Work Done Against Gravity MCQ - Practice Questions with Answers

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

$$W=\mathrm{\Delta }U=GMm[\frac{1}{r_1}-\frac{1}{r_2}]$$

Where $W\to$ work done
$\mathrm{\Delta }U\to$ change in Potential energy
$r_1,r_2\to$ distances

Putting r₁=R , and r₂=R+h

So

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

1. When ' $h$ ' is not negligible

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

2. When ' $h$ ' is very small

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

But $h$ is small compared to Earth's radius

$$\begin{array}{rl}& \frac{h}{R}\to 0\\ & {\text{ So }}_{0}W=mgh\end{array}$$

3. If $h=nR$ then

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