Question

The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K.
(a) U < K
(b) U > K
(c) U = K.

Answer 1

The kinetic energy of the moon with respect to the earth is K which is

U > K.

Option (B) is correct.

Explanation:

The orbital velocity of moon is:

V(om) = √ G Me / r  ----(1)

So kinetic energy of moon is K = 1 / 2 Mmvo^2m

K = GMeMm / 2r  ----(2)

U = - GMeMm / r ---(3)

Where r is the distance between the centres of both moon and earth. Therefore u is greater than K.

Thus the kinetic energy of the moon with respect to the earth is K which is

U > K.

Also learn more

Derive the formula for kinetic energy of the particle having mass m and velocity v using dimensional analysis.  ?

https://brainly.in/question/11246140

Answer 2

With respect to earth if the kinetic energy of the moon is K, then $|U|>K$

Explanation:

Moon’s orbital velocity is denoted by the formula,   $v_{0 m}=\sqrt{\frac{G M_{e}}{r}}$

So, moon's kinetic energy is $\mathrm{K}=1 / 2 \mathrm{M}_{\mathrm{m}} \mathrm{v}^{2}_{0 \mathrm{m}}$

$\mathrm{K}=\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{m}} / 2 \mathrm{r}$

$\mathrm{U}=\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{m}} / \mathrm{r}$

where r is the distance between the centres of moon and earth. Hence, $|U|>K$.

In the prograde direction, the earth is orbited by the moon. One revolution is completed in $27.32$ days with relation to stars. Relative to the sun, one revolution is completed in $29.53$ days.