# NCERT Solutions for Class 11 Physics Chapter 8

Class 11 is an important class for students as it prepares you for Class 12 and board exams. The students must be very regular in their studies in Class 11 as it forms the base for a lot of concepts and is also completely new. Extramarks provides NCERT Solutions for Physics Class 11 to help students understand the subject and concepts better.

## NCERT Solutions for Class 11 Physics Chapter 8 – Gravitation

Chapter 11 is based on Gravitation. It’s critical to understand that gravity and gravitation are not similar. Gravitation is the force acting between two bodies and gravity is a force that pulls a body to the earth’s centre. The chapter also tells us about escape speed and earth satellites. Students learn about the energy of an orbiting satellite and also geostationary and polar satellites. The chapter concludes by explaining the idea of weightlessness. The NCERT solution for class 11 Physics Chapter 8 gives a brief explanation of three vital laws proposed by Kepler and the universal law of gravitation. It comprises answers to all the questions that are listed in the practice section of Chapter 8 Gravitation.

Chapter 8 also includes several discussions on questions like the strength of the gravitational force existing on the earth and moon, Kepler’s law, escape velocity, etc. All topics preceding Ch 8 Physics class 11 NCERT solutions are based on the concepts you’re going to learn from this chapter. The first section defines the Gravitational force and teaches you the distinction between gravitation and gravity to enable you to understand the more complex subjects in the future. The most important laws of Gravitation are Kepler’s Laws. You will learn these laws in the second section which states the Universal Law of Gravitation, Gravitational Constant, what causes Acceleration, Acceleration due to Gravity of Earth, and how Gravitation works both above and below the surface of the earth.

NCERT Class 11 Physics Chapter 8 ignites your imagination with detailed information about various types of Satellites like Geostationary and Polar Satellites.

### Access NCERT Solutions for Class 11 Physics Chapter 8 – Gravitation

NCERT Solutions for Class 11 Physics Chapter 8 include all the important topics with a detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Chapter 8. NCERT Solutions for Class 11 Physics Chapter 8 Gravitation are prepared by expert teachers who have years of experience in teaching students, clearing their doubts and making concepts clear and understandable.

The chapter begins by elaborating on Kepler’s laws before moving on to the universal law of gravitation. We will also read about the gravitational constant and its equations. And about increasing speed due to gravity on earth and due to gravity below and above the earth’s surface.

The chapter also tells us about escape speed and earth satellites. We will study the energy of an orbiting satellite and also geostationary and polar satellites. The chapter concludes by explaining the idea of weightlessness.

The chapter also tells us about the Earth’s satellites and how they are related to Kepler’s laws. We are also taught to calculate the energy of these orbiting satellites. It also gives a brief explanation of geostationary and polar satellites. The chapter concludes by telling us about the phenomenon of weightlessness. This principle tells us how an object in free fall is weightless or in zero gravity. You will learn how far the vertically fired satellites go before getting attracted by the Law of Gravitation. Get to know about the movements of the international space station around the earth and how the Earth’s gravitational force influences it.

Points to remember.

1) Planets move around the Sun in elliptical orbits.

2) The speed of movement of planets around the Sun in elliptical orbits

3) The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Physics is also an interesting subject but it might also be quite difficult to master it. One of the most analytical subjects in science is Physics as it deals with the different properties of matter and nature. The Physics NCERT Solutions Class 11 aims at making the subject and the concepts easier for each student as all students do not have the same capacity. The main motive of the solutions is to make understanding the subject easier for the students and make it seem less doubting for the students, keeping up with the syllabus and pressure of having good marks so some students are not able to do both. Having a solutions book also makes it easier for the students to go through before the exam as all the syllabus is combined in one place for the students to quickly go through.

### Benefits Of NCERT Solution For Class 11 Physics Chapter 8

NCERT Solutions for Class 11 Physics Chapter 8 include interesting topics. This chapter will reveal the reasons behind the universal law of Gravitation. It is fascinating to know the various facts, and aspects about the force of gravitation on the earth apart from taking a glance at the functioning of satellites, space technology, etc. The topic of Gravitation is a complex one. With most of the topics being new, you find yourself unable to understand and remember the core level of the points of the chapter. This is where our NCERT solution for class 11 Physics Chapter 8 can help you. The complex concepts can break down into simple terms for your easy understanding of concepts. With the help of our NCERT Solutions, you will have a clear-cut understanding of the formula, concepts, and laws related to gravitation. So, without a doubt start reading our NCERT solutions.

