Year 12 Physics Module 5 ⏱ ~45 min 5 MC · 3 Short Answer Lesson 14 of 18

Gravitational Potential Energy

Track gravitational potential energy from one radius to another, keep work signs consistent, and decide when the familiar near-surface approximation is accurate enough.

Today's hook: A satellite moved outward gains gravitational potential energy even though both its initial and final values are negative. How can an energy increase still end below zero, and who does the required work?
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Connect and set the energy reference

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response
Before you read, predict

Why is gravitational potential energy negative? What does the negative sign mean physically?

Write your prediction before working through the lesson, you will return to it in the Review phase.

Warm-up, for a satellite orbiting Earth, which best describes the sign of its gravitational potential energy?

Learning Intentions
goals

Know, Define Gravitational PE

  • Define gravitational potential energy as $U = -GMm/r$
  • Explain the physical meaning of the negative sign
  • State the zero-at-infinity convention

Understand, Calculate Work Done

  • Calculate work done moving between points in a gravitational field
  • Use $\Delta U = GMm(1/r_1 - 1/r_2)$, $W_g=-\Delta U$ and quasistatic $W_{\rm ext}=+\Delta U$ correctly
  • Interpret positive and negative work physically

Can Do, Apply the Approximation

  • Show that $mgh$ is an approximation for $h \ll R$
  • Choose when the exact formula is required
  • Evaluate the limits of the near-surface approximation
Scan these before reading
vocab
Gravitational potential energy$U = -GMm/r$, the energy of the two-mass system due to their separation, relative to $U=0$ at infinity.
Zero-at-infinity conventionGPE is defined as zero when the masses are infinitely far apart; all finite separations have $U < 0$.
Separation energyFor two masses initially stationary at separation $r$, quasistatic separation to infinity requires external work $|U(r)|$. Orbit escape is a total-energy problem treated in Lesson 16.
Energy and work signs$\Delta U=U_f-U_i$; $W_g=-\Delta U$, while quasistatic $W_{\rm ext}=+\Delta U$.
Near-surface approximation$\Delta U \approx mgh$, valid only when $h \ll R_{\text{planet}}$.
Centre-to-centre distance$r = R_{\text{planet}} + h$, always measured from the centre of the central body, never from its surface.
Cross-lesson links: L12 introduced $U=-GMm/r$ for circular-orbit energy. L14 now owns radial gravitational potential energy, signed work between any two valid radii, and the conditions under which $\Delta U\approx mgh$. L15 will distinguish potential per unit mass and interpret potential gradients.
Misconceptions to fix
✗ Myth: A spacecraft must have positive gravitational potential energy to escape.
✓ Reality: With $U=0$ at infinity, $U=-GMm/r$ is negative at every finite $r$. Bound or unbound motion is classified using total mechanical energy, not the sign of $U$ alone.
✗ Myth: A negative value means the system has “lost” energy.
✓ Reality: Energy values depend on the chosen zero. Negative $U$ means the system is below the zero-at-infinity reference; positive energy must be supplied to separate the masses quasistatically to infinity.
✗ Wrong: $mgh$ works for any height.
✓ Right: $mgh$ is an approximation for $h \ll R$. For large altitudes (satellites, spacecraft), use the exact formula $\Delta U = GMm(1/r_1 - 1/r_2)$.

Gravitational potential energy can be positive for objects close to a planet.

The $mgh$ formula is an approximation that fails at large altitudes.

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Define radial gravitational potential energy

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Defining Gravitational Potential Energy
+5 XP

Why $U$ is negative at every finite separation for the chosen reference

Drop a ball near Earth's surface: as the separation decreases, gravitational potential energy falls and kinetic energy can rise. Move a satellite outward: its gravitational potential energy increases, becoming less negative. Both descriptions use the same radial model; $\Delta U\approx mgh$ is only the local constant-$g$ approximation.

Diagram showing gravitational potential energy as a function of distance, with the well becoming more negative as r decreases

Gravitational PE $U = -GMm/r$, the energy well becomes deeper (more negative) as distance $r$ decreases.

Derive $U(r)$ from the inverse-square force

Choose outward-positive $r$, so $F_r=-GMm/r^2$, and define $\Delta U=-\int F_r\,dr$.

$$U(r)-U(\infty)=-\int_{\infty}^{r}\left(-\frac{GMm}{r'^2}\right)dr'=-\frac{GMm}{r}$$

With $U(\infty)=0$: $\boxed{U(r)=-GMm/r}$.

