Two satellites of the same mass are orbiting round the earth at heights of R and $4R$ above the earth's surface: R being the radius of the earth. Their kinetic energies are in the ratio of

Let a satellite of mass $m$ revolve in an orbit at a height r from the Centre of the earth. If the speed of the satellite is $v$, then

$\frac{m{v}^2}{r}=\frac{GMm}{r^2}or m{v}^2=\frac{GMm}{r}$

where M is the mass of the earth. The kinetic energy of the satellite is given by

$\frac{1}{2}m{v}^2=\frac{GMm}{2r}$

The ratio of the kinetic energies of the two satellites is ( $\because$their masses are equal)

$\frac{{\displaystyle \frac{1}{2}}m{v_1}^2}{{\displaystyle \frac{1}{2}}m{v_2}^2}=\frac{GMm}{2r_1}·\frac{2r_2}{GMm}=\frac{r_2}{r_1}$

But $r_1=R+R=2R and r_2=R+4R=5R$

Therefore, the ratio $\raisebox{1ex}{$r_2$}\!\left/ \!\raisebox{-1ex}{$r_1$}\right.=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$. Hence the correct answer is (4).