Question

A body of weight 1 newton has a potential energy of 1 joule relative to the ground when it is at a height of:
a) 1m
b) 9.8m
c) $\dfrac{1}{9.8}m$
d) 0m

Answer 1

The gravitational potential energy of the Earth is a function of the distance from the centre of the Earth. The gravitational potential energy increases from the surface of the Earth as we go upwards. Hence taking potential energy of the body with respect to Earth’s surface to the height say ‘h’ of would yield the required answer such that potential energy with respect to Earth is 1 joule.
Formula used:
${{U}_{h}}=\dfrac{GMmh}{R(R+h)}$

Complete answer:
Let us say we have an object of mass ‘m’ such that it is taken above at a height ‘h’ with respect to ground. If ‘M’ is the mass of the Earth, ‘R’ is the radius of the Earth and G is the gravitational constant, then the potential energy acquired by the body with respect to Earth is,
${{U}_{h}}=\dfrac{GMmh}{R(R+h)}$
The acceleration due to gravity(g) is given by $g=\dfrac{GM}{{{R}^{2}}}$
Therefore substituting the value of ‘GM’ in the above equation we get,
$\begin{aligned} & {{U}_{h}}=\dfrac{GMmh}{R(R+h)} \\\ & \Rightarrow {{U}_{h}}=\dfrac{gRmh}{(R+h)} \\\ & {{U}_{h}}=\dfrac{mgh}{(1+\dfrac{h}{R})} \\\ \end{aligned}$
Since $h << R$, the above equation reduced to,
$\begin{aligned} & {{U}_{h}}=\dfrac{mgh}{(1+\dfrac{h}{R})} \\\ & \because h << R \\\ & {{U}_{h}}=mgh \\\ \end{aligned}$
The product of mass times the acceleration of the body is the weight of the body. In the question it is given that the weight of the body is 1 newton and the body has a potential energy of 1 joule relative to the ground. Hence the height ‘h’ at which the body is present is,
$\begin{aligned} & {{U}_{h}}=mgh \\\ & \Rightarrow 1Nm=1N\times h \\\ & \therefore h=1m \\\ \end{aligned}$

Therefore the correct answer of the above question is option a.

Note:
It is to be noted that the energy of the body is equal to the work done. The work done against the gravitational force is stored as potential energy of the body with respect to Earth. The work done is given by the dot product of force acting times the displacement.