3.7.2.3
Calculating gravitational potential: $V = -GM/r$
Gravitational Fields & Orbits — AQA A-Level Physics
$$V = -\frac{GM}{r}$$
- $V$: 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. (J kg⁻¹)
- $G$: Newton's gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²)
- $M$: mass producing the gravitational fieldA region of space in which a mass experiences a gravitational force. (kg)
- $r$: distance from the centre of the mass (m)
- As r decreases (moving towards the planet), V becomes more negative.
- As r increases (moving away), V becomes less negative, tending to zero at infinity.
- The 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. difference between two points is $\Delta V = V_{f} - V_{i}$.
Worked Example
A planet has diameter 7600 km and mass 3.5 × 10²³ kg. Calculate the 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. at 400 km above the surface.
Show Solution
1
Write the equation
$$V = -\frac{GM}{r}$$
2
Calculate r from the centre
Radius of planet = 7600 / 2 = 3800 km
$$r = 3800 + 400 = 4200 \text{ km} = 4.2 \times 10^6 \text{ m}$$3
Substitute values
$$V = -\frac{(6.67 \times 10^{-11}) \times (3.5 \times 10^{23})}{4.2 \times 10^6}$$
4
Evaluate
$$V = -5.6 \times 10^6 \text{ J kg}^{-1}$$
Answer
$V = -5.6 \times 10^6$ J kg⁻¹
Examiner Tips and Tricks
- Remember to calculate $r$ from the centre of the planet, not just the distance above the surface.
- Add the planet's radius to the altitude.
- $V$ is per unit mass. The total potential energy of a mass $m$ at that point is $E_p = mV$, also negative.
$V$ against $r$ for a radial field
Axes: $r$ horizontal (starting at $r = R$, surface), $V$ vertical (zero at top, negative below). Curve starts at large negative value at $r = R$ and rises asymptotically toward zero as $r \to \infty$. Shape is $V = -GM/r$. Mark the dashed line $V = 0$ at infinity.
Common Mistake
MEDIUM
Students often: Drop the minus sign or write $V = GM/r^2$ (confusing potential with field strength).
Instead: $V$ has $1/r$, not $1/r^2$. The minus sign is always part of the answer for an attractive field. Field strength $g$ uses $1/r^2$; potential $V$ uses $1/r$.
Instead: $V$ has $1/r$, not $1/r^2$. The minus sign is always part of the answer for an attractive field. Field strength $g$ uses $1/r^2$; potential $V$ uses $1/r$.