What is potential energy near a surface?

When the field is treated as uniform, lifting a mass m through a height h changes its gravitational potential energy by

§-Syllabus dot point

WAPhysicsSyllabus dot point

How do gravitational field strength and gravitational potential energy describe a field?

Define gravitational field strength and analyse changes in gravitational potential energy in a field

A focused answer to the WACE Year 12 Physics Unit 3 content point on gravitational fields and energy. Field strength as force per unit mass, field-line representation, near-surface and changing potential energy, and the work done moving a mass in a field.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WACE wants you to describe gravity as a field, calculate field strength, and analyse the energy changes when a mass moves within it. The field model lets you predict the force on any mass placed at a point without re-deriving it each time.

Field strength

The gravitational field strength at a point is the force per unit mass placed there:

g=\frac{F}{m}=\frac{GM}{r^2}\quad(\text{N kg}^{-1}).

Numerically this equals the free-fall acceleration in $\text{m s}^{-2}$ , which is why $g$ near Earth's surface is $9.8$ in both units. Field strength is a vector pointing toward the source mass.

Representing the field

Field lines show the direction a small test mass would be pushed and, by their spacing, the strength. Around a single planet the lines are radial and inward, spreading out with distance so the field weakens as $1/r^2$ . Close to a small patch of surface the lines are nearly parallel and evenly spaced, which is why we treat the field as uniform for everyday problems.

Potential energy near a surface

When the field is treated as uniform, lifting a mass $m$ through a height $\Delta h$ changes its gravitational potential energy by

\Delta E_p=mg\,\Delta h.

Raising a mass increases $E_p$ ; letting it fall converts that energy to kinetic energy. This near-surface model is what WACE expects for projectile and ramp problems, where $g$ barely changes over the heights involved.

Work done and energy conservation

The work done by you against gravity to raise a mass equals the increase in potential energy, and the work done by gravity on a falling mass equals the loss in potential energy. Combined with the kinetic energy $E_k=\tfrac{1}{2}mv^2$ , energy conservation gives launch and landing speeds directly. For a mass falling from rest through height $h$ , $mgh=\tfrac{1}{2}mv^2$ , so $v=\sqrt{2gh}$ , independent of mass.

Energy of a launched and falling mass

A 2.0 kg mass thrown upward at 15 metres per second

Take $g=9.8\ \text{m s}^{-2}$ . First find the maximum height using energy conservation, where all kinetic energy converts to potential energy:

\tfrac{1}{2}mv^2=mg\,\Delta h\;\Rightarrow\;\Delta h=\frac{v^2}{2g}=\frac{(15)^2}{2(9.8)}=11.5\ \text{m}.

Gravitational potential energy gained at the top, relative to launch:

\Delta E_p=mg\,\Delta h=2.0\times 9.8\times 11.5=225\ \text{J}.

This equals the initial kinetic energy $\tfrac{1}{2}(2.0)(15)^2=225\ \text{J}$ , confirming energy is conserved as the mass slows and rises.

Why the near-surface model works

The formula $\Delta E_p=mg\Delta h$ is only an approximation of the true field, which weakens as $1/r^2$ with distance from the planet's centre. It is accurate when the height change is tiny compared with the planet's radius, so $g$ barely changes. For Earth, climbing $1\ \text{km}$ changes $g$ by only about $0.03\%$ , so the uniform-field model is excellent for projectiles, ramps and buildings. It breaks down for satellites and deep-space probes, where the full field treatment and the inverse-square variation of $g$ must be used instead. Knowing exactly when the simple model is valid is itself an exam-worthy point.

Field strength versus potential energy

Keep the two ideas distinct: field strength is force per unit mass (a property of the field at a point), while potential energy depends on the mass placed there and its position. A region can have a strong field yet a chosen mass have little potential energy if it has barely moved.

Reading energy graphs

WACE often presents the energy story as graphs against position. For a mass dropped from rest, a graph of gravitational potential energy against height falls linearly (near a surface) while kinetic energy rises by the same amount, so the total mechanical energy stays a flat horizontal line, the signature of energy conservation with no losses. If the graph of total energy slopes downward, energy is being lost, usually to friction or air resistance as heat. Being able to read which curve is kinetic, which is potential and which is the conserved total, and to interpret a sloping total as a loss mechanism, is a recurring skill in extended-response questions.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20216 marksA space probe of mass

480\ \text{kg}

is lifted from the surface of the Moon (

g_{\text{Moon}}=1.6\ \text{N kg}^{-1}

) to a height of

250\ \text{m}

above the surface. (a) Calculate the gravitational potential energy gained. (b) The probe is then released from rest and falls back. Calculate its speed just before it lands, ignoring air resistance. (c) State one assumption made in part (b).

Show worked answer →

A 6 mark calculation rewards the energy change, an energy-conservation speed and a stated assumption.

(a) Energy gained: Treating the field as uniform over $250\ \text{m}$ , $\Delta E_p=mg\Delta h=480\times1.6\times250=1.92\times10^{5}\ \text{J}$ .
(b) Landing speed: All of this potential energy converts to kinetic energy: $\tfrac{1}{2}mv^2=mg\Delta h$ , so
$v=\sqrt{2g\Delta h}=\sqrt{2(1.6)(250)}=\sqrt{800}=28\ \text{m s}^{-1}.$
(c) Assumption: That $g$ is constant over the $250\ \text{m}$ fall (uniform field), and that no energy is lost (the Moon has no atmosphere, so this is reasonable).

Markers reward $\Delta E_p=mg\Delta h$ , the value $1.92\times10^{5}\ \text{J}$ , $v=\sqrt{2g\Delta h}$ giving $28\ \text{m s}^{-1}$ and a valid assumption.

WACE 20234 marksExplain the difference between gravitational field strength and gravitational potential energy, and explain why all objects in the same gravitational field fall with the same acceleration.

Show worked answer →

A 4 mark explanation needs the two definitions and the mass-cancellation argument.

Field strength: Gravitational field strength $g=F/m=GM/r^2$ is the force per unit mass at a point. It is a property of the field set by the source mass and distance, and does not depend on what test mass is placed there.
Potential energy: Gravitational potential energy depends on the actual mass placed in the field and how far it has moved, $\Delta E_p=mg\Delta h$ near a surface. It is a property of the mass-field system, not of the point alone.
Same acceleration: The force on a mass is $F=mg$ , so its acceleration is $a=F/m=g$ . The mass cancels, so every object accelerates at $g$ regardless of its mass.

Markers reward the per-unit-mass definition of $g$ , the mass-dependent $\Delta E_p$ and the $a=F/m=g$ cancellation.