# 4.8 Gravitational potential due to rigid body

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Gravitational potential to rigid body at a point is scalar sum of potentials due to elemental mass.

We have derived expression for gravitational potential due to point mass of mass, “M”, as :

$V=-\frac{GM}{r}$

We can find expression of potential energy for real bodies by considering the same as aggregation of small elements, which can be treated as point mass. We can, then, combine the potential algebraically to find the potential due to the body.

The derivation is lot like the derivation of gravitational field strength. There is, however, one important difference. Derivation of potential expression combines elemental potential – a scalar quantity. As such, we can add contributions from elemental parts algebraically without any consideration of direction. Indeed, it is a lot easier proposition.

Again, we are interested in finding gravitational potential due to a solid sphere, which is generally the shape of celestial bodies. As discussed earlier in the course, a solid sphere is composed of spherical shells and spherical shell, in turn, is composed of circular rings of different radii. Thus, we proceed by determining expression of potential from ring -->spherical shell -->solid sphere.

## Gravitational potential due to a uniform circular ring

We need to find gravitational potential at a point “P” lying on the central axis of the ring of mass “M” and radius “a”. The arrangement is shown in the figure. We consider a small mass “dm” on the circular ring. The gravitational potential due to this elemental mass is :

$dV=-\frac{Gdm}{PA}=-\frac{Gdm}{{\left({a}^{2}+{r}^{2}\right)}^{\frac{1}{2}}}$

We can find the sum of the contribution by other elements by integrating above expression. We note that all elements on the ring are equidistant from the point, “P”. Hence, all elements of same mass will contribute equally to the potential. Taking out the constants from the integral,

$⇒V=-\frac{G}{{\left({a}^{2}+{r}^{2}\right)}^{\frac{1}{2}}}\underset{0}{\overset{M}{\int }}dm$

$⇒V=-\frac{GM}{{\left({a}^{2}+{r}^{2}\right)}^{\frac{1}{2}}}=-\frac{GM}{y}$

This is the expression of gravitational potential due to a circular ring at a point on its axis. It is clear from the scalar summation of potential due to elemental mass that the ring needs not be uniform. As no directional attribute is attached, it is not relevant whether ring is uniform or not? However, we have kept the nomenclature intact in order to correspond to the case of gravitational field, which needs to be uniform for expression as derived. The plot of gravitational potential for circular ring is shown here as we move away from the center.

Check : We can check the relationship of potential, using differential equation that relates gravitational potential and field strength.

$⇒E=-\frac{dV}{dr}=\frac{d}{dr}\left\{\frac{GM}{{\left({a}^{2}+{r}^{2}\right)}^{\frac{1}{2}}}\right\}$

$⇒E=GMx-\frac{1}{2}X{\left({a}^{2}+{r}^{2}\right)}^{-\frac{1}{2}-1}X2r$

$⇒E=-\frac{GMr}{{\left({a}^{2}+{r}^{2}\right)}^{\frac{3}{2}}}$

The result is in excellent agreement with the expression derived for gravitational field strength due to a uniform circular ring.

## Gravitational potential due to thin spherical shell

The spherical shell of radius “a” and mass “M” can be considered to be composed of infinite numbers of thin rings. We consider one such thin ring of infinitesimally small thickness “dx” as shown in the figure. We derive the required expression following the sequence of steps as outlined here :

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