# Chapter 8 – Gravitation Questions and Answers: NCERT Solutions for Class 11 Physics

Class 11 Physics NCERT book solutions for Chapter 8 - Gravitation Questions and Answers.

Education Blogs Chapter 8 – Gravitation Questions and Answers: NCERT Solutions for Class 11...

Class 11 Physics NCERT book solutions for Chapter 8 - Gravitation Questions and Answers.

(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 Sun. Why?

(b) Yes. If the size of the spaceship is extremely large, then the gravitational effect of the spaceship may become measurable. The variation in g can also be detected.

(c) Tidal effect depends inversely on the cube of the distance, unlike force which depends inversely on the square of the distance. Since the distance of moon from the ocean water is very small as compared to the distance of sun from the ocean water on earth. Therefore, the tidal effect of Moon’s pull is greater than the tidal effect of the sun.

(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 – GMm (1/r2-1/r1) is more/less accurate than the formula mg (r2– r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the Earth.

(b) decreases

(c) mass of the body

(d) more

M = 2.5 x 1011solar mass = 2.5 x 1011x (2 x 1030) kg = 5.0 x 1041kg

We know that

(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.

(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.

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

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

(d) The escape speed of a body depends upon the height of the location from where the body is projected, because the escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also.

(b) Angular speed also varies slightly.

(c) Comet has constant angular momentum.

(d) Kinetic energy does not remain constant.

(e) Potential energy varies along the path.

(f) Total energy throughout the orbit remains constant.

(b) In the conditions of weightlessness, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face.

(c) Due to more blood supply to face, the astronaut may get headache.

(d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

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

Let at the point P, the gravitational force on the rocket due to Earth

R = 1.5 x108km = 1.5 x 1011m

Time period, T = 365.25 x 24 x 60 x 60 s

Let the mass of the Sun be M and that of Earth be m.

At distance r from centre of earth, kinetic energy becomes zero

.•. Change in kinetic energy = 1.25 x 107– 0 = 1.25 x 107m J

This energy changes into potential energy.

Initial potential energy at the surface of earth = GMem/’r

Initial potential between two stars, r =109km =1012m.

Initial potential energy of the system = -GMm/r

Total K.E. of the stars = 1/2Mv2+ 1/2Mv2

where v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r’ = 2 R.

:. Final potential energy of two stars = -GMm/2R

Since gain in K.E. is at the cost of loss in P.E

Suppose that the distance of either sphere from the mid-point of the line joining their centre is r. Then r=d/2=0.5 m. The gravitational field at the mid-point due to two spheres will be equal and opposite.

= 6400 + 36000 = 42400 km = 4.24 x 107m

Here M is the mass and R is the radius of the star.

The outward centrifugal force acting on a body of mass m at the equator of the star =mv2/R =mR w2——-(ii)

From equation (i), the acceleration due to the gravity of the star

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