two stars

$= Mv^{2} + (\frac{- GMM}{2 R})$

By mechanical energy conservation

$Mv^{2} - \frac{GMM}{2 R} = \frac{- GMM}{r}$

$v^{2} = \frac{- GM}{r} + \frac{GM}{2 R} = GM (- \frac{1}{r} + \frac{1}{2 R}) = 6.67 \times 1 0^{- 11} \times 2 \times 1 0^{30} [- \frac{1}{1 0 ^{12}} + \frac{1}{2 \times 1 0 ^{7}}] = 13.34 \times 1 0^{19} \times 5 \times 1 0^{- 8} = 6.67 \times 1 0^{12} v = 6.67 \times 1 0^{12} = 2.578 \times 1 0^{6} m / s$

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 midpoint of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable? Sol. $0; - 2.7 \times 1 0^{- 8} J / kg$ ; Yes; Unstable Explanation : The situation is represented in the given figure :

chap-7-class-11-exercise-ques-21-physics

Mass of each sphere, $M = 100 kg$ Separation between the spheres, $r = 1 m$ X is the midpoint between the spheres. Gravitational force at point X will be zero. This is because gravitational force exerted by each sphere will act in opposite directions. Gravitational potential at point X,

$= \frac{- GM}{( \frac{r}{2} )} - \frac{GM}{( \frac{r}{2} )} = - \frac{4 GM}{r} = \frac{- 4 \times 6.67 \times 1 0 ^{- 11} \times 100}{1} = - 2.67 \times 1 0^{- 7} J / kg$

Any object placed at X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Access chapter-wise NCERT Solutions for Class 11 Physics with easy explanations for every textbook question. Learn essential concepts, understand the solution steps, and improve your understanding of Physics from chapter to chapter.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 7

Clear Connection Between Gravitation and Electrostatics: Understand the similarities between gravitational and electrostatic forces, fields, and potential energy through simple comparisons.
Easy Understanding of Satellite Energy: Learn about the kinetic energy, potential energy, and total mechanical energy of a satellite and understand why the total energy of a bound satellite is negative.
Step-by-Step Kepler’s Laws: Understand the derivation of Kepler’s Third Law using Newton’s Law of Gravitation, with clear steps and explanations.
Variation of Acceleration Due to Gravity: Learn how acceleration due to gravity changes with height and depth, including the approximation used near Earth’s surface.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 7, prepared according to the latest NCERT syllabus.
Simple and Concept-Based Explanations: Understand gravitation, gravitational field, gravitational potential, escape velocity, satellites, and Kepler’s laws through clear explanations and step-by-step solutions.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 7 examples and exercise questions, covering the important concepts of Gravitation for CBSE, JEE, and NEET preparation.

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