Question

Two satellites A and B of equal mass move in the equatorial plane of the earth, close to earth's surface. Satellite A moves in the same direction as that of the rotation of the earth while satellite B moves in the opposite direction. Calculate the ratio of the kinetic energy of B to that of A in the reference frame fixed to the earth. ($g=9.8m{s}^{-2}$ and radius of the earth $=6.37\times {10}^6$km).

Answer 1

Let ${\omega }_A$ and ${\omega }_B$ be the absolute angular speeds of A and B. Since they are in the same orbit, their time periods must be the same, i.e., ${\omega }_A={\omega }_B$. Considering the dynamics of circular motion in the cases,
$m{\omega }_A^{2}R=\frac{GMm}{R^2}\Rightarrow {\omega }_A=\sqrt{\frac{g}{R}}$ $(\because GM=g{R}^2)$
Similarly, ${\omega }_B=\sqrt{\frac{g}{R}}$
And ${\omega }_A={\omega }_B=\sqrt{\frac{9.8}{6.37\times {10}^6}}=124\times {10}^{-5}$rad $s^{-1}$
Now ${\omega }_{AE}={\omega }_A-{\omega }_E=124\times {10}^{-5}-7.3\times {10}^{-5}$
$=116.7\times {10}^{-5}$rad $s^{-1}$
${\omega }_{BE}$(velocity of B relative to E)
$={\omega }_B-(-{\omega }_E)={\omega }_B+{\omega }_E$
$=131.1\times {10}^{-5}$rad $s^{-1}$
Therefore, $\frac{K_B}{K_A}=\frac{\frac{1}{2}m{\omega }_{BE}^2{r}^2}{\frac{1}{2}m{\omega }_{AE}^2{r}^2}=\frac{{131.3}^2}{{116.7}^2}=1.27$.