NCERT Solutions for Class 11 Physics Chapter 8 will also give a detailed explanation of the universal Law of Gravitation and vital laws proposed by Kepler. It provides you with detailed solutions to questions based on escape velocity, gains in momentum due to gravity, and more. Acceleration due to gravity increases with increased altitude. The first exercise includes theory questions, multiple-choice questions, and numerical-type questions on the topics of gravity, tidal effect of sun and moon, orbital size, gravitational intensity, comet orbits, speed of the projectile, equilibrium, etc. The second exercise includes questions on topics like geostationary satellites and related space technology topics.

Related Questions

The chapter is based on Gravitation. It’s critical to understand that gravity and gravitation are not similar.

What is Gravity?

Gravitation is the force acting between two bodies and gravity is a force that pulls a body to the earth’s centre. The NCERT solution for class 11 Physics Chapter 8 gives a brief explanation of three vital laws proposed by Kepler and the universal Law of Gravitation. The chapter begins by elaborating on Kepler’s laws before moving to the universal Law of Gravitation.

Which are the three laws explained in the NCERT solutions Chapter 8 of Physics Class 11 ?

The chapter begins by elaborating on Kepler’s laws before moving on to the universal law of gravitation. We will also read about the gravitational constant and its equations. And about increasing speed due to gravity on earth and due to gravity below and above the earth’s surface.

What does the second section of NCERT solutions Chapter 8 Physics Class 11 state?

The second section states the Universal Law of Gravitation, Gravitational Constant, what causes Acceleration, Acceleration due to Gravity of Earth, and how Gravitation works both above and below the surface of the earth.

(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. (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 sun. Why?

Ans.

(a) Gravitational influence of matter on nearby objects cannot be shielded by any means. This is due to the fact that gravitational force is independent of the nature of the material medium, however, it is not so in the case of the electrical forces. It is also independent of the presence of other objects.

(b)Yes, if the size of the space ship orbiting around the Earth is very large, then the astronaut can identify the change in Earth’s gravity (g).

(c) Tidal effect is inversely proportional to the cube of the distance. However, the gravitational force is inversely proportional to the square of the distance. The distance between the Moon and the Earth is very small in comparison to the distance between the Sun and the Earth. Therefore, the tidal effect of the Moon’s pull is more than the tidal effect of the Sun’s pull.

Q.2 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?

Ans.

$Since time taken by the Earth to revolve once around the Sun, T E = 1 year Orbital radius of Earth, R E = 1 AU Time taken by the given planet to revolve once around the Sun, T P = 1 2 T E = 0.5 year Let orbital radius of the given planet = R P According to Kepler’s third law of planetary motion, we have: R P = R E T P T E 2/3 R P = 1 0.5 year 1 year 2/3 = 0.63 AU ∴The orbital radius of the planet will be 0.63 times smaller than the orbital radius of Earth.$

Q.3 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Ans.

$Here,orbital period of a satellite of Jupiter, T 1 = 1.769 days = 1.769×24×60×60 s Orbital radius of satellite, r 1 = 4 .22×10 8 m G = Universal gravitational constant Orbital radius of the Earth, r = 1 A.U. = 1 .496×10 11 m Mass of the Jupiter is given by, M J = 4π 2 r 1 3 GT 1 2 M J = 4π 2 × 4 .22×10 8 m 3 G× 1.769×24×60×60 s 2 …(i) Now, orbital period of Earth around the Sun, T = 1 year = 365.25×24×60×60 s Mass of Sun is given by, M S = 4π 2 r 3 GT 2 M S = 4π 2 × 1 .496×10 11 m 3 G× 365.25×24×60×60 s 2 …(ii) Dividing equation (ii) by (i), we get M S M J = 4π 2 × 1 .496×10 11 m 3 G× 365.25×24×60×60 s 2 × G× 1.769×24×60×60 s 2 4π 2 × 4 .22×10 8 m 3 = 1045.04 ∴ M S M J ≈1000 ∴ M J ≈ 1 1000 M S ∴It can be concluded that the mass of Jupiter is about one-thousandth that of the Sun.$

Q.4 Let us assume that our galaxy consists of 2.5 × 1011 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 105 ly.

Ans.

