Question

Define the gravitational potential energy and derive the gravitational potential energy.

Answer 1

Hint: First we will define the gravitational potential energy i.e. Gravitational force pulls two bodies towards each other. Now, the energy stored by a mass by being pulled or attracted by another mass is stored in the form of potential energy. Now, using this definition we will find the gravitational potential energy i.e. $\text{gravitational potential energy}=m\centerdot g\centerdot h$

Formula used: $\text{gravitational potential energy}=m\centerdot g\centerdot h$

Complete step by step answer:
Now, as asked in the question, the gravitational potential energy can be defined as,
The energy stored in an object when it is taken in the direction against the gravitational force of the earth is called gravitational potential energy. It is expressed as $P.E{{.}_{\text{gravitational}}}$. The mathematical form of gravitational potential energy can be given by,
$\text{gravitational potential energy}=m\centerdot g\centerdot h$
Where, m is mass, g is gravitational acceleration of earth and h Is height of object from the surface of earth.
Now, using the definition we will derive the equation of gravitational potential energy of an object.
Now, consider the mass ‘M’ is placed at a point along the x-axis initially, a test mass ‘m’ is at infinite. Now, a small force F is applied to displace small mass ‘m’ from infinite to mass M, and the amount of work done for smallest distance travelled (dx) is given by,
$dw=Fdx$ ………………………(i)
Where, Force F can be given as, $F=\dfrac{GMm}{{{x}^{2}}}$.
Now, we know that the force F is the force of attraction of mass M on test mass m so the small mass travels in a negative direction towards mass M. So, F and dx are in the same direction. Then the work done is given as,
$dw=\left( \dfrac{GMm}{{{x}^{2}}} \right)dx$ ……………………(ii)
Now, we will integrate on both sides, on LHS we will integrate from $0-W$ and in RHS we will integrate from $\infty -r$. On integrating we will get,
$\int\limits_{0}^{W}{dw}=\int\limits_{\infty }^{r}{\left( \dfrac{GMm}{{{x}^{2}}} \right)dx}$
Now, integration of $\int{\dfrac{dx}{{{x}^{2}}}}$ can be given as,
 $\int{\dfrac{dx}{{{x}^{2}}}}=\int{{{x}^{-2}}}dx=\dfrac{{{x}^{-2+1}}}{-2+1}$
$\Rightarrow \int{\dfrac{dx}{{{x}^{2}}}}=\left( -1 \right){{x}^{-1}}=-\dfrac{1}{x}$
On, substituting this value in expression we will get,
\[\Rightarrow \left[ W-0 \right]=\mathop{-\left[ \dfrac{GMm}{x} \right]}_{\infty }^{r}\]
\[\Rightarrow W=-\left[ \dfrac{GMm}{r} \right]-\left( \dfrac{-GMm}{\infty } \right)\]
\[\Rightarrow W=-\dfrac{GMm}{r}\]
Now, as we know that the work done in bringing mass m from infinite to the point ‘r’ where is mass M is stored in the form of potential energy, so it is given as,
\[U=-\dfrac{GMm}{r}\]
Now, if a test mass moves from a point inside the gravitational field to the other point inside the same gravitational field of mass M then potential energy is given by,
 \[U=GMm\left[ \dfrac{1}{{{r}_{i}}}-\dfrac{1}{{{r}_{f}}} \right]\]
Now, if the particle is taken from surface of the earth to a point at height ‘h’ above the surface of the earth, ${{r}_{f}}=R$ and ${{r}_{f}}=R+h$, then potential energy is given by,
\[U=GMm\left[ \dfrac{1}{R}-\dfrac{1}{R+h} \right]\]
\[\Rightarrow U=GMm\left[ \dfrac{R+h-R}{\left( R \right)\left( R+h \right)} \right]\]
\[\Rightarrow U=GMmh\left[ \dfrac{1}{\left( R \right)\left( R+h \right)} \right]\] ………………(iii)
Now, when $h << R$, then $R+h=R$ and $g=\dfrac{Gm}{{{R}^{2}}}$.
Now, on substituting this in the above equation (iii), we get,
\[\Rightarrow U=\dfrac{GMmh}{{{R}^{2}}}\]
\[\Rightarrow U=gmh\].

Hence, the gravitational potential energy is given by, \[U=gmh\].

Note: In such a type of problem, students must remember the formulas and basic concept of gravitational force and energy. Students might make mistakes in considering the object inside the earth or outside the earth because if a student considers it outside the earth then, the sum may go wrong and might not derive the final answer.