Question

The time period of a geostationary satellite is 24 h, at a height 6R_E (R_E is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 R_E from surface will be,

• A. 6 $\sqrt{ 2}$ h
• B. 12 $\sqrt{ 2}$ h
• C. $\frac{24}{2.5} h$
• D. $\frac{12}{2.5} h$

Answer 1

Answer: (A)

Kepler’s Third Law :-
$T \propto r^{3/2}$
$\frac{T_{2}}{T_{1}} = \left(\frac{r_{2}}{r_{1}}\right)^{3/2}$
$=\left(\frac{R + 2.5R}{R + 6R}\right)^{3/2} = \frac{1}{2\sqrt{2}} $
$\Rightarrow T_{2} = \frac{24}{2\sqrt{2} }= 6\sqrt{2}$ hours