$Mass of our galaxy, M = 2 .5×10 11 solar mass Mass of Sun = 2 .0 × 10 36 kg ∴Mass of our galaxy, M = 2 .5×10 11 ×2×10 36 kg = 5×10 41 kg Diameter of our galaxy, d = 10 5 ly Radius of our galaxy, r = 5×10 4 ly 1 ly = 9 .46×10 15 m ∴ r = 5×10 4 ×9 .46×10 15 m = 4 .73 ×10 20 m As a star revolves around the galactic centre of our galaxy, its time period is, T = 4π 2 r 3 GT 2 T = 4π 2 r 3 GM 1/2 = 4× 3.14 2 × 4.73 3 ×10 60 m 3 6 .67×10 -11 Nm 2 kg -2 ×5×10 41 kg = 1 .12×10 16 s$

Q.5 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.

Ans.

(a) The total mechanical energy of a satellite is equal to the sum of its kinetic energy (always positive) and potential energy (maybe negative). The gravitational P.E. of the satellite is zero at infinity. The total energy of the satellite is negative as the Earth-satellite system is a bound system. Therefore, at infinity, the total energy of an orbiting satellite is equal to the negative of its kinetic energy.

(b) An orbiting satellite attains a definite amount of energy that enables it to revolve around the Earth. This energy is provided by the orbit of the satellite. It needs relatively lesser energy to move out of the effect of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

Q.6 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?

Ans.

$\begin{array}{l}\text{The}\text{\hspace{0.17em}}\text{escape velocity of a body from the Earth is given}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{as:}\text{\hspace{0.17em}}{\text{v}}_{\text{esc}}=\sqrt{2gR}\end{array}$

Here, g is the acceleration due to gravity and R is the radius of the Earth.

From the given relation, we observe that the escape velocity vesc is independent of the mass of the body and the direction of its projection. But, it depends on the location from where the body is projected. As the gravitational potential depends slightly on the height, therefore, escape velocity depends slightly on the height of the location from where the body is launched.

Q.7 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.

Ans.

For the given situation, the comet has constant values of angular momentum and total energy at all points of the orbit. But, its linear speed, angular speed, kinetic, and potential energy change with locations in the orbit.

Q.8 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet (b) swollen face, (c) headache, (d) orientational problem?

Ans.

The symptoms (b), (c), and (d) can afflict an astronaut in space.

(a) Legs carry the weight of the body in a standing position because of gravitational pull. The astronaut in space is in weightlessness state because of the absence of gravity.

Thus, swollen feet do not affect an astronaut in space.

(b) A swollen face is caused because of apparent weightlessness in space. Since the sense organs such as eyes, ears nose, and mouth constitute a person’s face, hence, the swollen face can affect to great extent the seeing, hearing, smelling and eating capabilities of an astronaut in space.

(c) Headache is caused due to mental strain. The headache will have the same effect on the astronaut in space as on a person on earth, therefore, it can affect the working of an astronaut in space.

(d) Space has several orientations and frames of reference in space. Hence, the orientation problems can seriously affect an astronaut in space.

Q.9 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 8.12) (i) a, (ii) b, (iii) c, (iv) 0.

Ans.

Since the gravitational potential (V) is constant everywhere inside a spherical shell, therefore, the gradient of gravitational potential

$\left(\frac{dV}{dr}\right)$

is zero everywhere inside the spherical shell. Since the gravitational intensity is the negative of the gradient of gravitational potential, therefore, the gravitational intensity is also zero everywhere inside the spherical shell. This shows that gravitational forces acting at a point inside a spherical shell are symmetric.

If we remove the upper half of a spherical shell, then the net gravitational force acting on a particle located at centre C will act downwards.

Since the gravitational intensity at a point is the gravitational force per unit mass at that point, it will also act downwards. Therefore, the gravitational intensity at the centre C will be along arrow c. Hence, the option (iii) is correct.

Q.10 Choose the correct answer from among the given ones:
For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Ans.

Since the gravitational potential (V) is constant everywhere inside a spherical shell, therefore, the gradient of gravitational potential

$\left(\frac{dV}{dr}\right)$

is zero everywhere inside the spherical shell. Since the gravitational intensity is equal to the negative of the gradient of gravitational potential, therefore, the gravitational intensity is also zero everywhere inside the spherical shell. This shows that gravitational forces acting at a point inside a spherical shell are symmetric.

If we remove the upper half of a spherical shell, then the net gravitational force acting on a particle located at centre C will act downwards.

Since the gravitational intensity at a point is the gravitational force per unit mass at that point, it will also act downwards. Therefore, the gravitational intensity at the centre P will be along arrow e.

Hence, (ii) is correct.

Q.11 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 ×1030 kg, mass of the earth = 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

Ans.

