Question

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

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

Answer 1

The correct option is C

$\frac{1}{2}$

Step 1

We know, Kinetic energy of a satellite is half of its magnitude of potential energy.

The potential energy of satellite $=\frac{-GMm}{r}$

Kinetic Energy of satellite $=\frac{GMm}{2r}$

where, G is gravitational constant

M is mass of the planet and m is the mass of the satellite

r is radius

Step 2

Required ratio

$$\begin{gathered} \frac{K.E}{P.E}=\frac{\left({\displaystyle \raisebox{1ex}{$Gm$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}\right)}{\left({\displaystyle \raisebox{1ex}{$GMm$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)} \\ =\frac{1}{2} \end{gathered}$$

The ratio of kinetic energy to potential energy is $\frac{1}{2}$.

Hence, the correct answer is option (c) $\frac{1}{2}$.