Question

Two satellites $A$ and $B$, of equal mass, move in the equatorial plane of earth close to the 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. Determine the ratio of the kinetic energy of $B$ to that of $A$ in the reference frame fixed to earth.

• A. $6$
• B. $5$
• C. $1.2$
• D. $2.6$

Answer 1

Answer: (C)

$1.2$

Let ${\omega }_A$ and ${\omega }_B$ be the absolute angular speed of A and B. Since they are in same orbit, their time period must be same i.e., ${\omega }_A={\omega }_B$

Considering, the dynamics of circular motion in cases $m{\omega }_A^{2}R=\frac{GMm}{R^2}$

${\omega }_A=\sqrt{\frac{g}{R}}$ $(\because \frac{GM}{R^2}=8)$

similarly

${\omega }_B=\sqrt{\frac{g}{R}}$

So, ${\omega }_A={\omega }_B=\sqrt{\frac{9.8}{6.37\times {10}^6}}=124\times {10}^{-5}$ rad/s

Now

${\omega }_{AE}={\omega }_A-{\omega }_E=124\times {10}^5-7.3\times {10}^{-5}$ rad/s

$=116.7\times {10}^5$ rad/s

and

${\omega }_{BE}$ (angular velocity of B relative to E)

$={\omega }_B-(-{\omega }_E)$

$={\omega }_B+{\omega }_E$

${\omega }_{BE}=131.1\times {10}^{-5}$ rad/s

So,

$\frac{k_B}{k_A}=\frac{\frac{1}{2}m{\omega }_{BE}^2{r}^2}{\frac{1}{2}m{\omega }_{AE}^2{r}^2}={\left(\frac{131.3}{116.7}\right)}^2$

$=1.27$

$\therefore \frac{K_B}{K_A}=1.27$.