## NCERT Solutions Class 11 Physics Gravitation – Free PDF Download

***According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.**

**NCERT Solutions for Class 11 Physics Chapter 8 Gravitation** are vital resources for students to refer to score good marks in the Class 11 exam. These solutions are similar to referring to a number of textbooks by the experts. These solutions strictly adhere to the latest update on the CBSE Syllabus 2023-24. They present students with the answers to the questions in the textbooks, important questions from previous years’ question papers and sample papers. The NCERT Solutions for Class 11 Physics Chapter 8, designed by the faculty at BYJU’S helps students frame the answers for complex questions that appear in the exam.

In our lives, we know that all material objects are attracted towards the earth. Anything which we throw up falls down. Climbing a hill is more tiring than going downhill. Want to know more about Gravitation? Not able to answer the questions from NCERT Textbook? Is understanding the concepts difficult? BYJU’S is the one-stop solution for all students’ needs. For more assistance, students can download the Class 11 Physics Chapter 8 PDF of NCERT Solutions from the link provided here.

## NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

### Access answers of NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

**Q.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 spaceship 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. (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?**

**Solution:**

**(a).** **No**, as of now, no method has been devised to shield a body from gravity because gravity is independent of the medium, and it is the virtue of each and every matter. So the shield would exert gravitational forces.

**(b).** **Yes**, if the spaceship is large enough, then the astronaut will definitely detect Mars’s gravity.

**(c).** Gravitational force is inversely proportional to the square of the distance, whereas Tidal effects are inversely proportional to the cube of the distance. So as the distance between the earth and the moon is smaller than the distance between the earth and the sun, the moon will have a greater influence on the earth’s tidal waves.

**Q.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 the mass of the earth/mass of the body.
(d) The formula –G M m(1/r **

_{2}– 1/r

_{1}) 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.

**Solution:**

(a) decreases

(b) decreases

(c) mass of the body

(d) more

**Q.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?**

**Solution:**

Time taken by the earth for one complete revolution, T_{E }= 1 Year

The radius of Earth’s orbit, R_{E }= 1 AU

Thus, the time taken by the planet to complete one complete revolution

T_{P }=

_{E }=

Let the orbital radius of this planet = R_{P}

Now, according to Kepler’s third law of planetary motion,

Therefore, the radius of the orbit of this planet is 0.63 times smaller than the radius of the orbit of the Earth.

**Q.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 × 10 ^{8} m. Show that the mass of Jupiter is about one-thousandth that of the sun.**

**Solution:**

Given,

Orbital period of Io, T_{I0} = 1.769 days = 1.769 × 24 × 60 × 60 s

Orbital radius of Io, R_{I0} = 4.22 × 10^{8} m

We know the mass of Jupiter

M_{J} = 4π^{2}R_{I0}^{3} / GT_{I0}^{2} . . . . . . . . . . . . . . . (1)

Where;

M_{J} = Mass of Jupiter

G = Universal gravitational constant

Also,

The orbital period of the earth

T_{E }= 365.25 days = 365.25 × 24 × 60 × 60 s

The orbital radius of the earth, R_{E }= 1 AU = 1.496 × 10^{11 }m

We know that the mass of the sun is

M_{S} =

Therefore,

Now, on substituting the values, we will get

=

Therefore,

M_{S }~ 1000 × M_{J }

**Q.5. Let us assume that our galaxy consists of 2.5 × 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. **

**Solution:**

Mass of our galaxy, M=2.5×10 ^{11} solar mass

1 Solar mass = Mass of Sun =2×10^{30} kg

Mass of our galaxy, M=2.5×10^{11}×2×10 ^{30}=5×10^{41}kg

Diameter of Milky Way,d=10^{5} ly

Radius of Milky Way,r=5×10^{4 }ly

1 ly=9.46×10^{15}m

Therefore, r=5×104×9.46×10^{15}=4.73×10^{20}m

The time taken by the star to revolve around the galactic centre is given by the relation

T=(4π^{2}r^{3}/GM)^{1/2}

=

= 1.12 x 10^{16} s

=

= 3.55 x 10^{8} years

**Q.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 the 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 the earth’s influence.**

**Solution:**

(a) If the zero potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.

