**Questions on Gravitation**

(1) Acceleration due to gravity at a height ‘h’ is given by

**, with usual notations.**

**g’ = GM/(R+h)**^{2}Surface value of acceleration due to gravity,

**g = GM/R**

^{2}If ‘h’ is small compared to the radius ‘R’ of the earth,

**g' = g(1-2h/R)**

(2) Acceleration due to gravity at a depth ‘d’ is given by

**g'' = g (1-d/R)**

Note that this is true for all values of ‘d’.

(3) Gravitational potential energy of a mass ‘m’ at a height ‘h’ is given by

**U= -GMm/(R+h)**

This can be written as

**U = -GMm/r**where ‘r’ is the distance from the centre of the earth.

**v**

Escape velocity from a height ‘h’ =

_{e}= √(2GM/R) = √(2gR)**√[2GM/(R+h)] = √[2g'(R+h]**

(5) Kinetic energy and total energy of a satellite are equal in magnitude. But K.E. is positive where as total energy is negative. The potential energy of a satellite is negative and is equal to twice the total energy.( Note that this is true in all central field motion under inverse square law force, as for example, the energy of the electron in the hydrogen atom.)

In the case of a satellite of mass ‘m’ in an orbit of radius ‘r’:

**Potential energy = -GMm/r**

Kinetic energy = +GMm/2r

Total energy = -GMm/2r

Kinetic energy = +GMm/2r

Total energy = -GMm/2r

**T**

(7) Orbital speed ‘v’ of a satellite in an orbit of radius ‘r’ is independent of its mass and is given by

^{2}α r^{3}**v = √(GM/r) = √(g'r)**where g' is the acceleration due to gravity at the orbit and M is the mass of the earth (or planet).

Let us now discuss the following question which appeared in the Kerala Medical Entrance Test paper of 2002:

**The escape velocity of a body on an imaginary planet which has thrice the radius of the earth and twice the mass of the earth is (where v**

(a) √(2/3).v

We have v

_{e}is the escape velocity on the earth)(a) √(2/3).v

_{e}(b) √(3/2).v_{e}(c) √(2).v_{e}/3 (d) 2v_{e}/√3 (e) 2v_{e}/3_{e}= √(2GM/R). Replacing R with 3R and M with 2M, we obtain the answer as√(2/3).ve [option (a)].

Consider now the following question which may confuse some of you:

**The orbital velocity of an artificial satellite near the surface of the moon is increased by 41.4%. The satellite will**

(a) move in an orbit of radius greater by 41.4% (b) move in an orbit of radius twice the original value (c) move in an elliptical orbit (d) fall down (e) escape into outer space

The correct option is (e). The orbital speed of any satellite moving round any heavenly body is √(GM/r) where as the escape velocity is √(2GM/r). This means that the escape velocity is √2 times the orbital speed or 1.414 times the orbital speed. Therefore, when the orbital speed is increased by 41.4% the satellite will escape into outer space.

(a) move in an orbit of radius greater by 41.4% (b) move in an orbit of radius twice the original value (c) move in an elliptical orbit (d) fall down (e) escape into outer space

Consider now the question which appeared in the Kerala Engineering Entrance Test paper of 2001:

**The orbital speed of an artificial satellite very close to the surface of the earth is V0. Then the orbital speed of another artificial satellite at a height equal to 3 times the radius of the earth is**

(a) 4V0 (b) 2V0 (c) V0 (d) 0.5V0 (e) 2V0/3

We have V0 =√(GM/R). At a height equal to three times the radius of the earth, the orbital velocity is obtained by replacing R with R+3R = 4R. The answer is 0.5R [option (d)].

(a) 4V0 (b) 2V0 (c) V0 (d) 0.5V0 (e) 2V0/3

The following simple question appeared in the IIT 2001 test paper:

**A simple pendulum has a time period T1 when on earth’s surface, and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T2/T1 is**

(a) 1 (b) √2 (c) 4 (d) 2

The required ratio is [2π√(L/g’)] / [2π√(L/g)] = √(g/g’). But g = GM/R^2 and g’ = GM/(2R)^2 so that g/g' = 4. The answer therefore is 2 [option(d)].

(a) 1 (b) √2 (c) 4 (d) 2

**If a body of mass ‘m’ is raised from the surface of the earth to a height ‘h’ which is comparable to the radius of the earth R, the work done is**

(a) mgh (b) mgh[1-(h/R)] (c) mgh[1+(h/R)] (d) mgh/[1+(h/R)]

Note that ‘g’ is the acceleration due to gravity on the surface of the earth. The work done for raising the body is the difference between the gravitational potential energies at the height ‘h’ and at the surface. Therefore, work done, W = -GMm/(R+h) – (-GMm/R) where M is the mass of the earth. Therefore, W = GMm/R - GMm/(R+h) = mgR – mgR/[1+(h/R)], on substituting g=GM/R

(a) mgh (b) mgh[1-(h/R)] (c) mgh[1+(h/R)] (d) mgh/[1+(h/R)]

^{2}.

Thus, W = mgR[1- 1/1+(h/R)] = mgh/[1+(h/R)]