Question

For the satellite moving in an orbit around the earth, the ratio of its potential energy to kinetic energy is:
(A) $\dfrac{1}{{\sqrt 2 }}$
(B) $2$
(C) $\sqrt 2$
(D) $\dfrac{1}{2}$

Answer 1

Hint The satellite rotates around the earth with the centripetal force equal to that of the gravitational force. Use the formula below and calculate the potential energy and the kinetic energy and divide the both to find the ratio of the potential to the kinetic energy.

Useful formula
(1) The formula of the centripetal force is given by
${F_c} = \dfrac{{m{v^2}}}{r}$
Where ${F_c}$ is the centripetal force, $m$ is the mass of the satellite, $v$ is the velocity of the rotation and $r$ is the radius of the rotation.
(2) The formula of the gravitational kinetic force is given by
${F_g} = \dfrac{{GMm}}{{{r^2}}}$
${F_g}$ is the gravitational force, $G$ is the gravitational constant and $M$ is the earth.
(3) The formula of the kinetic energy is given by
${E_k} = \dfrac{1}{2}m{v^2}$
Where ${E_k}$ is the kinetic energy.
(4) The formula of the gravitational potential energy is given by
${E_g} = \dfrac{{GMm}}{r}$
${E_{g\,}}$ is the gravitational potential energy.

Complete step by step solution
We know that the movement of the satellite around the earth with the centripetal force equal to that of the gravitational force.
${F_c} = d{F_g}$
Substituting the formula of the gravitational force and the kinetic energy in it.
$\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$ -------------(1)
By cancelling the similar terms in left and the right hand side of the equation,
${v^2} = \dfrac{{GM}}{r}$
Multiplying the equation (1) by half on both sides,
$\dfrac{1}{2}\dfrac{{m{v^2}}}{r} = \dfrac{1}{2}\dfrac{{GMm}}{{{r^2}}}$
From the formula of the kinetic energy,
${E_k} = \dfrac{1}{2}\dfrac{{GMm}}{r}$ --------------(2)
Let us take the formula of the gravitational potential energy,
${E_g} = \dfrac{{GMm}}{r}$ ---------------(3)
By dividing the equation (1) by (2),
$\dfrac{{{E_k}}}{{{E_g}}} = \dfrac{{\dfrac{{GMm}}{{2r}}}}{{\dfrac{{GMm}}{r}}}$
By simplifying the above equation,
$\dfrac{{{E_k}}}{{{E_g}}} = \dfrac{1}{2}$

Thus the option (D) is correct.

Note The formula of the gravitational force is not only applicable for the area inside the earth surface, it is applicable to all the areas within the universe. The rocket is provided with the fuel for the escaping velocity and then it rotates around the earth with centripetal force.