Physics

Gravitation - 2 Session

Session Opener How much velocity do you need to impart a stone such that it escapes the gravitational field of the earth?

Session Objectives

Session Objective 1.Energy Conservation in Planetary & Satellite Motion 2.Escape Velocity 3.Orbital Velocity of satellite 4.Variation of g with height & depth 5.Calculation of gravitational field due to some common systems 6.Satellite Motion and weightlessness

Energy Conservation in Planetary & Satellite Motion Gravitational force is a conservative force. Therefore, work done by gravitational force is a independent of the path followed by a body under the force. Thus mechanical energy is conserved. Work done by gravitational force is independent of the path followed

Escape Velocity It is the minimum velocity with which a body must be projected vertically upward in order that it may just escape the gravitational pull of earth. Important Fact : Escape velocity depends on mass of planet and radius of planet and not on mass of the object which has to be given this velocity.

P.E.=0 K.E.=0 From Conservation of Energy K.E i +P.E i =K.E f +P.E. f ReRe Escape Velocity

Orbital Velocity of satellite (V orb ) Orbital velocity is the velocity which is given to an artificial earth’s satellite so that it may start revolving round the earth in a fixed, circular orbit.

r m For motion in an orbit F a =F g Orbital Velocity of Satellite (V orb )

F M R m ReRe h R =R e +h Height ‘h’ above the surface of the earth. Depth ‘d’ below the surface of the earth. g With Depth and Height

r t=0 t=T Hence, [Kepler’s Third Law] Time Period of Satellite (T)

Expressions of gravitational field for different bodies Gravitational field E due to a spherical shell of mass M and radius R at a point distant r from the centre. (a) When r > R(b) When r = R (c) When r < R E = 0 (d) When r = 0 E = 0 r R E

Expressions of gravitational field for different bodies Gravitational field E due to a solid sphere of radius R and mass M at a point distant r from the centre. (a) When r > R(b) When r = R (c) When r < R(d) When r = 0 E = 0 E Rr

Satellite Motion and weightlessness The motion of satellite is based on the gravitational force. The planet applies necessary centripetal force by applying gravitational force. Inside the satellite astronauts feel weightlessness because the centrifugal force there is balanced by gravitational force. Important Facts : Neither the speed nor the time period depend upon the mass, size and shape of the satellite. Speed and time period depend upon radius (r)

Class Test

Class Exercise - 1 A planet revolves in an elliptical orbit around the sun. Then out of the following physical quantities the one which remains constant is (a) velocity(b) kinetic energy (c) total energy(d) potential energy

Solution As there is no loss of energy in the motion of the planet in the orbits so total energy remains constant Hence answer is (c).

Class Exercise - 2 Energy supplied to change the radius of the orbit of a satellite of mass m from r to 2r is (a) (b) (c) zero(d) None of these

Solution Initial energy Final energy Energy supplied E = E f – E i Hence, answer is (a).

Class Exercise - 3 A planet with uniform density and radius r shrinks in size to a new radius of, but same density. What is the ratio of the new and old escape velocities? (a) 2 : 1(b) (c) 1 : 2(d)

Solution So So (V e ) new : (V e ) old = 1 : 2. Hence, answer is (c).

Class Exercise - 4 The escape velocity of a body when projected vertically upward from the surface of earth is 11.2 km/s. If the body is projected in a direction making an angle of 30 o with the vertical, then its escape velocity will be (a) km/s (b) km/s (c) km/s (d) 11.2 km/s

Solution Escape velocity does not depend on angle projection. Hence, answer is (d).

Class Exercise - 5 The escape speed for a projectile in the case of earth is 11.2 km/s. A body is projected from the surface of the earth with a velocity which is equal to twice the escape speed. The velocity of the body, when at infinite distance from the centre of the earth, is (a) 11.2 km/s(b) 22.4 km/s (c) (11.2) km/s(d) (11.2) km/s

Solution Total initial energy is Total final energy or V = = (11.2) Hence, answer (d).

Class Exercise - 6 R is the radius of the earth. At a height h above the surface of the earth the acceleration due to gravity is of its value on the surface of the earth. Then h is (a) h = 100R (b) h = 99R (c) h =10R(d) h = 9R

Solution g at surface is At height h or 10R = R + h h = 9R Hence, answer is (d).

Class Exercise - 7 As we go from equator to poles, the value of g (a) remains unchanged (b) decreases (c) increases (d) decreases and then increases

Solution So g´ increases. Hence, answer is (c).

Class Exercise - 8 If the earth suddenly stops rotating, then acceleration due to gravity would not change at (a) equator (b) poles (c) a latitude 45 o (d) None of these

Solution We know that So g´ = g for any value of. Hence, answer is (b).

Class Exercise - 9 A body of mass m is released from the surface of earth into a tunnel passing through the centreof the earth. What is the kinetic energy of the particle as it passes through the centre? (a) (b) (c) (d)

Solution Potential energy at the surface At the centre of earth it has and K.E. = Hence, answer is (b).

Class Exercise - 10 The gravitational field in a region is given by. What is the work done by an external agent to slowly shift a particle of mass 2 kg from point (0, 0) to point (5 m, 4 m)? (a) –180 J(b) 180 J (c) 20 J(d) –20 J

Solution Work done Hence, answer is (b).

Thank you