Gravitational Potential Energy

$U = -\dfrac{GMm}{r}$

$G=6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}$; $M$ and $m$ are masses; $r$ is centre-to-centre separation. Use point masses or non-overlapping spherically symmetric bodies.

The negative sign follows from attractive gravity together with the convention $U(\infty)=0$. Positive external work is required for a slow outward move. This means:

  • As $r$ decreases, $U$ becomes more negative: the system lies farther below the zero-at-infinity reference
  • As $r$ increases (object rises), $U$ becomes less negative closer to zero
  • At $r = \infty$, $U = 0$ by convention, the reference point for zero GPE
Key Insight

For the chosen reference, $U$ is negative at all finite separations and approaches zero from below as $r$ increases. Do not use the sign of $U$ alone to classify a trajectory: that requires the system's total mechanical energy.

Why Zero at Infinity?

Setting $U = 0$ at $r = \infty$ is a physically meaningful convention. At infinite separation there is no gravitational interaction. This choice means:

  • Moving an object from finite $r$ to infinity requires positive work (energy must be supplied)
  • An object at finite $r$ has less energy than at infinity, hence $U < 0$
  • For an initially stationary two-mass configuration, quasistatic separation to infinity requires external work $|U|$; an orbiting body's escape requirement also depends on its kinetic energy
Worked example 1 · GPE of a satellite

Find the gravitational potential energy of a 1000 kg satellite at $r=8.0\times10^6$ m from Earth's centre.

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Given: $m=1000$ kg, $r=8.0\times10^6$ m, $G=6.67\times10^{-11}$ N m² kg⁻² and $M_E=5.97\times10^{24}$ kg.
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Model and reference: Earth is treated as spherically symmetric, $r$ is centre-to-centre and $U=0$ at infinity.
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Equation: $U=-GMm/r$.
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Substitute: $$U=-\frac{(6.67\times10^{-11})(5.97\times10^{24})(1000)}{8.0\times10^6}=-4.97\times10^{10}\text{ J}$$
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Answer and interpretation: $U=-5.0\times10^{10}$ J to two significant figures. The negative value places the system below the zero-at-infinity reference.

Gravitational PE: $U=-GMm/r$ (J), for point masses or spherically symmetric bodies with $r$ measured centre-to-centre and $U=0$ at infinity. At finite $r$, $U<0$ and approaches zero from below as $r$ increases.

Pause, copy the highlighted GPE definition and reference convention into your book before moving on.

A 500 kg satellite moves from $r_1 = 7 \times 10^6$ m to $r_2 = 9 \times 10^6$ m. How does its GPE change?

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Separate $\Delta U$, work by gravity and external work

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Work Done in Gravitational Fields
+5 XP

Calculating energy changes when moving between orbital radii

We just saw that gravitational PE is $U=-GMm/r$. That raises a question: how do we distinguish the system's energy change from work done by gravity or by an external agent? This card answers it → calculate $\Delta U=U_f-U_i$ first, then apply the appropriate work sign.

Potential energy is defined so that work done by the conservative gravitational force is the negative of the potential-energy change. For a slow move with negligible change in kinetic energy, an external agent supplies the positive of that change.

Signed energy and work ledger

$\Delta U=U_f-U_i$

$\Delta U = \left(-\dfrac{GMm}{r_2}\right) - \left(-\dfrac{GMm}{r_1}\right)$

$\Delta U=GMm\!\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$

$W_g=-\Delta U$   ·   for a quasistatic move, $W_{\rm ext}=+\Delta U$

Physical Interpretation

  • Moving outward ($r_2>r_1$): $\Delta U>0$, so $W_g<0$. For a quasistatic lift, $W_{\rm ext}>0$.
  • Moving inward ($r_2<r_1$): $\Delta U<0$, so $W_g>0$. If no other interaction removes energy, kinetic energy increases.
  • Do not write bare $W$. State whose work is being calculated and whether the move is quasistatic.
HSC Tip

Always write $r = R_{\text{Earth}} + h$ explicitly. The most common error is using altitude $h$ instead of centre-to-centre distance $r$.

Worked example 2 · Signed work between radii

A 1000 kg satellite is moved quasistatically from $r_1=7.0\times10^6$ m to $r_2=1.0\times10^7$ m. Find $\Delta U$, $W_g$ and $W_{\rm ext}$.

