GRAVITATION

1.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 ve 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

2.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) escape into outer space

3.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

4.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

5.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)]


ANSWERS

1.  correct option (a).
    We have v_e = √(2GM/R). Replacing R with 3R and M with 2M, we obtain the answer as√(2/3).v_e

2. correct option is (d).
    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.

3. correct option is (d).
   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.

4. correct option is (d).
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 .

5.correct option is (c).
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^2.

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