## Derivation of Gravitational Potential Energy due to Point mass and on the Earth

Definition of Gravitational Potential Energy:
When an object is brought from infinity to a point in the gravitational field then work done acquired by the gravitational force is stored in the form of potential energy which is called gravitational potential energy.

Let us consider, An object of mass $m$ brought from infinity to a point in the gravitational field. If work done acquired by force is $W$ then gravitational potential energy

$U=W_{\infty \rightarrow r}$

Derivation of Gravitational Potential energy due to a Point mass:

Let us consider,
The mass of the point object (i.e point mass)=$m$
The mass of the object that produces the gravitational field = $M$

If the point mass $m$ is at a distance $x$ then the gravitational force between the objects is

$F=G \frac{M \: m}{x^{2}} \qquad(1)$

If the point mass moves a very small distance element $dx$ that is at distance $x$ from point $O$ then the work done to move the point object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M \: m}{x^{2}}.dx$

Therefore, the work done to bring the point mass from infinity to point $P$ that is at distance $r$ from point $O$ then work done to move the point object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M \: m}{x^{2}}.dx$

$[w]_{0}^{W} = G\: M\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx$

$[W-0] = G\: M\: m [-\frac{1}{x} ]_{\infty}^{r}$

$W = G\: M\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$W = -\frac{G\: M\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$W = -\frac{G\: M\: m }{r}$

This work done by the force is stored in the form of potential energy i.e

$U=W$

$U=-\frac{G\: M\: m }{r}$

Thus the above equation represents the gravitational potential energy of an object at point $P$

Gravitational Potential Energy on Earth:

Let us consider, The mass of Earth = $M_{e}$
The radius of earth = $R_{e}$
The mass of the object = $m$
The distance from centre $O$ of the earth to point $P$ = $r$
The distance from the surface of the earth to point $P$ = $h$

If the object is at a distance $x$ then the gravitational force is

$F=G \frac{M_{e} \: m}{x^{2}} \qquad(1)$

If the object moves a very small distance element $dx$ that is at distance $x$ from centre point $O$ of the earth then the work done to move an object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M_{e} \: m}{x^{2}}.dx$

Therefore, the work done to bring the object from infinity to point $P$ that is at distance $r$ from centre point $O$ of the earth then the work done to move an object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M_{e} \: m}{x^{2}}.dx$

$[w]_{0}^{W} = G\: M_{e}\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx$

$[W-0] = G\: M_{e}\: m [-\frac{1}{x} ]_{\infty}^{r}$

$W = G\: M_{e}\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$W = -\frac{G\: M_{e}\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$W = -\frac{G\: M_{e}\: m }{r}$

The above equaton shows that the work done by force is stored in the form of gravitational potential energy i.e.

$U=W$

$U = -\frac{G\: M_{e}\: m }{r}$

Where $r=R_{e}+h$, then above equation can be written as

$U = -\frac{G\: M_{e}\: m }{R_{e}+h} \qquad(2)$

This is the equation of the gravitational potential energy at point $P$. The other form of the above equation i.e

$U = -\frac{g R_{e}^{2} \: m }{R_{e}+h} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

If the object is placed on the surface of the earth then $h=0$. So gravitational potential energy on the surface of the earth
$U=-\frac{G \: M_{e} \: m}{R_{e}}$

This is the equation of the gravitational potential energy of an object placed on the surface of the earth.

$U=-\frac{g R_{e}^{2} m}{R_{e}} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

$U=-g R_{e} m$ This is another form of the gravitational potential energy of an object placed on the surface of the earth.

Note:

We know that the gravitational potential energy at any point from above the surface of the earth

$U = -\frac{G\: M_{e}\: m }{R_{e}+h}$

$U = V m \qquad \left( \because V= -\frac{G\: M_{e}\: }{R_{e}+h} \right)$

$U = Gravitational \: Potential \times \: mass \: of \: an \: object$