Question

A thin rod of length $L$ is bent to form a semicircle. The mass of the rod is $M$. What will be the gravitational potential at the center of the circle?
(A) $\dfrac{{ - GM}}{L}$
(B) $\dfrac{{ - GM}}{{2\pi L}}$
(C) $\dfrac{{ - \pi GM}}{{2L}}$
(D) $\dfrac{{ - \pi GM}}{L}$

Answer 1

The gravitational potential is given by the formula, $V = \dfrac{{ - GM}}{r}$. So here the radius of the semicircle is used in the formula, So we can substitute the length of the rod in place of the radius as, $L = \pi r$ and get the answer.
Formula used: In the solution we will be using the following formula,
$\Rightarrow V = \dfrac{{ - GM}}{r}$
where $V$ is the gravitational potential
$G$ is the universal gravitational constant
$M$ is the mass of the body
and $r$ is the distance.

Complete step by step answer:
The gravitational potential due to any point mass is given by the formula,
$\Rightarrow V = \dfrac{{ - GM}}{r}$ where $M$ is the mass and $r$ is the distance from the point mass where gravitational potential is being calculated.
Now in the case of the semi circle we need to calculate the gravitational potential due to it at the center of the circle. Now, all the mass off the rod is concentrated only at the perimeter of the circle. So, all the mass of the rod is located at a constant distance from the center of the circle which is given by the radius of the circle.
Therefore, we can write the gravitational potential due to the semicircle at the center of the circle as,
$\Rightarrow {V_C} = \dfrac{{ - GM}}{r}$
Now, the rod is of the length $L$. So as the rod is bent in the form of a semicircle, the length of the rod will be half of the perimeter of the circle of the radius $r$.
Hence we can write,
$\Rightarrow L = \dfrac{{2\pi r}}{2}$ where $2\pi r$ is the perimeter of the circle of the radius $r$.
Hence, the length is given as,
$\Rightarrow L = \pi r$
So we can write the length in the form of the radius as,
$\Rightarrow r = \dfrac{L}{\pi }$
So substituting this value of the radius in the formula for the gravitational potential we get,
$\Rightarrow {V_C} = \dfrac{{ - GM}}{{\dfrac{L}{\pi }}}$
We can simplify this as,
$\Rightarrow {V_C} = \dfrac{{ - \pi GM}}{L}$
So the gravitational potential due to the mass of a rod bent in the shape of a semicircle at the center is given as, $\dfrac{{ - \pi GM}}{L}$
So the option (D) is correct.

Note:
The gravitational potential at a particular location is given by the gravitational energy at that point divided by the mass of the body. It is defined as the work done per unit mass to move an object to that location from some other point.