Gravitational potential is always negative

Gravitational Fields & Orbits — AQA A-Level Physics

Key Definition

Gravitational potential — The work done per unit mass in bringing a test mass from infinity to a defined point. Symbol: V. Units: J kg⁻¹.

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

Gravitational potentialThe work doneEnergy transferred when a force moves an object. In electrical circuits, W = QV (chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). times potential difference). per unit mass in bringing a small test mass from infinity to that point. Always negative. Measured in J kg⁻¹. is defined as zero at infinity.
Since gravitational force is attractive, work must be done on a mass to move it to infinity. This means V is negative at all finite distances from a mass.
Closer to a mass, V is more negative (lower). Further away, V becomes less negative (higher), approaching zero at infinity.
V is a scalar quantityA quantity that has magnitude (size) only. It has no direction.. This is different from g, which is a vector.
Near the Earth's surface, the familiar $GPE = mgh$ applies. For large distances, use the full gravitational potential equation.

Common Mistake MEDIUM

Students often: Forget the negative sign in gravitational potential calculations.
Instead: $V$ is always negative near a mass. Only drop the sign if the question asks for "change in" potential as a magnitude. Keep the negative sign when stating potential at a point.

V = -\frac{GM}{r}

$V$ is gravitational potential in $\text{J kg}^{-1}$, $M$ is the source mass, $r$ is distance from the centre. The minus sign means $V$ is always negative for an attractive field.

Examiner Tips and Tricks

When asked to "explain why $V$ is negative", quote the definition: work is done by the field as a mass moves from infinity inward, so the work done per unit mass against the chosen reference (zero at infinity) is negative.
$V$ at infinity is the agreed reference: $V_\infty = 0$. State this explicitly in derivation answers.
$V$ is a scalar so add potentials algebraically (with signs). Do not add them as vectors.