Question

If gravitational potential is taken as zero, at the surface of earth, then kinetic energy of a satellite of mass m revolving in a circular orbit of radius r will be [M is mass of earth]:
A) $\dfrac{{GMm}}{{2r}}$.
B) $\dfrac{{GMm}}{r}$.
C) $\dfrac{{3GMm}}{{2r}}$.
D) $\dfrac{{2GMm}}{r}$.

Answer 1

Gravitational potential is defined as energy which is gained by the object in raising a certain height above the surface of the earth. The centripetal force is defined as the force which is required for the object to stay at a particular radius while revolving around any planet.
Formula used: The formula of the kinetic energy is given by,
$\Rightarrow K \cdot E = \dfrac{1}{2}m{v^2}$
Where the mass is m and the velocity is v of the object.
The formula of the centripetal force is given by,
$\Rightarrow {F_c} = \dfrac{{m{v^2}}}{r}$
Where mass is m, the velocity is v and the radius is r.
The formula of the gravitational force is given by,
$\Rightarrow {F_G} = \dfrac{{GMm}}{{{r^2}}}$
Where gravitational constant is G, the mass of the earth is M, the mass of the object is m and the orbit radius is r.

Complete step by step solution:
It is given in the problem that if gravitational potential is taken as zero, at the surface of earth, then we need to find the kinetic energy of a satellite of mass m revolving in a circular orbit of radius r,
At some height the centripetal force will be equal to the gravitational force,
$\Rightarrow {F_G} = {F_c}$
$\Rightarrow \dfrac{{GMm}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}$
$\Rightarrow \dfrac{{GM}}{r} = {v^2}$
$\Rightarrow {v^2} = \dfrac{{GM}}{r}$………eq. (1)
The kinetic energy of the satellite is given by,
$\Rightarrow K \cdot E = \dfrac{1}{2}m{v^2}$
Where mass is m and velocity is v.
$\Rightarrow K \cdot E = \dfrac{1}{2}m{v^2}$
Replacing the value of ${v^2}$ from the equation (1) in the above equation.
$\Rightarrow K \cdot E = \dfrac{1}{2}m{v^2}$
$\Rightarrow K \cdot E = \dfrac{1}{2}m\left( {\dfrac{{GM}}{r}} \right)$
$\therefore K \cdot E = \dfrac{{GMm}}{{2r}}$
The kinetic energy of the satellite is equal to$K \cdot E = \dfrac{{GMm}}{{2r}}$.
The correct answer for this problem is option A.

Note: The students are advised to understand and remember the formula of the gravitational force, the formula of the centripetal force and the formula of the kinetic energy. The kinetic energy is the energy by the virtue of the motion of the object.