Work done moving a mass in a gravitational field: $\DeltaW = m\DeltaV$

Gravitational Fields & Orbits — AQA A-Level Physics

\Delta W = m \Delta V

$ΔW$: 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). / change in GPE (J)
$m$: mass being moved (kg)
$ΔV$: change in 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)

Work is done when a mass moves against the gravitational fieldA region of space in which a mass experiences a gravitational force. lines (away from the planet).
This 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). equals the change in gravitational potential energyThe capacity to do work. Measured in joules (J).The energyThe capacity to do work. Measured in joules (J). an object possesses due to its position in a gravitational fieldA region of space in which a mass experiences a gravitational force...
When $V = 0 (at$ infinity), $GPE = 0$.

Full GPE expression for large distances

\Delta \text{GPE} = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

$M$: mass producing the field (kg)
$m$: mass moving in the field (kg)
$r₁$: initial distance from centre of M (m)
$r₂$: final distance from centre of M (m)

This comes from the difference: GPE at r₂ minus GPE at r₁.
If the mass moves away from the planet (r₂ > r₁), ΔW is positive: work is done against gravity.
Do not confuse this with $GPE = mg\Delta h$, which only applies near the surface where g is constant.

Worked Example

A spacecraft of mass 300 kg leaves the surface of Mars to an altitude of 700 km. Calculate the work done. Radius of Mars = 3400 km. Mass of Mars = 6.40 × 10²³ kg.

Show Solution

Write the GPE equation

$$\Delta \text{GPE} = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$

Determine r₁ and r₂

$r_1 = 3400 \text{ km} = 3.4 \times 10^6 \text{ m}$ (surface)

$r_2 = 3400 + 700 = 4100 \text{ km} = 4.1 \times 10^6 \text{ m}$

Substitute values

$$\Delta \text{GPE} = (6.67 \times 10^{-11}) \times (6.40 \times 10^{23}) \times 300 \times \left(\frac{1}{3.4 \times 10^6} - \frac{1}{4.1 \times 10^6}\right)$$

Evaluate

$$\Delta \text{GPE} = 640 \times 10^6 \text{ J} = 640 \text{ MJ (2 s.f.)}$$

Answer

$\Delta W = 640$ MJ

Common Mistake MEDIUM

Students often: Using $GPE = mg\Delta h$ for objects far from the surface.
Instead: $GPE = mg\Delta h$ only works near the surface where g is approximately constant. For large distances, use the full expression with 1/r₁ - 1/r₂.