Question

A geostationary satellite is orbiting the earth at a height of 4R above the earth’s surface. The time period of another satellite at a height of 2R is _____

(a) 24 x (3/5)^1/2 hr

(b) 12 x (3/5)^1/2 hr

(c) 6 x (3/5)^1/2 hr

(d) 72 x (3/5)^1/21 hr

The question was asked during an interview for a job.

I'd like to ask this question from Gravitation in section Gravitation of Physics – Class 11

Answer 1

Correct choice is (a) 24 x (3/5)^1/2 hr

Easiest explanation: The time period of a satellite is given by;

T = 2 x pi x (R+h)^3/2 / (G x M)^1/2

For a geostationary satellite;

24 = (5R)^3/2 x (constant); [As per giver information]

Required;

t = (3R)^2/2 x (constant)

t = Time period of the satellite at a height of 2.5R

t/24 = (3R/5R)^3/2

 = (3/5)^3/2

t = 24 x (3/5)^1/2 hr.