$Here, mass of sun, M S = 2×10 30 kg Mass of earth, M E = 6×10 24 kg Orbital radius of Earth, r = 1 .5×10 11 m Let mass of the rocket be m If x is the distance from the centre of the Earth where the gravitational force acting on satellite Q becomes equal to zero. Using Newton’s law of gravitation, we can equate gravitational forces acting on satellite Q under the effect of the Sun and the Earth as: GmM S r-x 2 = G mM E x 2 r-x x 2 = M S M E ∴ r-x x = 2 × 10 30 kg 6 × 10 24 kg 1 2 = 577.35 1 .5×10 11 m – x = 577.35x 578.35x = 1 .5×10 11 m ∴x = 2 .59×10 8 m$

Q.12 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km.

Ans.

$Here, orbital radius of the Earth, r = 1 .5×10 11 m Period of revolution of Earth around the Sun, T = 1 year = 365.25 days = 365.25×24×60×60 s Universal gravitational constant, G = 6 .67×10 –11 Nm 2 kg –2 Let mass of Sun be M and mass of Earth be m Gravitational force acting on Earth due to Sun, F g = GMm r 2 Let angular velocity of Earth around Sun be ω Centripetal force acting on Earth, F c = mrω 2 = mr 2π T 2 = mr 4π 2 T 2 Since gravitational pull of Sun on Earth provides the necessary centripetal force, ∴ GMm r 2 = mr 4π 2 T 2 ∴M = 4π 2 r 3 GT 2 M = 4× 3.14 2 × 1 .5×10 11 m 3 6 .67×10 –11 Nm 2 kg –2 × 365.25×24×60×60 s 2 = 2 .0×10 30 kg ∴ Mass of the Sun = 2×10 30 kg$

Q.13 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50×108 km away from the sun?

Ans.

$Here, seperation of the earth from the sun, r E = 1 .5×10 8 km = 1 .5×10 11 m Let time period of the earth be T E Time period of saturn, T s = 29 .5 T E Let distance of saturn from the s un be r S Using Kepler’s third law of planetary motion, we have T = 4π 2 r 3 GM 1/2 In case of saturn and sun, we have: r S 3 r E 3 = T S 2 T E 2 ∴ r S = r E T S T E 2/3 r S = 1 .5×10 11 m× 29 .5 T E T E 2/3 r S = 1 .5×10 11 m× 29.5 2/3 = 14 .32×10 11 m ∴Distance between saturn and sun = 14 .32×10 11 m$

Q.14 n 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?

Ans.

$Here, weight of body, W = 63 N Let radius of Earth be R E Acceleration due to gravity at height h from the Earth’s surface is given as: g’ = g 1+h R E 2 …(i) Here, g = Acceleration due to gravity on the surface of Earth Here, height, h = R E 2 …(ii) Substituting the value of h from equation (ii) in equation (i), we obtain: g’ = g R E +h R E 2 = gR E 2 R E + R E 2 2 g’ = g 1+ 1 2 2 = 4 9 g Weight of a body of mass m at height h, W’ = mg’ W’ = m× 4 9 g = 4 9 W = 4 9 ×63 N = 28 N$

Q.15 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?

Ans.

$Let mass of the body be m Given, weight of the body at the surface of Earth, W = mg = 250 N Here, R E = Radius of the Earth Acceleration due to gravity at depth d is given as: g’ = 1 – d R E g The body is located at depth, d = 1 2 R E ∴g’ = 1 – R E 2×R E g = 1 2 g Weight of the body at depth d is given as: W’ = mg’ = 1 2 mg = 1 2 W W’ = 1 2 ×250 N = 125 N$

Q.16 A rocket is fired vertically with a speed of 5 kms–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 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G= 6.67 × 10–11 Nm2kg–2.

Ans.

$Here, velocity of the rocket, v = 5 kms -1 = 5×10 3 ms -1 Mass of Earth, M E = 6 .0×10 24 kg Radius of Earth, R E = 6 .4×10 6 m Height attained by rocket mass, m be h At the Earth’surface, Total energy of rocket, E T = Kinetic energy + Potential energy E T = 1 2 mv 2 + -GM E m R E At highest point h, velocity of rocket, v = 0 At highest point h, Potential energy of rocket = – GM E m R E +h Total energy of rocket = 0 + -GM E m R E +h = – GM E m R E +h According to the law of conservation of energy, we have Total energy of rocket at the Earth’s surface = Total energy at height h 1 2 mv 2 + -GM E m R E = – GM E m R E +h 1 2 v 2 = GM E 1 R E – 1 R E +h 1 2 v 2 = gR E h R E +h Here, g = GM R E 2 =9.8 ms -2 ∴ v 2 R E + h = 2gR E h h = R E v 2 2gR E -v 2 h = 6 .4×10 6 m× 5 × 10 3 ms -1 2 2×9 .8 ms -2 ×6 .4×10 6 m – 5 × 10 3 ms -1 2 = 1 .6×10 6 m Height attained by the rocket w.r.t. the centre of the Earth, H = R E +h H = 6 .4×10 6 m +1 .6×10 6 m = 8 .0×10 6 m$

