Question

For a satellite moving in an orbit around the earth, the ratio of K.E to P.E is:

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

Answer 1

The correct option is A $-\frac{1}{2}$

The kinetic energy of a satellite revolving at a height '$h$' from Earth's surface is:
$KE=\frac{GMm}{2(R+h)}$

Potential energy is:
$PE=-\frac{GMm}{(R+h)}$

So, $\frac{KE}{PE}=-\frac{1}{2}$