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

The correct option is A $\frac{1}{R}$

Gravitational force provides the required centripetal force.
The gravitational force provides the required centripetal force in orbit of earth.
$\therefore \frac{G{M}_{e}M}{R^2}=\frac{m{v}_0^2}{R}$
$v_0=\sqrt{\frac{G{M}_e}{R}}$
Kinetic energy $=\frac{1}{2}m{v}_0^2$
$\therefore KE=\frac{1}{2}m{\left(\frac{G{M}_e}{R}\right)}^{2/2}$
$=\frac{1}{2}\frac{mG{M}_e}{R}$
$KE\propto \frac{1}{R}$.