Radius of orbit of satellite of earth is $R$. Its kinetic energy is proportional to

• Correct answer
  A

  $\frac{1}{R}$
• B

  $\frac{1}{\sqrt{R}}$
• C

  $R$
• D

  $\frac{1}{R^{3 / 2}}$

Solution

Gravitational force provides the required centripetal force. The gravitational force provides the required centripetal force in orbit of earth.
$\therefore \quad \frac{G M_e M}{R^2}=\frac{m v_0^2}{R}$ $v_0=\sqrt{\frac{G M_e}{R}}$ kinetic energy $=\frac{1}{2} m v_0^2$ $\therefore \quad \mathrm{KE}=\frac{1}{2} m\left(\frac{\mathrm{G} M_{\mathrm{e}}}{R}\right)^{2 / 2}$ $=\frac{1}{2} \frac{m G M_e}{R}$ $K E \propto \frac{1}{R}$

Hence, the correct answer is option: 'A'.