(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.

**Q. 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?**

**Solution:**

The escape speed is given by the expression

(a) The escape speed of a body from the Earth does not depend on the mass of the body.

(b) The escape speed of a body from the Earth does not depend on the location from where a body is projected.

(c) The escape speed does not depend on the direction of the projection of a body.

(d) The escape speed of a body depends upon the height of the location from where the body is launched since the escape velocity depends on the gravitational potential at the point from which it is launched. This potential, in turn, depends on the height.

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

**Solution:**

(a) According to Kepler’s second law, the linear speed of the comet keeps changing. When the comet is near the sun, its speed will be the fastest, and when it is far away from the sun, its speed will be the least.

(b) Angular speed also varies slightly.

(c) Comet has constant angular momentum.

(d) Kinetic energy changes.

(e) Potential energy changes along the path.

(f) Total energy will remain constant throughout the orbit

**Q9. Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.**

**Solution:**

**(a).** In zero gravity, the blood flow to the feet isn’t increased, so the astronaut does not get **swollen feet.**

**(b).** There is more supply of blood to the face of the astronaut. Therefore, the astronaut will have a **swollen face.**

**(c).** Due to increased blood supply to their faces, astronauts can be affected by **headaches****.**

**(d).** Space has different orientations, so **orientational problems** can affect an astronaut.

**Q.10: In the following two exercises, 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 Fig ) (i) a, (ii) b, (iii) c, (iv) 0**

** **

**Solution: (iii) c**

**Reason:**

Inside a hollow sphere, gravitational forces on any particle at any point are symmetrically placed. However, in this case, the upper half of the sphere is removed. Since gravitational intensity is gravitational force per unit mass, it will act in a direction point downwards along ‘c’.

**Q.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.**

**Solution: (ii) e**

**Reason:**

Making use of the logic/explanation from the above answer, we can conclude that the gravitational intensity at P is directed downwards along e.

**Q.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 x 10 ^{30} kg, and the mass of the earth = 6 x 10^{24} kg. Neglect the effect of other planets, etc. (Orbital radius = 1.5 x 10^{11} m)**

**Solution:**

Mass of Sun, M_{sun}= 2 x 10^{30} kg

Mass of Earth, M_{earth} = 6 x 10^{24} kg

Orbital radius, r = 1.5 x 10^{11} m

Mass of the rocket = m

Let P be the point at a distance x at which the gravitational force on the rocket is due to Earth.

^{11}– x = 577.35x

^{11}/578.35 = 2.59 x 10

^{8}m

**Q.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 × 10 ^{8} km.**

**Solution:**

Given:

Earth’s orbit, r = 1.5 × 10^{11} m

Time taken by the Earth for one complete revolution

T = 1 year = 365.25 days

i.e. T = (365.25 × 24 × 60 × 60) seconds

Since, Universal gravitational constant G = 6.67 × 10^{–11} Nm^{2} kg^{–2}

Therefore, mass of the Sun M =

^{30}Kg.

**Q. 14. A Saturn year is 29.5 times the Earth year. How far is Saturn from the Sun if the Earth is 1.50 x 10 ^{8} km away from the Sun?**

**Solution:**

According to Kepler’s third law of planetary motion,

From the above equation, we get T^{2} ∝r^{3}

T_{s} and r_{s} are the orbital periods and the mean distance of Saturn from the Sun, respectively.

T_{e} and r_{e} are the orbital periods and the mean distance of Earth from the Sun, respectively.

