Gravitational Potential Due to Point Mass and Uniform Ring

Gravitational Potential Due to Point Mass and Uniform Ring

In the main Gravitation article, we have seen the expressions for the gravitational potential for different situations. Here we will rigorously derive expressions for the potential due to a Point Mass system and a Uniform Ring. The main idea for the calculation of gravitational potential is the calculation of work done to move a unit mass from a point of reference to the field of influence and we will employ this concept here.

Gravitational Potential due to a Point Mass

Simplest of the problem is calculating gravitational potential due to a point mass system. For that, we shall consider a point mass M at point A. We will find out the potential at a point P which is r distance away. By definition the potential is the work done while moving a unit mass from a point of reference, here infinity to the point r within the gravitational field of M. Thus the potential at point P is

gravitational potential due to point mass

    \[ V(r)=\frac{U(r)-U(\infty)}{1} \]

but U(\infty) is 0. Thus

    \[ \boxed{V(r)=-\frac{GM}{r}} \]

We will use this expression in the following section to calculate the potential due to uniform ring.
In the diagram below we have plotted gravitational potential due to a point charge as a function of distance from the point mass.
gravitational potential due to point mass as a function of distance from the point mass

Alternative derivation

Gravitational field due to a point mass is given by  -\frac{GM}{r^3}\vec{r}. We now use the main definition to calculate the gravitational potential

    \[  V(r)= -\int_{\infty}^{r}\vec{E}.d\vec{r} =GM \int_{\infty}^{r}\frac{dr}{r^2} \]

    \[  V(r)= -\frac{GM}{r} \]

Notice that the directions of the gravitational field and d\vec{r} are opposite to each other.

Gravitational Potential due to a Uniform Ring at a Point on its Axis

Here we will calculate the gravitational potential at point P due to a uniform ring of mass M, centered at O. From the figure shown below, the circumference of the ring makes an angle \theta with the line OP drawn from the center of the ring. We will consider a very small element of the circumference with a mass of dM. Now to calculate the potential at point P due to the mass element dM we will calculate the amount of work done to move a unit mass from \infty to the point P.

    \[ dV(r)=-\frac{GdM\times 1}{z}=-\frac{GdM}{\sqrt{a^2+r^2}} \]

gravitational potential due to uniform ring

To calculate the potential at point P due to the whole ring, we have to integrate dV over the whole ring. Potential being a scalar quantity, this can be done very easily. Thus by integrating, we find

    \[ V(r)=-\int{\frac{GdM}{\sqrt{a^2+r^2}}} \\=-\frac{G}{\sqrt{a^2+r^2}}\int{dM} \]

    \[ \boxed{V(r)=-\frac{GM}{\sqrt{a^2+r^2}}} \]

This is the final expression for the potential due to a uniform ring at a point on its axis. In the next article we will use this expression to calculate the Gravitational Potential due to Spherical Shell.

Gravitational Potential due to a Uniform Ring at a Point on its Axis as a function of distance from the center of the ring

In the diagram above we have shown the gravitational potential due to a uniform ring at a point on its axis as a function of distance from the center of the ring.

Alternative derivation

Gravitational field due to a uniform ring at a point on its axis is given by  -\frac{GMr}{(a^2+r^2)^{\frac{3}{2}}}\hat{r}}. We now use the main definition to calculate the gravitational potential

    \[  V(r)= -\int_{\infty}^{r}\vec{E}.d\vec{r} = GM\int_{\infty}^{r}\frac{rdr}{(a^2+r^2)^{\frac{3}{2}}} \]

    \[  V(r)= -\frac{GM}{\sqrt{a^2+r^2}} \]

 

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