Work Done Against Gravity MCQ - Practice Questions with Answers

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

Is given by

$$
U_h=-\frac{m g R}{1+\frac{h}{R}}
$$

So at the surface of the earth put $h=0$
We get $U_s=-m g R$
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 $h$.

$$
W=\Delta U=G M m\left[\frac{1}{r_1}-\frac{1}{r_2}\right]
$$

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

$$
W=\Delta U=G M m\left[\frac{1}{R}-\frac{1}{R+h}\right]
$$

1. When ' $h$ ' is not negligible

$$
W=\frac{m g h}{1+\frac{h}{R}}
$$

2. When ' $h$ ' is very small

$$
W=\frac{m g h}{1+\frac{h}{R}}
$$

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

$$
\begin{aligned}
& \frac{h}{R} \rightarrow 0 \\
& \text { So }_0 W=m g h
\end{aligned}
$$

3. If $h=n R$ then

$$
W=m g R * \frac{n}{n+1}
$$