Question

The orbital velocity of a satellite orbiting the earth is half the escape velocity of the earth. What is the height above the surface of the earth at which it is orbiting? (Let the radius of the earth (R) = 6400 km).

(a) 6400 km

(b) 3200 km

(c) 9600 km

(d) 4800 km

I got this question in class test.

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

Answer 1

Right answer is (a) 6400 km

Best explanation: Orbital velocity (v) = [(G x M) / (R+h)]^1/2

Escape velocity (v’) = [(2 x G x M) / R]^1/2

We know; v = v’/2

v^2 = v’^2/4

[(G x M) / (R+h)] = [(2 x G x M) / R]/4

1 / (R+h) = 1 / (2R)

2R = R + h

h = R

Therefore; h = 6400km.