Question

The escape velocity of the earth is “v”. If an object is thrown vertically upwards with a velocity “kv”, what is the speed of the object at infinity?

(a) v/(k^2-1)^1/2

(b) v(k^2-1)^1/2

(c) v^2/(k^2-1)

(d) v^2 (k^2-1)

This question was addressed to me in examination.

My question is taken from Gravitation topic in chapter Gravitation of Physics – Class 11

Answer 1

The correct option is (b) v(k^2-1)^1/2

Easy explanation: From the law of conservation of energy, we have;

(m x (kv) ^2)/2 – (G x M x m)/R = (m x p^2)/2

m = Mass of object

M = Mass of earth

R = Radius of the earth

p = Velocity of the object at infinity

(kv)^2/2 –(g x R) = p^2/2

(kv)^2/2 – v^2/2 = p^2/2; v = (2 x g x R) ^½

(k^2-1)v^2 = p^2

Therefore; p = v(k^2-1) ^1/2.