Question

The period of revolution of an earth's satellite close to the surface of earth is $90$ minutes. The period of another earth's satellite in an orbit at a distance of three times earth's radius from its surface will be

• A. 90 minutes
• B. $90\times \sqrt{8}$ minutes
• C. 270 minutes
• D. 720 minutes

Answer 1

Answer: (D)

720 minutes

Time period, $T^2=\frac{4{\pi }^2}{GM}{R}^3$
i.e $T^2\propto R^3$
$\Rightarrow {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3$

Here, $R_1=R_e$ and $R_2=R_e+3R_e=4R_e$
$\therefore {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_e}{4R_e}\right)}^3$
$\Rightarrow \frac{{90}^2}{T_2^2}=\frac{1}{64}$

$\Rightarrow T_2=90\times {64}^{1/2}=90\times 8=720$ minutes