The time period of Saturn,T_{s} = 29.5 T_{e}

⇒r_{s}^{3}/r_{e}^{3} = T_{s}^{2}/T_{e}^{2}

r_{s}=r_{e}(T_{s}/T_{e})^{2/3}

= 1.5 x 10^{8 }x 29.5^{2/3}

= 14.3 x 10^{8 }km

**Q.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?**

**Solution:**

Given:

Weight of the man, W = 63 N

We know that acceleration due to gravity at height ‘h’ from the Earth’s surface is

g’ =

And, R

_{e}= Radius of the Earth

For h =

Also, the weight of a body of mass ‘m’ kg at the height of ‘h’ meters can be represented as

W’ = mg

= m ×

=

=

**Q. 16. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the earth if it weighed 250 N on the surface?**

**Solution:**

Weight of a body on the Earth’s surface, W=mg=250 N

Radius of the Earth = R_{e}

Let d be at a distance halfway to the centre of the earth, d=R_{e}/2

Acceleration due to gravity at d is given by the relation

g_{d}=(1− d/R_{e})g

g_{d}=(1−R_{e}/2R_{e})g

=g/2

Weight of the body at depth d

W′=mg_{d}

=mg/2 = W/2 = 250/2

=125 N

**Q.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 × 10^{24} kg; mean radius of the earth = 6.4 × 10^{6} m; G = 6.67 × 10^{–11} N m^{2} kg^{–2 }**

**Solution:**

Given:

Velocity of the missile, v = 5 km/s = 5 × 10^{3} m/s

Mass of the Earth, M_{E} = 6 × 10^{24} kg

Radius of the Earth, R_{E} = 6.4 × 10^{6} m

Let the height reached by the missile be ‘h’ and the mass of the missile be ‘m’.

Now, at the surface of the Earth

The total energy of the rocket at the surface of the Earth = Kinetic energy + Potential energy

T_{E1 }=

Now, at the highest point, ‘h’

Kinetic Energy = 0 [Since, v = 0] And, Potential energy =

Therefore, the total energy of the missile at the highest point ‘h’

T_{E2} = 0 +

_{E2}=

According to the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface T_{E1} = Total energy at height ‘h’ T_{E2}

v^{2} = GM_{E}h/R_{E}^{2} +R_{E}h

Where,

^{-2}

V^{2} = 2gR_{E}^{2}h/R_{E}^{2}+R_{E}h

Therefore, v^{2} (R_{E} + h) = 2gR_{E}h

^{2}R

_{E}= h (2gR

_{E}– v

^{2})

^{6 }m

Therefore, the missile reaches a height of 1.6 × 10

^{6 }m from the surface.

Now, the distance from the center of the earth = h + R_{E }= 1.6 × 10^{6 }+ 6.4 x 10^{6 }= 8.0 x10^{6} m

Therefore, the greatest height the missile can attain before falling back to the earth is 1.6 × 10^{6} ^{ }m, and the final distance of the missile from the centre of the earth is 8.0 x10^{6} m.

**Q.18. 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.**

**Solution:**

**Solution:**

^{30}kg

^{4}km = 10

^{7}m

^{9}km = 10

^{12}m

Total K.E. of the stars = 1/2Mv

^{2}+ 1/2Mv

^{2 }= Mv

^{2}

Therefore, the final potential energy of two stars when they are close to each other = -GMm/2R

^{2 }-GMm/2R

^{2 }-GMm/2R = -GMm/r

^{2}=(Gm/2R) – (Gm/r)

^{-11}x 2 x 10

^{30}[(1/2 x 10

^{7}) – (1/10

^{12})]

^{2}= 6.67 x 10

^{12}

^{6}m/s

**Solution:**

Given:

Radius of spheres, R = 0.10 m

Distance between two spheres, r = 1.0 m

Mass of each sphere, M = 100 kg

From the above figure, ‘A’ is the mid-point since each sphere will exert the gravitational force in the opposite direction. Therefore, the gravitational force at this point will be zero.

Gravitational potential at the midpoint (A) is

U=

U=

U=

^{-7}J /kg

Therefore, the gravitational potential and force at the mid-point of the line connecting the centres of the two spheres is = -2.668 x 10^{-7} J /kg

The net force on an object placed at the mid-point is zero. However, if the object is displaced even a little towards any of the two bodies, it will not return to its equilibrium position. Thus, the body is in an unstable equilibrium.

