NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

NCERT solutions Class 11 Physics Chapter 8 Gravitation is a vital resource you must refer to score good marks in the Class 11 examination. This solution is the result of referring to a number of textbooks by experts. These solutions are prepared as per the latest CBSE syllabus 2020-21. They present you the answers to the questions in the textbooks, important questions from previous year question papers and sample papers.

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From the above figure, ‘A’ is the mid-point and 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= [GMr2+GMr2]left [ frac{-GM}{frac{r}{2}}+frac{-GM}{frac{r}{2}} ight ]

U= [4GMr]left [ frac{-4GM}{r} ight ]

U= [4×(6.67×1011)×(1000)1.0]left [ frac{-4 imes (6.67 imes 10^{-11}) imes (1000)}{1.0} ight ] Rightarrow U= -2.668 x 10-7 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 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 = 6.0×1024 kg, radius = 6400 km



Radius of the Earth, R = 6400 km = 0.64 × 107 m

Mass of Earth, M = 6 x 1024 kg

Height of the geostationary satellite from earth’s surface, h = 36000 km = 3.6 x 107 m

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

GP = GMR+hfrac{-GM}{R+h}
Rightarrow GP = (6.67×1011)×(6×1024)(0.64×107)+(3.6×107)frac{-(6.67 imes 10^{-11}) imes (6 imes 10^{24})}{(0.64 imes 10^{7})+(3.6 imes 10^{7})}
Rightarrow GP = 40.02×10134.24×107frac{-40.02 imes 10^{13}}{4.24 imes 10^{7}}= -9.439 × 106 J/Kg

Therefore, the gravitational potential due to Earth’s gravity on a geostationary satellite orbiting earth is -9.439 × 106 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 = 2×1030 kg).   


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

Gravitational force, fG = GM  mR2frac{GM;m}{R^{2}} [Neglecting negative sign]

M = Mass of the star = 2.5× 2 × 1030 = 5 × 1031 kg
m = Mass of the object
R = Radius of the star = 12 km = 1.2 ×104 m
Therefore, fG = (6.67×1011)×(5×1031)  m(1.2×104)2=1.334×1013m  Nfrac{(6.67 imes 10^{-11}) imes (5 imes 10^{31});m}{(1.2 imes 10^{4})^{2}}=1.334 imes 10^{13}m;N

Now, Centrifugal force, fc= m r ω2
Here, ω = Angular speed = 2πν
ν = Angular frequency = 10 rev s–1
fc = m R (2πν)2

fc = m × (1.2 x 104) × 4 × (3.14)2 × (1.2)2 = (6.82 ×105m) N

As fG > fC, the object will remain stuck to the surface of 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? Mass of the space ship = 1000 kg; mass of the sun = 2×1030 kg; mass of mars = 6.4×1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 ×10km; G = 6.67×10-11 N m2kg–2.


Mass of the spaceship, ms =1000 kg
Mass of the Sun, M=2×1030kg
Mass of Mars, mm =6.4×1023kg
Radius of orbit of Mars, R=2.28×1011 m
Radius of Mars, r=3395×103m
Universal gravitational constant, G=6.67×10−11Nm2kg−2

The potential energy of the spaceship due to the gravitational attraction of the Sun, Us =−GMms/R
Potential energy of the spaceship due to the gravitational attraction of Mars, Um=−Gmmms/r

Total energy of the spaceship, E= Um+Us =[−GMms/R]+[-Gmmms/r]

The negative sign indicates that the satellite is bound to the system.
Energy required to launch the spaceship out of the solar system =−(total energy of the spaceship)

– E = [GMms/R]+[Gmmms/r]

= Gms(M/R + mm/r)

=  6.67×1011×103×(2×10302.28×1011+6.4×10233.395×106)6.67 imes 10^{-11} imes 10^{3} imes left ( frac{2 imes 10^{30}}{2.28 imes 10^{11}}+frac{6.4 imes 10^{23}}{3.395 imes 10^{6}} ight )

= 5. 91 x 1011 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? Mass of mars = 6.4×1023 kg; radius of mars = 3395 km; G = 6.67×10-11 N m2 kg–2.


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

Mass of the rocket = m
Mass of the Mars ( M) = 6.4 × 10²³ Kg
Radius of the Mars (R) = 3395 km = 3.395 × 106 m

Initial Potential energy = – GMm/R
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 Law of conservation of energy.
– GMm/(R+h) =  – GMm/R + (0.4) mv²

GM/(R+h) =  1/R [ GM – 0.4Rv2]
(R+h/R) = GM/[ GM – 0.4Rv2]

h/R = {GM/[ GM – 0.4Rv2]} – 1

h/R = (0.4Rv2/GM – 0.4Rv2)

h = (0.4R2v2/GM – 0.4Rv2)

h=0.4×(2×103)×(3.395×106)26.67×1011×6.4×10230.4×(2×103)2×3.395×106h = frac{0.4 imes (2 imes 10^{3}) imes (3.395 imes 10^{6})^{2}}{6.67 imes 10^{-11} imes 6.4 imes 10^{23}-0.4 imes (2 imes 10^{3})^{2} imes 3.395 imes 10^{6}}

h = 495 km

NCERT solutions help your preparation with worksheets, exemplar problems, MCQs (multiple choice questions), HOTS (high order thinking skills), short answer questions, NCERT tips and tricks. Solving this solution repeatedly will help you in understanding 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 be knowing 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’s or the moon’s.

We find various complex but easy to understand topics in this chapter such as acceleration due to gravity, finding potential energy difference 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 an interplanetary motion such as a planet revolving around the sun with a speed double than that of earth, this question will show you the use of Kepler’s laws of motion. We will be seeing 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 shall see questions related to a space shuttle escape velocity and its dependency on factors like an object’s mass, location, 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 it. We will be seeing problems with gravitational intensity within a hemisphere. We will be knowing the mass of the sun in one of the questions asked below. Do you know the distance between Uranus and earth and the time taken to reach there? You 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. We will see the international space station’s motion around the earth and the influence of the earth’s gravitational field upon it. Do you 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 in mars are the type of questions you will find below.

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

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Explain gravitational force covered in the Chapter 8 of NCERT Solutions for Class 11 Physics.

According to Newton’s universal law of gravitation, the force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. Students who wish to understand the concepts covered in this chapter are advised to download the PDF of solutions from CoolGyan’S. The solutions are created by a set of highly experienced subject matter experts with the aim of helping students to score well in the board exams.

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