Gravitational potentialAQA A-Level Physics: Revision notes

Gravitational potential, $V$, at a point is the work done per unit mass in bringing a small mass from infinity to that point. Potential is defined as zero at infinity. Its unit is J kg⁻¹.

Because gravity is attractive, the field does the work as the mass comes in from infinity, so potential is negative everywhere. The more negative the value, the closer to the mass. The sign shows that energy must be supplied to move the mass back out to infinity.

Outside a point mass or uniform sphere of mass $M$, at distance $r$ from the centre,

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

The negative sign says the potential is lower than at infinity. As $r$ increases, $V$ increases towards zero. At the Earth's surface $V = -6.25\times10^{7}$ J kg⁻¹.

Potential adds as a scalar when more than one mass contributes.

The work done in moving a mass $m$ between two points is

$W = m\Delta V$

where $\Delta V$ is the potential difference. Moving outwards, $\Delta V$ is positive, so work must be done on the mass.

Worked example: lifting 250 kg from the Moon's surface ($-2.82\times10^{6}$ J kg⁻¹) to 1.00 × 10⁶ m above ($-1.79\times10^{6}$ J kg⁻¹): $W = 250\times1.03\times10^{6} = 2.6\times10^{8}$ J.

An equipotential surface joins points of equal potential. Around a sphere they are concentric spheres, and they are always perpendicular to the field lines.

No work is done moving along an equipotential, because $\Delta V = 0$ so $W = m\Delta V = 0$. The force is perpendicular to the displacement.

Plotted against $r$ (outside the surface), $V = -GM/r$ is negative, rising towards zero as $r$ increases. $g = GM/r^2$ is positive in magnitude and falls as $1/r^2$.

The two are related by

$g = -\dfrac{\Delta V}{\Delta r}$

so $g$ is the negative of the gradient of the V–r graph. Also, the change in potential $\Delta V$ between two radii equals the area under the g–r graph between those radii.