Question

A geo-stationary satellite orbits around the earth in a circular orbit of radius $36,000\text{km}$. Then, the time period of a spy satellite orbiting a few hundred km above the earth’s surface $(R_{\text{earth}}=6,400\text{km})$ will approximately be-
(Assume ${\left(\frac{8}{45}\right)}^{1.5}=0.08$)

• A. $\frac{1}{2}\text{hr}$
• B. $1\text{hr}$
• C. $2\text{hr}$
• D. $4\text{hr}$

Answer 1

The correct option is C $2\text{hr}$

Kepler's law
From $T^2\propto r^3$
$\Rightarrow \frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}$
Here $T_1=24\text{hrs}$
$r_1=36000\text{km}$, $r_2=6400\text{km}$
$T_2=?,T_2^2=T_1^2\times \frac{r_2^3}{r_1^3}$
Puting the values here,
$T_2^2={24}^2\times {\left(\frac{6400}{36000}\right)}^3$
$T_2=24\times {\left(\frac{6400}{36000}\right)}^{3/2}$
$=24\times 0.08$
$=1.96\simeq 2\text{hr}$