Question

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

• A. 2
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

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, $\frac{K}{|U|}=\frac{1}{2} \frac{G M_{e} m}{R_{e}} \times \frac{R_{e}}{G M_{e} m}=\frac{1}{2}$