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Set the ledger: $\Delta U=U_f-U_i$, $W_g=-\Delta U$ and, for this quasistatic move, $W_{\rm ext}=+\Delta U$.
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Equation: $\Delta U=GMm(1/r_1-1/r_2)$.
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Substitute: $$\Delta U=(6.67\times10^{-11})(5.97\times10^{24})(1000)\left(\frac1{7.0\times10^6}-\frac1{1.0\times10^7}\right)$$
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Calculate: $\Delta U=+1.71\times10^{10}$ J. The positive sign is expected because $r_2>r_1$.
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Answer: $W_g=-1.71\times10^{10}$ J and $W_{\rm ext}=+1.71\times10^{10}$ J. The three quantities have matching magnitudes here because the move is quasistatic.

Energy and work signs: $\Delta U=GMm(1/r_1-1/r_2)$; $W_g=-\Delta U$; for a quasistatic move, $W_{\rm ext}=+\Delta U$. Always state whose work is meant and use centre-to-centre radii.

Add the highlighted work formula and sign rules to your notes before the check below.

Moving a satellite to a higher orbit increases its gravitational PE (makes it less negative).

When gravity does positive work on a falling object, the object's GPE increases.

For an outward quasistatic move, $GMm(1/r_1-1/r_2)$ is the positive change in $U$ and equals the external work.

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Read the energy graph and connect it to work

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What the $U$–$r$ curve tells you
+5 XP

Use signed vertical differences and gradients without losing the reference convention

On a graph of $U=-GMm/r$ against outward radius $r$, the curve stays below zero and rises toward zero. Between two marked radii, the signed vertical change is $\Delta U=U_2-U_1$. The curve is steep near the central mass and flattens with distance, matching the decreasing magnitude of the gravitational force.

Graph and area meanings

$\text{vertical change on the }U\text{–}r\text{ graph}=\Delta U$

With outward-positive $r$: $F_r=-\dfrac{dU}{dr}$

$W_g=\displaystyle\int_{r_1}^{r_2}F_r\,dr=-\Delta U$

The integral is the signed area under a force-component-versus-radius graph. For an outward displacement, $F_r<0$, so that area and $W_g$ are negative while $\Delta U$ is positive. Lesson 15 develops the related potential-gradient idea per unit mass.

Interactive · Gravitational potential energy graph Open fullscreen ↗

Set $r_2>r_1$. Compare $U_1$, $U_2$ and $\Delta U$, then toggle the local $mgh$ approximation. Watch how the exact curve flattens as radius increases.

On an outward-positive radial axis, a probe moves from $r_1$ to a larger $r_2$. Which signed graph statement is correct?

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Test the near-surface approximation

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Comparison with $mgh$, The Near-Surface Approximation
+5 XP

The near-surface approximation derived from the exact formula

We just saw that $\Delta U = GMm(1/r_1 - 1/r_2)$ is the exact formula for work in a gravitational field. That raises a question: how does the familiar $mgh$ formula relate to this, and when does it fail? This card answers it → $mgh$ is derived by assuming $h \ll R$, making $R + h \approx R$; it breaks down at satellite altitudes.

For small heights above Earth's surface, the exact gravitational PE formula reduces to the familiar $mgh$. Understanding how this approximation is derived, and when it breaks down, is essential for the HSC.

Deriving $mgh$ from the Exact Formula

$\Delta U = GMm\!\left(\dfrac{1}{R} - \dfrac{1}{R + h}\right) = GMm\!\left(\dfrac{h}{R(R + h)}\right)$

For $h \ll R$: $R + h \approx R$, so $\Delta U \approx \dfrac{GMmh}{R^2}$

Since $g = GM/R^2$: $\boxed{\Delta U \approx mgh}$

The $mgh$ formula is an approximation valid only when the height $h$ is much smaller than Earth's radius $R \approx 6.37 \times 10^6 \text{ m}$. For large altitudes (satellites, spacecraft), the exact formula must be used.

Important

In the exact formula, $g = GM/R^2$ is the gravitational field strength at Earth's surface. At altitude $h$, the field strength is $g' = GM/(R+h)^2$, which is less than $g$. The $mgh$ formula assumes $g$ is constant, which fails at large $h$.

Worked example 3 · Exact result versus $mgh$

For a 1 kg mass raised 100 m above Earth's surface, compare the exact $\Delta U$ with $mgh$.

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Radii: $r_1=R_E=6.37\times10^6$ m and $r_2=R_E+h=6.3701\times10^6$ m.
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Exact: $$\Delta U=GMm\left(\frac1{r_1}-\frac1{r_2}\right)=981.3\text{ J}$$
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Approximate: $\Delta U\approx mgh=(1)(9.8)(100)=980$ J.
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Percentage error: $\lvert980-981.3\rvert/981.3\times100\%=0.14\%$.
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Conclusion: $h/R_E\approx1.6\times10^{-5}$, so the constant-$g$ approximation is excellent. Rounding $g$ to $9.8\text{ m s}^{-2}$ contributes to the small difference.