Q.17 The escape speed of a projectile on the earth’s surface is 11.2 kms–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.

Ans.

$Here, escape speed of the projectile from Earth, v esc = 11.2 kms -1 Velocity of projection of the projectile, v P = 3v esc Let mass of the projectile be m Let velocity of the projectile far away from the Earth be v f ∴Total energy of the projectile on the Earth = 1 2 mv P 2 – 1 2 mv esc 2 Gravitational potential energy of the projectile far away from the Earth = 0 ∴Total energy of the projectile far away from the Earth = 1 2 mv f 2 According to the law of conservation of energy, we have 1 2 mv P 2 – 1 2 mv esc 2 = 1 2 mv f 2 v f = v P 2 – v esc 2 v f = 3v esc 2 – v esc 2 = 8 v esc = 31.68 kms -1$

Q.18 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 = 6.0×1024 kg; radius of the earth = 6.4×106 m; G = 6.67×10–11 Nm2kg–2.

Ans.

$Given, height of satellite, h = 400 km = 4×10 5 m = 0 .4×10 6 m Mass of Earth, M = 6 .0×10 24 kg Mass of satellite, m = 200 kg Radius of Earth, R = 6 .4×10 6 m Universal gravitational constant, G = 6 .67×10 –11 Nm 2 kg –2 Total energy of the satellite at height h, E T = PE + KE E T = -GMm R+h + 1 2 mv 2 Since orbital velocity of the satellite, v = GM R+h ∴Total energy of the orbiting satellite at height, h = 1 2 m GM R+h – GMm R+h = – 1 2 GMm R+h This is also called as the bound energy of the satellite. Energy required to move the satellite out of its orbit = 1 2 GM e m R e +h = 1 2 × 6 .67 × 10 –11 Nm 2 kg –2 ×6 .0 × 10 24 kg×200 kg 6 .4 × 10 6 m+0 .4 ×10 6 m = 5 .9× 10 9 J$

Q.19 Two stars each of one solar mass (= 2×1030 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).

Ans.

$Here, mass of each star, M = 2 × 10 30 kg Radius of each star, R = 10 4 km = 10 7 m Initial distance between the stars, r = 10 9 km = 10 12 m For negligible speeds, v = 0 ∴Total initial energy of two stars = -GMM r + 1 2 mv 2 = -GMM r + 0 …(i) In case the stars are about to collide: Let velocity of the stars be v Separation between the centers of the stars = 2R Final kinetic energy of both stars = 1 2 Mv 2 + 1 2 Mv 2 = Mv 2 Final potential energy of both stars = -GMM 2R Total final energy of the two stars = Mv 2 – GMM 2R …(ii) According to the law of conservation of energy, we have Mv 2 – GMM 2R = -GMM r v 2 = -GM r + GM 2R v 2 = GM – 1 r + 1 2R v 2 = 6 .67×10 -11 Nm 2 kg -2 ×2×10 30 kg× – 1 10 12 m + 1 2×10 7 m ≈ 6 .67×10 12 m 2 s -2 v = 6 .67×10 12 m 2 s -2 = 2 .58×10 6 ms -1$

Q.20 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 centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Ans.

$The situation is shown in the given figure. Here, mass of each sphere, M = 100 kg Distance between the spheres, r = 1 m Here, we assumed S to be the midpoint between the spheres. Since gravitational force applied by each sphere at the midpoint will be equal and opposite, ∴Gravitational force at midpoint S = 0 Gravitational potential at midpoint S, V G = -GM r 2 – GM r 2 = -4 GM r V G = 4×6 .67×10 -11 Nm 2 kg -2 ×100 kg 1 m = -2 .67×10 -8 Jkg -1 Since effective force acting on an object placed at the mid-point is zero, therefore, any object placed at point S will be in equilibrium. However, the equilibrium is unstable, because, any variation in the position of the object will change the effective force in that direction.$

Q.21 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.

