For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is -
Answer & explanation
Correct answer: option 4
Potential Energy : U = \(-\frac{GMm}{r}\)
Kinetic Energy : K = \(\frac{1}{2} mv^2\) ; velocity : v = \(\sqrt{\frac{GM}{r}}\)
⇒ Kinetic Energy : K = \(\frac{1}{2}\frac{GMm}{r}\)
∴ Ratio : \(\frac{K}{U} = \frac{1}{2}\)