Question

The altitude of a geostationary satellite is nearly $6$ times the radius of the Earth. The period of revolution of an identical satellite revolving at an altitude $0.75$ times the radius of the Earth will be :

• A. $4hrs$
• B. $3hrs$
• C. $12hrs$
• D. $2hrs$

Answer 1

The correct option is B $3hrs$

Time period is related to the orbit radius as:
$T^2\propto R^3$
$\Rightarrow \frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}$
Given $R_1=7R_e$ and $R_2=1.75R_e$
$\frac{T_1^2}{T_2^2}=(\frac{7R_e}{1.75R_e}{)}^3=4^3$
$\Rightarrow T_2=\frac{T_1}{\sqrt{4^3}}=\frac{24}{8}=3$ hours