# Work done in Gravitational Field

### How to Calculate Work Done in Gravitational Field ?

Gravitational Field as a Conservative Field Consider the gravitational field of a point particle of mass M located at the origin. A test particle of mass m is placed at the point P (x, y, z)

The position vector of P ,

$\displaystyle \vec{OP} = \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

$\displaystyle \vec{OP} = r = \sqrt{x^2 + y^2 +z^2}$

The gravitational force acting on the particle at P,

$\displaystyle \vec{F} = -\frac{G M m}{r^2}\hat{r} = -\frac{G M m}{r^3}\vec{r}$

The work done by the gravitational field for a small displacement ( $\displaystyle \vec{dr} = dx\hat{i} + dy\hat{j} + dz\hat{k}$ ) is

$\displaystyle dW = \vec{F}.\vec{dr} = -\frac{G M m}{r^3}\vec{r}.\vec{dr}$

$\displaystyle dW = -\frac{G M m}{r^3} (xdx + ydy + zdz)$

$\displaystyle dW = -\frac{G M m}{r^3} r dr$

$\displaystyle dW = -\frac{G M m}{r^2} dr$

For a closed loop,

$\displaystyle dW = \oint\vec{F}.\vec{dr} = -\oint\frac{G M m}{r^2} dr = [\frac{G M m}{r}]_{r_p}^{r_p} = 0$

The gravitational field of a point particle is conservative. But what about an arbitrary gravitational field ?
we use the superposition principle.

Here, we use the superposition principle:

$\displaystyle \oint\vec{F}.\vec{dr} = \oint ( \vec{F_1} + \vec{F_2} + \vec{F_3} + ……) .\vec{dr}$

Where $\vec{F_1}$ , $\vec{F_2}$ ,$\vec{F_3}$ . . . are the gravitational forces due to each point particle constituting an extended object, which exerts the net force $\vec{F}$

$\oint\vec{F_1}.\vec{dr}+ \oint\vec{F_2}.\vec{dr} + \oint\vec{F_3}.\vec{dr} + …. = 0 + 0 + 0 + …= 0$

Gravitational field is a conservative force field. This means that one can define potential energy for this field