Question

Derive the expression for gravitational potential energy.

Answer 1

Hint:The work done in bringing a mass from infinite to a point in the gravitational field is known as gravitational energy.
It is inversely proportional to the distance and directly proportional to the product of masses of source mass and test mass.

Complete step by step answer:
Suppose we bring a body of mass m from $\infty $ to a point in the gravitational field having distance r from the centre of earth. Then the force of attraction between the earth and the object when an object is at the distance x from the centre of the earth.
$F = \dfrac{{GMm}}{{{x^2}}}$ …..(1)
Where
M $ = $ mass of earth
m $ = $ mass of body
Now consider a small work done for a small distance dx then
$dW = Fdx$ …..(2)
So, the total work done in bringing the mass m from $\infty $ to a point which is at distance r from the centre of the earth.
$W = \int\limits_\infty ^r {\dfrac{{GMm}}{{{x^2}}}dx} $ (From equation 1 and 2)
Now integrating it from $\infty $ to r because body is taken from $\infty $ and bring it at r distance from the centre of earth.
$W = \int\limits_\infty ^r {\dfrac{{GMm}}{{{x^2}}}dx} $
$W = GMm\int\limits_\infty ^r {\dfrac{{dx}}{{{x^2}}}} $
$W = GMm\int\limits_\infty ^r {{x^{ - 2}}dx} $
$W = GMm\left[ {\dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_\infty ^r$
$W = GMm\left[ { - \dfrac{1}{x}} \right]_\infty ^r$
So, $W = GMm\left[ { - \dfrac{1}{r} + \dfrac{1}{\infty }} \right]$
$W = \dfrac{{ - GMm}}{r}$
This work done is also known as gravitational potential energy. So the gravitational potential energy is
${U_g} = \dfrac{{ - GMm}}{r}$

Note:
In many problems in which particle is moving upward and downward under gravity then following expression is used
$U = mgh$
Where
m $ = $ mass of body
g $ = $ gravitational acceleration
h $ = $ height from which body is coming