Physics

Gravitation - 2

Session

Session OpenerHow 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 VelocityIt 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 radiusof planet and not on mass of the object which has to begiven this velocity.

P.E.=0K.E.=0

From Conservation of EnergyK.Ei+P.Ei=K.Ef+P.E.f

e

e

e

e2

RGM2

V

0R

mGMmV

21

Re

Escape Velocity

2

ee

1K.E. mV2GM mP.E. R

ee

2GMEscape Velocity R

Orbital Velocity of satellite (Vorb)

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

mr

mVF2

a

2g rGMmF

rVa2

For motion in an orbitFa=Fg

Orbital Velocity of Satellite (Vorb)

2e2

e

GM m mVrr

GMV r

eorb

GMOrbital Velocity V r

e2ee

GM 2hg 1 for h RRR

F

MR

m

Re h

R =Re+h

Height ‘h’ above the surface of the earth.

Depth ‘d’ below the surface of the earth.

2ee

GM dg 1 RR

g With Depth and Height

2GMmF

(R h)

h 2

F GMg m (R h)

2 2 2GM g

R (1 h / R) (1 h / r)

r

t=0t=T

Hence,[Kepler’s Third Law]

Time Period of Satellite (T)

orb

Dis tance 2 r 2 rT Velocity V GMr

3rT 2 GM

22 34T rGM

Expressions of gravitational field for different bodiesGravitational 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

2

GMEr

(b) When r = R

2GMER

(c) When r < RE = 0

(d) When r = 0

E = 0

rR

E

Expressions of gravitational field for different bodiesGravitational 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

2

GMEr

2

GMER

(c) When r < R (d) When r = 0

E = 0

3GMrER

E

R r

Satellite Motion and weightlessnessThe motion of satellite is based on the gravitational force. The planet appliesnecessary centripetal force by applyinggravitational 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 - 1A 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

SolutionAs 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 - 2Energy 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

GMm4r

GMm2r

SolutionInitial energy

iGMm 1 GME m

r 2 r

GMm2r

Final energy

fGMm 1 GME m2r 2 2r

GMm

4r

Energy suppliedE = Ef – Ei

GMm4r

Hence, answer is (a).

Class Exercise - 3A 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)

R2

2 : 1

1: 2

Solution

eGMVR

34 RG G 43 RR R 3

So So (Ve) new : (Ve)old = 1 : 2.

eV R

Hence, answer is (c).

Class Exercise - 4The 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 30o with the vertical, then its escape velocity will be(a) km/s (b) km/s

(c) km/s (d) 11.2 km/s

11.23

11.22

311.2

2

Solution

Escape velocity does not depend onangle projection.

Hence, answer is (d).

Class Exercise - 5The escape speed for a projectile inthe case of earth is 11.2 km/s. Abody is projected from the surface ofthe earth with a velocity which isequal to twice the escape speed. Thevelocity of the body, when at infinitedistance 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/s3 2

SolutionTotal initial energy is

2i e

GMm 1(T.E.) m(2v )R 2

GMm GMm2R R

GMm

R

Total final energy2

f1(T.E.) = 0 mv2

21mv2

2e

GMm m VR

or V = = (11.2)e2v 2

Hence, answer (d).

Class Exercise - 6R 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

1100

Solutiong at surface is 2

GMgsR

At height h 2 2GM 1 GMgh

100(R h) R

or 10R = R + h h = 9R

Hence, answer is (d).

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

Solution

2 2g ́= g - R cos

cos decreases

So g´ increases.

Hence, answer is (c).

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

Solution

We know that2 2g ́= g - cos

At poles, 2

So g´ = g for any value of .

Hence, answer is (b).

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

(a) (b)

(c) (d)

3 mgr2

1 mgr2

5 mgr2

1 mgr3

SolutionPotential energy at the surface

GMmP.E.R

At the centre of earth it has 3 GMmP.E.2 R

and K.E. = 21 mv2

2GMm 3 GMm 1Then mvR 2 R 2

21 1 GMm 1mv mgR2 2 R 2

Hence, answer is (b).

Class Exercise - 10The 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

E 10 i j N/kg

Solution

Work doneˆ ˆ ˆ ˆW mE (5i 4 j 0i 0 j)ˆ ˆ ˆ ˆ2 10(i j) (5i 4 j)

20(5 4 ) 180 Joule

Hence, answer is (b).

Thank you