Question

A geostationary satellite orbits around the earth in a circular orbit of radius $36000km$. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth's surface will approximately be $($ Given: $R_{Earth}=6400$ km $)$

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

Answer 1

Answer: (C)

$2h$

For a satellite of mass $m$ moving with a velocity $v$ in a circular orbit of radius $r$ around the earth of mass $M$, we have,
$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$ or $v=\sqrt{\frac{GM}{r}}$
But $v=\frac{2\pi r}{T}$.
Thus $\frac{2\pi r}{T}=\sqrt{\frac{GM}{r}}$ or $T\propto r^{\frac{3}{2}}$
$\therefore \frac{T_2}{T_1}={\left(\frac{r_2}{r_1}\right)}^{\frac{3}{2}}$ ...................$\left(1\right)$
Given $r_2=6400$ km and $r_1=36000$ km. For a geostationary satellite $T_1=24$ h. Using these values in $\left(1\right)$, we have get,
$T_2=24\times {\left(\frac{64}{360}\right)}^{\frac{3}{2}}=1.8$ hours.
Hence the closest choice is (c).