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 $V = -GM/r = -1.57 \times 10^7$ J/kg. At Earth's surface, $V = -6.26 \times 10^7$ J/kg. The difference, $\Delta V = 4.69 \times 10^7$ J/kg, is the energy per kilogram required to lift any object (regardless of its mass) from the surface to that orbit. This is what makes gravitational potential useful: it's a property of the location, not of what sits there.

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 (work done against gravity). Work per unit mass between two points: $W/m = \Delta V = V_2 - V_1$. 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 strength $g$? This card answers it → $g = -dV/dr$; a steep slope in the $V$ vs $r$ graph means strong field.

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:

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

The magnitude is the familiar $|g| = GM/r^2$, directed inward.

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 $|g| \propto 1/r^2$ — more rapidly than $|V| \propto 1/r$. The slope of the $V$ vs $r$ graph at any point equals $-g$.

Potential gradient: $g = -dV/dr$ (N/kg). Negative sign → $g$ points toward decreasing $V$ (inward). From $V = -GM/r$: differentiating gives $dV/dr = GM/r^2$, so $g = -GM/r^2$ (magnitude $GM/r^2$). For finite differences: $g \approx -\Delta V/\Delta r$. Steep $V$-vs-$r$ slope = strong field.

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

Surfaces of constant gravitational potential

We just saw that field strength equals the negative potential gradient ($g = -dV/dr$). That raises a question: what do the surfaces of constant potential look like, and why is no work done moving along them? This card answers it → equipotentials are concentric spheres; $\Delta V = 0$ along them, so work done per unit mass is zero; field lines are always 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 $\Delta V = 0$, work done per unit mass $= \Delta V = 0$
• 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. No work done moving along an equipotential ($\Delta V = 0 \Rightarrow W/m = 0$). Field lines are always perpendicular to equipotentials. Close spacing = strong field; wide spacing = weak field.

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

Misconceptions — final check

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