JEE Main Syllabus 2025 PDF - Subject Wise Topics

# Work Done Against Gravity - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Work Done Against Gravity is considered one of the most asked concept.

• 4 Questions around this concept.

## Solve by difficulty

If  g  is the acceleration due to gravity on the earth's surface,  the gain in the potential energy of an object of mass  m   raised from the surface of the earth to  a height equal to the radius  R   of the earth is :

## Concepts Covered - 1

Work Done Against Gravity

The gravitational potential energy at height 'h' from the earth's surface

Is given by $U_{h}=-\frac{mgR}{1+\frac{h}{R}}$

So at the surface of earth put h=0

We get $U_s=-mgR$

So if the body of mass m is moved from the surface of earth to a point at height h from the earth's surface

Then there is a change in its potential energy.

And this change in its potential energy is known as work done against gravity to move the body from earth surface to height h.

$W=\Delta U=GMm\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]$

Where $W \rightarrow$ work done

$\Delta U \rightarrow$ change in Potential energy

$r_{1},r_{2}\rightarrow$ distances

Putting r1=R , and r2=R+h

So $W=\Delta U=GMm\left [ \frac{1}{R}-\frac{1}{R+h} \right ]$

1. when 'h' is not negligible

$W=\frac{mgh}{1+\frac{h}{R}}$

1. when 'h' is very small

As $W=\frac{mgh}{1+\frac{h}{R}}$

But $h$ is small as compared to earth's radius

$\frac{h}{R}\rightarrow 0$

So $W=mgh$

1. If h = nR then

$W=mgR*\frac{n}{n+1}$

## Study it with Videos

Work Done Against Gravity

"Stay in the loop. Receive exam news, study resources, and expert advice!"