# Variation of Gravitational Potential with Distance

Gravitation potential is defined as work done, which is negative, to bring a point originally at infinite distance from a planet to a point close to it; either on its surface or at some distance exterior to it, or in the planets interior.

For sake of simplicity we will consider the planet to be completely spherical, so by symmetry the gravitational field lines are all directed radially inward.

For a point exterior to the planet but near to its surface, the gravitation potential

$V_p=-\frac{GM}{r} \cdots where~(r > R)$

By the similar equation, we can substitute r with R for all points on the surface of the planet, so we get

$V_s=-\frac{GM}{R} \cdots for~points~on~surface$

So far so good!

But for the point interior to the planet, where (x < R), the fact that the planet is not modelled as hollow sphere, we can’t take potential at interior point to be zero.

In this case we will have to work-out the formula to calculate potential using the first principle. By the definition, the potential is the work done to move a unit mass over a distance (dx) in a field (g). So the work done is ($\vec{g} . \vec{dx}$). By integrating this expression from $\infty$ to x, we get

$V_x=-\int_\infty^x{\vec{g}.\vec{dx}}$

First the mass is first brought from $\infty$ to surface and then from surface to distance (x). We split the expression in two as below:

We get $V_x=-\left[\int_\infty^R{\vec{g_o}.\vec{dx}}+\int_R^x{\vec{g_i}.\vec{dx}}\right] \\ where \\ \\ g_o~\cdots~gravitational~field~outside~the~planet, \\ \\ and \\ \\ g_i~\cdots~gravitational~field~inside~the~planet.$

So, we get $V_x=-\left[\int_\infty^R{\frac{GM}{x^2}dx}+\int_R^x{\frac{GMx}{R^3}dx}\right]$

This will give,

$V_x=-\left[\frac{GM}{R} + \frac{GM}{2R^3}(x^2 - R^2)\right]$

And the resulting equation is $V_x=-\frac{GM}{2R^3}[3R^2 - x^2]$

Using this formula we can workout the potential at centre by substituting x = 0

$V_c=-\frac{3}{2}\frac{GM}{R}=-\frac{3}{2}V_s$

This means that potential at centre is 50% more than that at the surface!

So, in summary,

Location Potential Nature of Graph of Variation
Outside the planet $V_o=-\frac{GM}{x}$ Rectangular hyperbola
On Surface $V_s=-\frac{GM}{R}$ Constant
Inside the planet $V_x=-\frac{GM}{2R^3}[3R^2 - x^2]$ Parabola
At Centre $V_c=-\frac{3}{2}\frac{GM}{R}$ Constant

## What is Gravitational potential?

Gravitational potential at any point is the work done on 1 kg of mass to move it from infinity to that point. Gravitational potential is always negative; meaning work is not done ON the body, but rather body does the work in moving from infinity to a finite point. Gravitational potential at infinity is zero!

## 3 thoughts on “Variation of Gravitational Potential with Distance”

1. Ivanimbro says:

Infinity being anywhere beyond the leading edge of the first gravity wave? So work happens only within this universe since beyond it potential=zero?

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