Question

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

• A. $\frac{1}{R}$
• B. $\frac{1}{\sqrt{R}}$
• C. $R$
• D. $\frac{1}{R^{3/2}}$

Answer 1

Answer: (A)

Gravitational force provides the required centripetal force.
The gravitational force provides the required centripetal force in orbit of earth.

$\therefore \frac{G{M}_{e}m}{R^2}=\frac{m{v}_0^2}{R}$
$\Rightarrow v_o=\sqrt{\frac{G{M}_e}{R}}$
Kinetic energy $=\frac{1}{2}m{v}_o^2$
$\therefore KE=\frac{1}{2}m{\left(\frac{G{M}_e}{R}\right)}^{2/2}$
$=\frac{1}{2}\frac{mG{M}_e}{R}$
$\Rightarrow KE\propto \frac{1}{R}$