Question

A satellite orbiting in a circular orbit of radius R completes one revolution in 3 hours. If orbital radius of a geostationary satellite is 42000 km then the value of R is???

Answer 1

answer : 10500 km

explanation : according to Kepler's law,

$T^2\propto r^3$, means square of time period of object is directly proportional to cube of radius of circular orbit of that object.

so, $\left(\frac{T_1}{T_2}\right)^2=\left(\frac{r_1}{r_2}\right)^3$

here, $T_1$ = 3 hours and $r_1$ = R km

$T_2$ = 24 hours and $r_2$ = 42000 km

now, (3/24)² = (R/42000)³

or, (1/8)² = (R/42000)³

or, (1/64) = (R/42000)³

or, (1/4)³ = (R/42000)³

or, (1/4) = (R/42000)³

or, R = 42000/4 = 10500 km

hence, radius of circular orbit of satellite is 10500 km.