Near-surface approximation: $\Delta U \approx mgh$ (valid only when $h \ll R$). Derivation: $\Delta U = {G}{M}{m} \cdot h/(R(R+h)) \approx {G}{M}{m}h/R^2 = mgh$ when $R+h \approx R$. When $h$ is a substantial fraction of $R$, the error becomes significant and can approach or exceed 100%; use the exact radial formula.

Pause, write the highlighted approximation condition and derivation into your book before moving on.

Three of these statements about the $mgh$ approximation are correct. Pick the odd one out.

Essential formulae, Gravitational Potential Energy

GPE at distance $r$: $U = -GMm/r$

Energy change: $\Delta U = GMm(1/r_1 - 1/r_2)$

Work: $W_g=-\Delta U$; quasistatic $W_{\rm ext}=+\Delta U$

Approximation: $\Delta U \approx mgh$, only for $h \ll R$

Distance: $r = R_{\text{planet}} + h$, always use centre-to-centre distance

Fill the gap. The potential-energy change from $r_1$ to $r_2$ is $\Delta U=GMm(\text{\_\_\_\_\_})$. The missing expression is _____.

Real world, Launching to Geostationary Orbit

Consider a 1000 kg satellite moved from Earth's surface ($R_E = 6.37 \times 10^6 \text{ m}$) to geostationary orbit ($r = 4.22 \times 10^7 \text{ m}$):

$$\Delta U = (6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)\!\left(\frac{1}{6.37 \times 10^6} - \frac{1}{4.22 \times 10^7}\right) \approx 5.29 \times 10^{10} \text{ J}$$

This 53 GJ is the increase in gravitational potential energy only. A real launch also changes kinetic energy and includes atmospheric, propulsion and trajectory losses, so $\Delta U$ is not the total launch-energy requirement.

Why is $\Delta U$ positive when a satellite is moved outward even though both $U_1$ and $U_2$ are negative?

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Apply the signed radial-energy model

Activities

Activity 1, Calculation Drills
ApplyBand 4

Practise calculating $U$, $\Delta U$, $W_g$ and quasistatic $W_{\rm ext}$

  1. Find the gravitational potential energy of a 500 kg satellite at $r = 6.5 \times 10^6$ m from Earth's centre.
  2. Calculate $\Delta U$, $W_g$ and the quasistatic $W_{\rm ext}$ when a 2000 kg satellite is moved from Earth's surface to $r=10^7$ m.
  3. A 1500 kg spacecraft is at altitude 400 km above Earth. Calculate (a) its GPE and (b) $\Delta U$, $W_g$ and quasistatic $W_{\rm ext}$ for a move to 1000 km altitude. ($R_E=6.37\times10^6$ m)

A 1000 kg satellite is moved from $r = 7.0 \times 10^6$ m to $r = 8.0 \times 10^6$ m. Using $GM_E = 3.98 \times 10^{14} \text{ N m}^2\text{/kg}$, the energy required is closest to:

Activity 2, The Meaning of Negative GPE
UnderstandBand 5

Explain the physical significance of the negative sign

A student says: "If GPE is negative, energy has been taken away from the object." Explain why this statement is misleading. Discuss (a) why potential energy belongs to the two-mass system, (b) what $U=0$ at infinity means and (c) why positive external work is needed for a quasistatic move to infinity.

Which correctly describes why gravitational potential energy is defined as zero at infinity?

Wrap-up, Summary & Common Errors

Misconceptions, final check

✗ "I used altitude $h$ instead of $r = R + h$ in the formula."
✓ $r$ is always the centre-to-centre distance. For a satellite at 400 km altitude above Earth: $r = 6.37 \times 10^6 + 4.0 \times 10^5 = 6.77 \times 10^6 \text{ m}$. Forgetting to add $R_E$ is the single most common error in this topic.
✗ "I dropped the negative sign from $U = -GMm/r$."
✓ The negative sign is required by $U=0$ at infinity and attractive gravity. Bound/unbound classification uses total mechanical energy, not $U$ alone. In $\Delta U=GMm(1/r_1-1/r_2)$, the subtraction of the two negative endpoint values has already been completed.