Ans.

$Here, mass of the Earth, M = 6 .0×10 24 kg Radius of the Earth, R = 6400 km = 6 .4×10 6 m Height of the geostationary satellite from the surface of Earth, h = 36000 km = 3 .6×10 7 m Universal Gravitational constant, G = 6 .67×10 -11 Nm 2 kg -2 Gravitational potential energy at height h due to Earth’s gravity is given as: V = -GM R+h V = -6 .67×10 -11 Nm 2 kg -2 ×6 .0 × 10 24 kg 3 .6×10 7 m + 0 .64×10 7 m = -9 .4× 10 6 Jkg -1$

Q.22 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)

Ans.

$The body will stuck to the surface of a star if the inward gravitational force is more than the outward centrifugal force caused by the rotation of the star. Here, radius of the star, R = 12 km = 1 .2 ×10 4 m Mass of the star, M = 2 .5×2×10 30 = 5×10 30 kg Let mass of the body = m Gravitational force, f g = GMm R 2 Here, G = 6 .67×10 -11 Nm 2 kg -2 ∴ f g = 6 .67×10 -11 Nm 2 kg -2 ×5 × 10 30 kg×m 1 .2 ×10 4 m 2 = 2 .31×10 11 mN Centrifugal force, f c = mrω 2 Angular speed, ω = 2πν f c = mR 2πν 2 Here, angular frequency, ν = 1 .2 revs –1 ∴ f c = m×(1 .2 ×10 4 m)×4× 3.14 2 × 1 .2 revs –1 2 = 1 .7×10 5 mN As f g > f c , the body will remain stuck to the star.$

Q.23

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 × 108 kg; G = 6.67 × 10–11 Nm2kg–2.

Ans.

$Here, mass of spaceship, m S = 1000 kg Mass of Sun, M = 2×10 30 kg Mass of Mars, m M = 6 .4×10 23 kg Radius of the orbit of Mars, R = 2 .28×10 8 km = 2 .28 ×10 11 m Radius of Mars, r = 3395 km = 3 .395×10 6 m Universal gravitational constant, G = 6 .67×10 –11 Nm 2 kg –2 P.E. of the spaceship because of the gravitational attraction of the Sun, (PE) S = -GMm S R Potential energy of the spaceship because of the gravitational attraction of Mars, (PE) M = -Gm M m S r As the spaceship is stationed on Mars, its velocity =0 ∴Kinetic energy of the spaceship = 0 Total energy of the spaceship, E T = (PE) S + (PE) M E T = -GMm S R – -Gm M m S r = -Gm S M R + m M r The negative sign shows that the system is in bound state. Energy required for launching the spaceship from the solar system = – Total energy of the space ship = – Gm S M R + m M r = Gm S M R + m S r = 6 .67×10 –11 Nm 2 kg –2 ×1000 kg× 2 × 10 30 kg 2 .28 ×10 11 m + 6 .4 × 10 23 kg 3 .395 × 10 6 m = 5 .98×10 11 J$

Q.24 A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 kms–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 Nm2kg2.

Ans.

$Here, mass of the Mars, M = 6 .4×10 23 kg Initial velocity of the rocket, v = 2 kms -1 = 2×10 3 ms -1 Radius of the Mars, R = 3395 km = 3 .395×10 6 m Universal gravitational constant, G = 6 .67×10 –11 Nm 2 kg –2 Mass of the rocket be m Initial K.E. of the rocket = 1 2 mv 2 Initial P.E. of the rocket = -GMm R Total initial energy of the rocket = 1 2 mv 2 – GMm R If 20 % of initial kinetic energy is lost because of Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height. ∴Total amount of initial energy available = 80 100 × 1 2 mv 2 – GMm R = 0.4 mv 2 – GMm R Let maximum height achieved by the rocket be h At maximum height, the velocity of the rocket = 0 ∴At maximum height, the K.E. of the rocket = 0 At height h, total energy of the rocket = – GMm R+h According to the law of conservation of energy for the rocket, we have: 0.4 v 2 = GM R – GM R+h R+h h = GM 0.4 v 2 R h = 0.4 R 2 v 2 GM-0.4 v 2 R h = 0.4× 3 .395 × 10 6 m 2 × 2× 10 3 ms -1 2 6 .67×10 –11 Nm 2 kg –2 ×6 .4 × 10 23 kg -0.4× 2× 10 3 ms -1 2 × 3 .395 × 10 6 m = 495 km$

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