**Q.22: As you have learnt in the text, a geostationary satellite orbits the earth at the height of nearly 36,000 km from the surface of the earth. What is the potential due to the earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero.) Mass of the earth = 6.0×10 ^{24} kg, radius = 6400 km**

**Solution:**

Given:

Radius of the Earth, R = 6400 km = 0.64 × 10^{7 }m

Mass of Earth, M = 6 x 10^{24 }kg

Height of the geostationary satellite from the earth’s surface, h = 36000 km = 3.6 x 10^{7} m

Therefore, gravitational potential at height ‘h’ on the geostationary satellite due to the earth’s gravity

G_{P }=

_{P}=

_{P }=

^{6}J/Kg

Therefore, the gravitational potential due to Earth’s gravity on a geostationary satellite orbiting earth is -9.439 × 10^{6} J/Kg

**Q.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. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category.) Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = 2×10 ^{30} kg) **

**Solution:**

Any matter/object will remain stuck to the surface, if the outward centrifugal force is lesser than the inward gravitational pull.

Gravitational force, f_{G} =

Here,

M = Mass of the star = 2.5× 2 × 10^{30} = 5 × 10^{31 }kg

m = Mass of the object

R = Radius of the star = 12 km = 1.2 ×10^{4} m

Therefore, f_{G} =

Now, Centrifugal force, f_{c}= m r ω^{2}

Here, ω = Angular speed = 2πν

ν = Angular frequency = 10 rev s^{–1}

f_{c} = m R (2πν)^{2}

f_{c} = m × (1.2 x 10^{4}) × 4 × (3.14)^{2} × (1.2)^{2} = (6.82 ×10^{5}m) N

As f_{G} > f_{C}, the object will remain stuck to the surface of the black hole.

**Q. 24. A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? The mass of the spaceship = 1000 kg; the mass of the sun = 2×10 ^{30} kg; the mass of Mars = 6.4×10^{23} kg; the radius of Mars = 3395 km; the radius of the orbit of Mars = 2.28 ×10^{8 }km; G = 6.67×10^{-11} N m^{2}kg^{–2}**

**.**

**Solution:**

The mass of the spaceship, m_{s} =1000 kg

the mass of the Sun, M=2×10^{30}kg

the mass of Mars, m_{m} =6.4×10^{23}kg

The radius of the orbit of Mars, R=2.28×10^{11 }m

The radius of Mars, r=3395×10^{3}m

Universal gravitational constant, G=6.67×10^{−11}Nm^{2}kg^{−2}

The potential energy of the spaceship due to the gravitational attraction of the Sun, Us =−GMm_{s}/R

The potential energy of the spaceship due to the gravitational attraction of Mars, U_{m}=−Gm_{m}m_{s}/r

Total energy of the spaceship, E= U_{m}+U_{s} =[−GMm_{s}/R]+[-Gm_{m}m_{s}/r]

The negative sign indicates that the satellite is bound to the system.

The energy required to launch the spaceship out of the solar system =−(total energy of the spaceship)

– E = [GMm_{s}/R]+[Gm_{m}m_{s}/r]

= Gm_{s}(M/R + m_{m}/r)

=

= 5. 91 x 10^{11} J

**Q.25. A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 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? The mass of mars = 6.4×10**.

^{23}kg; the radius of mars = 3395 km; G = 6.67×10^{-11}N m^{2}kg^{–2}**Solution:**

Speed of the rocket fired from the surface of mars ( v) = 2 km/s

The mass of the rocket = m

The mass of Mars ( M) = 6.4 × 10²³ Kg

The radius of Mars (R) = 3395 km = 3.395 × 10^{6} m

The initial Potential energy = – GMm/R

The initial kinetic energy of the rocket = 1/2 mv²

Total initial energy = 1/2 mv² – GMm/R

Due to atmospheric resistance, 20% of the kinetic energy is lost by the rocket.

