Question

An earth satellite X is revolving around earth in an orbite whose radius is one -fourth the radius of orbit of a communication satellite .Time period revolution of X is ? ​

Answer 1

Answer:

3 hrs

Explanation:

According to the Kepler's laws of planetary motion ,

$t {}^{2} = kr {}^{3}$

Where t is time period , k is constant and r is the radius .

the time period of communication satellite is 24 hrs . then

( look into the inserted diagram )