Two KEAM (Engineering) 2007 Multiple Choice Questions on Gravitation

The following questions which appeared in KEAM 2007 question paper are typical and of the type repeatedly asked in various entrance examinations:

(1) The change in potential energy when a body of mass ‘m’ is raised to a height nR from earth’s surface is (R = radius of the earth)

(a) mgR(n/n–1) (b) mgR (c) mgR(n/n+1)

(d) mgR(n²/n² +1) (e) mgR/n

This question appears often in entrance question papers. There may be slight change in the wording (as for example, “what is the work done in lifting a body of mass m, from the earth’s surface, through a height nR?”).

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

U = – GMm/(R+h) where M is the mass of the earth and G is the Gravitational constant.

Since h = nR, the change in potential energy is

– GMm/(R+nR) – (–GMm/R) = (GMm/R)[1 – 1/(1+n)] = (GMm/R)[n/(1+n)]

Since g = GM/R², the change in potential energy becomes mgRn/(n+1), given in option (c).

(2) The escape velocity of a body on the surface of the earth is 11.2 km/s. If the mass of the earth is doubled and its radius is halved, the escape velocity becomes

(a) 5.6 km/s (b) 11.2 km/s (c) 22.4 km/s

(d) 44.8 km/s (e) 67.2 km/s

The escape velocity (v_e) of a body on the surface of the earth is given by

v_e = √(2GM/R)

Therefore we have √(2GM/R) = 11.2 km/s

If the mass (M) of the earth is doubled and its radius (R) is halved, the escape velocity becomes √(2×4GM/R) = 2×√(2GM/R) = 22.4 km/s.

Two Questions Involving Gravitational Potential Energy & Escape Velocity

The following question will be simple if you have a clear idea of the potential energy in a gravitational field:
Two bodies of masses m₁ and m₂ are initially at rest and and at infinite distance apart. When a very feeble momentary push is given to one of them, they move towards each other because of the gravitational force between them. When the separation between them is ‘r’, their relative velocity of approach is (if G is the gravitational constant)
(a) [2Gr(m₁+m₂)]^½ (b) [2Gr(m₁–m₂)]^½ (c) [1/2Gr(m₁+m₂)]^½

(d) [2G (m₁+m₂)/r]^½ (e) [2G(m₁ – m₂)/r]^½
Imagine that initially the masses are at a distance ‘r’ apart. The gravitational potential energy of the system then is –Gm₁m₂/r. If the separation between the masses is to be increased to infinity, kinetic energy equal to Gm₁m₂/r is to be supplied to the system. (At infinite separation, kinetic energy as well as potential energy is zero). When the masses are allowed to approach, they lose potential energy (which becomes negative) and gain kinetic energy and at a separation ‘r’, the kinetic energy is +Gm₁m₂/r.
If v₁ and v₂ are the velocities of the masses, we have
½ m₁v₁ ² + ½ m₂v₂ ² = Gm₁m₂/r.
Since the momenta are equal (in magnitude), m₁v₁ = m₂v₂ so that v₂ = m₁v₁/m₂.
Substituting this in the energy equation above, v₁ = [2G/(m₁ + m₂)r]^½ ×m₂.
Similarly, v₂ = [2G/(m₁ + m₂)r] ^½ ×m1.
Since the masses are moving in opposite directions, their relative velocity is
v₁ + v₂ =[2G/(m₁ + m₂)r]^½ × (m₂ + m₁) = [2G(m₁+m₂)/r]^½
[The solution to this problem has been edited. Thank you Dr. Ravikrisnan, for pointing out the mistake in the solution which I had posted yesterday].

Now, consider the following question, which will be simple if you have a clear idea of the velocity of escape in a gravitational field:
The surface value of acceleration due to gravity on a planet of radius ‘R’ is ‘g’. If a body of mass ‘m’ at infinite distance from the planet is given a feeble momentary push so that it moves towards the planet, what will be its kinetic energy when it strikes the surface of the planet? (Assume that the gravitational field of the planet only is significant and the effect of atmosphere on the motion of the body is negligible).
(a) Infinite (b) mgR² (c) ) mgR³ (d) 2mgR (e) mgR
The body will strike the surface of the planet with a velocity equal to the escape velocity which is √(2gR). The kinetic energy with which it will strike the surface is therefore ½ m[√(2gR)]² = mgR.

Questions on Gravitation

You should definitely remember the following relations to ensure good score in gravitation:
(1) Acceleration due to gravity at a height ‘h’ is given by

g’ = GM/(R+h)², with usual notations.
Surface value of acceleration due to gravity, g = GM/R²
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.

(4) Escape velocity from the surface of earth (or any planet or star),

v_e = √(2GM/R) = √(2gR)
Escape velocity from a height ‘h’ = √[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

(6) As per Kepler’s law, T² α r³
(7) Orbital speed ‘v’ of a satellite in an orbit of radius ‘r’ is independent of its mass and is given by 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_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
We have v_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.
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)].
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)].

Consider now the following question which appeared in H.P.P.M.T.2005:
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/R².
Thus, W = mgR[1- 1/1+(h/R)] = mgh/[1+(h/R)]