Question

The magnitude of gravitational potential energy of earth-moon systems is U which is zero at infinite separation. If K is the kinetic energy of the moon with respect to the Earth, then
(A) $\left| U \right| = K$
(B) $\left| U \right| < K$
(C) $\left| U \right| > K$
(D) Either B or C

Answer 1

Hint
Kinetic energy is the energy possessed by a body in motion, while the gravitational potential energy is the energy acquired by an object due to the change in its position. The kinetic energy depends directly on the gravitational potential energy of the body.

Complete step by step answer
In this question we are required to find the relation between absolute gravitational potential and the kinetic energy of the Earth-moon system. Gravitational potential energy is possessed by all objects that undergo a change in position under the influence of a gravitational field. Let us assume the following variables:
Mass of the Earth is ${M_e}$
Mass of the moon is${M_m}$
The distance between the two is r
Based on these, the gravitational potential energy of the system is given as:
$U = - \dfrac{{G{M_e}{M_m}}}{r}$
Taking the absolute of both sides gives us:
$\left| U \right| = \left| { - \dfrac{{G{M_e}{M_m}}}{r}} \right| = \dfrac{{G{M_e}{M_m}}}{r}$ [Eq. 1]
We know that the orbital velocity of the moon is given by:
${v^2} = \dfrac{{G{M_e}}}{r}$
And the kinetic energy will be:
$K = \dfrac{1}{2}{M_m}{v^2} = \dfrac{{G{M_e}{M_m}}}{{2r}}$ [Eq. 2]
On comparing Eq. 1 and Eq. 2, we can see that
$\left| U \right| > K$ because of a factor of 2 being present in the denominator of K.
Hence, the correct answer is option (C).

Note
Since the zero of gravitational potential energy is at a measurable distance from the infinity, all values become negative. As this distance would increase, the gravitational potential energy would tend to zero and vice-versa.