The remaining Kinetic energy= 80% of 1/2mv²

= 2/5 mv²= 0.4 mv² ——–(1)

When the rocket reaches the highest point, at the height h above the surface, the kinetic energy will be zero, and the potential energy is equal to – GMm/(R+h)

Applying the Law of conservation of energy,

– GMm/(R+h) = – GMm/R + (0.4) mv²

GM/(R+h) = 1/R [ GM – 0.4Rv^{2}]
(R+h/R) = GM/[ GM – 0.4Rv^{2}]

h/R = {GM/[ GM – 0.4Rv^{2}]} – 1

h/R = (0.4Rv^{2}/GM – 0.4Rv^{2})

h = (0.4R^{2}v^{2}/GM – 0.4Rv^{2})

h = 495 km

Also Access |

NCERT Exemplar for Class 11 Physics Chapter 8 |

CBSE Notes for Class 11 Physics Chapter 8 |

NCERT Solutions help students prepare for their final examination with worksheets, exemplar problems, MCQs (multiple choice questions), HOTS (high order thinking skills), short answer questions, and NCERT tips and tricks. Solving these solutions repeatedly will help them understand the types of questions asked in Class 11 examinations and entrance exams for the under-graduation course.

## Class 11 Physics NCERT Solutions for Chapter 8 Gravitation

Gravitation is a very popular subject for Class 11 students, as most of the topics you study in the future are based on this phenomenon. We should know the difference between gravitation and gravity in order to understand more complex subjects. This chapter consists of questions belonging to various important concepts, such as how can a body be shielded from gravitational influences of any nearby matter whose gravitational force is greater on earth, the sun or the moon.

We find various complex but are easy to understand topics in this chapter, such as acceleration due to gravity and finding potential energy differences between two points which are at a certain distance from the earth’s centre. We can even know that the acceleration due to gravity increases/decreases when the depth increases or altitude decreases. We will be finding questions on interplanetary motion, such as a planet revolving around the sun with a speed double that of the earth; this question will show you the use of Kepler’s laws of motion. We will see questions on Jupiter’s satellite radius of revolutions.

This chapter talks about the distance and speed of light among stars which are far away in a different galaxy and the time taken for the stars to complete one revolution. Here, we will also see questions related to a space shuttle’s escape velocity and its dependency on factors like an object’s mass, location, the direction of projection and the gravitational influences on it. We will be finding the angular speed, kinetic energy, total energy, potential energy, linear speed and angular momentum of an asteroid revolving around a star.

We will be going through the medical problems faced by astronauts in space and the reason behind them. We will see problems with gravitational intensity within a hemisphere. We will know the mass of the sun in one of the questions asked below. Do students know the distance between Uranus and earth and the time taken to reach there? They will get to know it by studying this solution.

We will see here the maximum height attained by a missile when it is fired vertically upwards before falling down on the earth’s surface. Further, we will see the international space station’s motion around the earth and the influence of the earth’s gravitational field upon it. Do students know what will happen when two planets collide with each other, what will be their speed? We will see it here. How a star turns into a black hole and the energy required to launch a space station on Mars are the type of questions they will find below.

In order to help students to cover this topic in a detailed and comprehensive way by solving problems, we have provided Physics Chapter 8 NCERT Solutions for Class 11 in PDF format.

### Important Concepts of NCERT Solutions for Class 11 Physics Chapter 8

Chapter 8 provides students with all the required knowledge about Gravitation and related important concepts. **NCERT Solutions** provides a clear understanding of the fundamental concepts. Some of the important concepts explained in Chapter 8 are

8.1 Introduction

8.2 Kepler’s laws

8.3 Universal law of gravitation

8.4 The gravitational constant

8.5 Acceleration due to the gravity of the earth

8.6 Acceleration due to gravity below and above the surface of the earth

8.7 Gravitational potential energy

8.8 Escape speed

8.9 Earth satellites

8.10 Energy of an orbiting satellite

8.11 Geostationary and polar satellites

8.12 Weightlessness

**Disclaimer – **

**Dropped Topics – **

8.11 Geostationary and Polar Satellites

8.12 Weightlessness

Exercises 8.3–8.5, 8.22–8.25

Appendix 8.1

## Comments