Name: Work Done on a Block
Uploaded: 2019-07-19
Duration: 3 min 57 s
Description: Check out this solved illustration based on calculating work done on an object by forces acting at different angles. Just click this video to explore!

Question

Derive an expression for gravitational potential energy and prove that it is independent of the path followed.

Accepted 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.

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

Where
$M =$ Mass of earth
$m =$ Mass of body
Now consider a small work done for a small distance $d x$ then

$d W = F d x \dots . . (2)$

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.

$W = \int_{\infty}^{r} \frac{G M m}{x^{2}} d x$ (From equation 1 and 2)

Now integrating it from $\infty$ to $r$ because body is taken from $\infty$ and bring it to $r$ distance from the centre of earth.

$W = \int_{\infty}^{r} \frac{G M m}{x^{2}} d x W = G M m \int_{\infty}^{r} \frac{1}{x^{2}} d x W = G M m \int_{\infty}^{r} x^{- 2} d x W = G M m {[\frac{x^{- 2 + 1}}{- 2 + 1}]}_{\infty}^{r} W = G M m {[- \frac{1}{x}]}_{\infty}^{r} W = G M m [- \frac{1}{r} + \frac{1}{\infty}] W = \frac{- G M m}{r}$

This work done is also known as gravitational potential energy. So the gravitational potential energy is

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

The gravitational potential energy is independent of the path followed as it depends only on the vertical heights of the initial and the final positions of the object.

When we throw an object vertically above to the height of $h$ metre then the work done against the gravity is

$W' = - m g h$

Where,

Mass of the object= $m$

Height= $h$

Acceleration due to gravity= $- g$

And when the object falls back from the same height the work done is

$W " = m g h$

Here the total work done,

$W = W' + W " W = - m g h + m g h = 0$

Hence, the total work done in travelling certain height and falling back to same place is zero, this is the property of conservative force as the total work done depends only on the initial and final position of the object. Since the distance travelled is zero, the work done must be zero which is the same here.
That's why gravitational force is a conservative force.

Derive an expression for gravitational potential energy and prove that it is independent of the path followed.

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