Question

The work done to rise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:
A. $mgR$
B. $2mgR$
C. $\dfrac{1}{2}mgR$
D. $\dfrac{3}{2}mgR$

Answer 1

Hint: Work done to perform this task will be equal to the change in potential energy. Gravitational acceleration will not be the same at such high heights.

Formula Used:
Gravitational acceleration at the surface of the earth is given by,
$g=\dfrac{GM}{{{R}^{2}}}$

The potential energy in a gravitational system is given by,
$U=-\dfrac{GMm}{R}$

Where M is the Mass of Earth,
m is the mass of the object
And R is the radius of Earth,
r is the distance between two objects
G is the gravitational constant.

Complete step by step answer:

First, let us talk about why we need to perform work to take the object to a higher height.

When the object is on the surface, it is at the lowest potential state possible for the object. The weight of the object is working vertically downwards, you need to work against that force.

As you take the object away from the surface, the value of the gravitational acceleration also changes. As a result, we cannot perform this simple calculation to get the correct result,


The best way to calculate work in these cases is by calculating the potential energy difference.

Hence, the potential energy of the object when it was on the surface is,
${{U}_{1}}=-\dfrac{GMm}{R}$

The potential energy at the new location is,
${{U}_{2}}=-\dfrac{GMm}{2R}$

Hence the change in potential energy is given by,
$\Delta U={{U}_{2}}-{{U}_{1}}$
$\Rightarrow \Delta U=-\dfrac{GMm}{2R}-(-\dfrac{GMm}{R})$
$\Rightarrow \Delta U=\dfrac{GMm}{R}-\dfrac{GMm}{2R}$
$\Rightarrow \Delta U=\dfrac{GMm}{2R}$
$\Rightarrow \Delta U=\dfrac{1}{2}(m)(\dfrac{GM}{{{R}^{2}}})(R)$
$\Rightarrow \Delta U=\dfrac{1}{2}mgR$
Hence, the amount of work that we need to perform is,
$W=\Delta U=\dfrac{1}{2}mgR$
So, the correct option is - (C).

Additional Information:
You can find the gravitational acceleration using equation (1) as well.

Note: The most common mistake while doing this kind of problem is considering that the gravitational acceleration remains the same. However, that is not the case. As you can see, the value of g has changed significantly at that height.