Q. For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

1607 191 JIPMERJIPMER 2008Gravitation Report Error

A

2

B

$\frac{1}{2}$

C

$\frac{1}{2}$

D

$2$

Solution:

Key Idea Kinetic energy of satellite is half of its potential energy.
Potential energy of satellite
$U = - \frac{G M _{e} m}{R _{e}}$
where $R_{e}$ is radius of earth, $M_{e}$ the mass of earth, $m$ the mass of satellite and $G$ the gravitational constant.
$∣ U ∣ = \frac{G M _{e} m}{R _{e}}$
Kinetic energy of satellite
$K = \frac{1}{2} \frac{G M _{e} m}{R _{e}}$
Thus, $\frac{K}{∣ U ∣} = \frac{1}{2} \frac{G M _{e} m}{R _{e}} \times \frac{R _{e}}{G M _{e} m} = \frac{1}{2}$