Question:
A geostationary satellite is orbiting around an arbitary planet ' $\mathrm{P}$ ' at a height of $11 \mathrm{R}$ above the surface of ' $P$ ', $R$ being the radius of ' $P$ '. The time period of another satellite in hours at a height
of $2 \mathrm{R}$ from the surface of ' $\mathrm{P}$ ' is has the time period of 24 hours.
Correct Option:
Solution:
(3)
$\mathrm{T} \propto \mathrm{R}^{3 / 2}$
$\frac{24}{\mathrm{~T}}=\left(\frac{12 \mathrm{R}}{3 \mathrm{R}}\right)^{3 / 2} \Rightarrow \mathrm{T}=3 \mathrm{hr}$