Gravitational Potential

GPE per unit mass, a scalar field quantity

A GPS satellite orbits at 20,200 km altitude, where Earth's gravitational potential is $V = -GM/r = -1.51 \times 10^7$ J/kg. At Earth's surface, $V = -6.26 \times 10^7$ J/kg. The difference, $\Delta V = 4.75 \times 10^7$ J/kg, is the increase in gravitational potential energy per kilogram between those locations. For a quasistatic transfer with no kinetic-energy change, it is also the minimum external work per kilogram. This is what makes gravitational potential useful: it is a property of location, not of the test mass.

Gravitational potential $V = -GM/r$ as a function of distance $r$ from the central mass.

Equipotential surfaces (spheres) and gravitational field lines (radial arrows) around a spherical mass.

Key properties of gravitational potential:

• Scalar quantity potential has magnitude only, no direction (unlike field strength $\vec{g}$, which is a vector)
• Negative for all finite $r$ the negative sign reflects that gravity is attractive; work must be done to move a mass to infinity
• Zero at infinity by convention, $V = 0$ at $r = \infty$, where there is no gravitational interaction
• Units J/kg (joules per kilogram), equivalent to m$^2$/s$^2$

Gravitational potential: $V = U/m = -GM/r$ (J/kg, scalar). Always negative for finite $r$; zero at infinity. Moving to larger $r$ means $V$ becomes less negative. For a quasistatic transfer, $W_{\rm ext}/m = \Delta V$; work done by gravity is $W_g/m=-\Delta V$. Always use $r = R + h$.

Pause, copy the highlighted potential definition and work-per-unit-mass rule into your book before moving on.

Connecting the scalar potential to the vector field

We just saw that gravitational potential $V = -GM/r$ is a scalar energy-per-unit-mass. That raises a question: how does the scalar potential connect to the vector field? This card answers it → $g_r = -dV/dr$; a steep slope in the $V$ vs $r$ graph means a large field magnitude.

The gravitational field strength $g$ is the negative of the rate at which the potential changes with distance. Where the potential changes rapidly (steep slope), the field is strong; where it barely changes (flat slope), the field is weak.

For $V = -GM/r$, differentiating gives:

$$\frac{dV}{dr} = \frac{d}{dr}\!\left(-\frac{GM}{r}\right) = \frac{GM}{r^2}$$

Therefore:

$$g_r = -\frac{dV}{dr} = -\frac{GM}{r^2}$$

Thus $\vec g=g_r\hat r$ and the magnitude is the familiar $|\vec g| = GM/r^2$.

Graphs of $V$ vs $r$ and $g$ vs $r$

• $V$ vs $r$: $V = -GM/r$ is always negative, approaching zero from below as $r \to \infty$. Steeper near the mass, flatter far away.
• $g$ vs $r$: Magnitude falls as $|\vec g| \propto 1/r^2$, more rapidly than $|V| \propto 1/r$. The $V$-graph slope is $dV/dr=|\vec g|$ for a point mass, while $g_r=-dV/dr$.

Potential gradient: $g_r = -dV/dr$ (N/kg) for outward-positive $r$. From $V = -GM/r$, $dV/dr = GM/r^2$, so $g_r = -GM/r^2$ while $|\vec g|=GM/r^2$. For a finite interval, $\bar g_r \approx -\Delta V/\Delta r$; keep the signs of both changes. A steep $V$-vs-$r$ slope means a strong field.

Add the highlighted potential-gradient relationship to your notes before the check below.

Surfaces of constant gravitational potential

We just saw that the signed radial field is the negative potential gradient ($g_r = -dV/dr$). That raises a question: what do surfaces of constant potential look like? This card answers it → equipotentials are concentric spheres around one spherical mass; $\Delta V = 0$ along them, so both external and gravitational work are zero; field lines are perpendicular.

An equipotential surface is a surface on which the gravitational potential has the same value everywhere. Moving along an equipotential requires no work, all the action happens when you move between equipotentials (different $V$ values).

• Spherical around a point mass concentric spheres centred on the mass, each sphere at a different value of $r$ and hence $V$
• Closer together near the mass the spacing indicates field strength: closely spaced means strong field; widely spaced means weak field
• No work done moving along an equipotential since $\Delta V = 0$, both $W_{\rm ext}/m=\Delta V$ for a quasistatic move and $W_g/m=-\Delta V$ are zero
• Field lines are perpendicular to equipotentials if $\vec{g}$ had a component along the surface, work would be done; since $W = 0$, $\vec{g}$ must be purely normal

Equipotential surfaces: $V = \text{const}$ on each surface; concentric spheres around a point mass. Along an equipotential, $\Delta V=0$, so $W_{\rm ext}=W_g=0$. Field lines are perpendicular to equipotentials. Close spacing means a strong field; wide spacing means a weak field.

Pause, write the highlighted equipotential rules into your book before moving on.

Misconceptions, final check

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