Two satellites of equal mass are launched in circular orbits at heights ' $R$ ' and ' $2 R$ ' above the…
Two satellites of equal mass are launched in circular orbits at heights ' $R$ ' and ' $2 R$ ' above the surface of the earth. The ratio of their kinetic energies is ( $\mathrm{R}=$ radius of the earth)
1:3
3:2
4:9
9:4
Solution
Kinetic energy of a satellite is given by
$\mathrm{K} \cdot \mathrm{E}=\frac{\mathrm{GMm}}{2 \mathrm{r}}$
The two satellites are of the same mass
$\begin{aligned}
& \therefore \mathrm{K} \cdot \mathrm{E} \propto \frac{1}{\mathrm{r}} \\
& \frac{(\mathrm{K} \cdot \mathrm{E})_1}{(\mathrm{~K} \cdot \mathrm{E})_2}=\frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=\frac{3}{2}
\end{aligned}$