Copy into your books

Key Definitions

  • GPE: $U = -GMm/r$, negative for all finite $r$
  • Zero of GPE: defined at $r = \infty$
  • $|U|$: quasistatic separation work only for an initially stationary configuration

Key Formulae

  • $U = -GMm/r$
  • $\Delta U = GMm(1/r_1 - 1/r_2)$
  • $W_g=-\Delta U$; quasistatic $W_{\rm ext}=+\Delta U$
  • $\Delta U \approx mgh$ (for $h \ll R$ only)

Important Points

  • Always use $r = R + h$ (centre-to-centre)
  • Negative $U$ means below the zero-at-infinity reference
  • $mgh$ is an approximation only

Common Errors

  • Using $h$ instead of $r = R + h$
  • Dropping the negative sign
  • Using $mgh$ for orbital altitudes
  • Writing $W$ without stating $W_g$ or $W_{\rm ext}$

Three statements are correct. Pick the lie.

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Complete the independent assessment

Quick recall, Gravitational Potential Energy
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 10 marks
+5 XP

ApplyBand 4(3 marks) 1. Calculate the gravitational potential energy of a 1500 kg spacecraft at altitude 400 km above Earth, then calculate $\Delta U$ for a move to 1000 km altitude and state the quasistatic external work. ($R_E=6.37\times10^6\text{ m}$, $M_E=5.97\times10^{24}\text{ kg}$)

1 mark: correct $r_1$ and $r_2$ · 1 mark: correct $U_1$ · 1 mark: correct signed $\Delta U=W_{\rm ext}$ with quasistatic condition

AnalyseBand 5(3 marks) 2. Show that for small heights $h$ above Earth's surface, $\Delta U \approx mgh$. Start from the exact expression $\Delta U = GMm(1/R - 1/(R+h))$ and use the approximation $h \ll R$.

1 mark: correct algebraic manipulation to $\Delta U = GMmh/R(R+h)$ · 1 mark: applying $h \ll R$ so $R+h \approx R$ · 1 mark: substituting $g = GM/R^2$

EvaluateBand 6(4 marks) 3. Evaluate the $mgh$ approximation by calculating the percentage error when applied to a 1 kg mass raised to: (a) 100 m, (b) 1000 km, (c) 36,000 km (geostationary altitude). Discuss the validity of the approximation in each case.

1 mark each: correct calculation and conclusion for (a) and (b) · 1 mark: correct calculation for (c) · 1 mark: overall discussion of validity

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (3 marks): $r_1=6.77\times10^6\text{ m}$ and $r_2=7.37\times10^6\text{ m}$ (1 mark). $U_1=-(6.67\times10^{-11})(5.97\times10^{24})(1500)/(6.77\times10^6)=-8.82\times10^{10}\text{ J}$ (1 mark). $U_2=-8.10\times10^{10}\text{ J}$, so $\Delta U=U_2-U_1=+7.18\times10^9\text{ J}$ and, for a quasistatic move, $W_{\rm ext}=+\Delta U$ (1 mark).

Q2 (3 marks): $\Delta U = GMm(1/R - 1/(R+h)) = GMm \cdot h / [R(R+h)]$ (1 mark). For $h \ll R$: $R + h \approx R$, so $\Delta U \approx GMmh/R^2$ (1 mark). Since $g = GM/R^2$: $\Delta U \approx mgh$ (1 mark).

Q3 (4 marks): Using the stated rounded constants: (a) $h=100\text{ m}$: exact $\Delta U=981.3\text{ J}$, $mgh=980\text{ J}$, error $\approx0.14\%$, excellent approximation (1 mark). (b) $h=1000\text{ km}$: exact $\Delta U\approx8.48\times10^6\text{ J}$, $mgh=9.8\times10^6\text{ J}$, error $\approx15.5\%$, poor at orbital altitude (1 mark). (c) $h=36{,}000\text{ km}$: exact $\Delta U\approx5.31\times10^7\text{ J}$, $mgh\approx3.53\times10^8\text{ J}$, error $\approx564\%$, invalid (1 mark). The approximation is reliable only while $h\ll R_E$ and $g$ changes negligibly (1 mark).

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Retrieve, review and choose remediation

Check what actually stuck
Arcade practice · Asteroid Blaster

Use this optional retrieval game to rehearse radial distance, gravitational-energy signs and equation conditions. Lesson 16 owns the escape-speed derivation.

How did your thinking change?

At the start you were asked how an outward move can increase gravitational potential energy while both endpoint values remain negative, and who does the work.

Answer using a signed ledger: state the $U=0$ reference, compare $U_1$ and $U_2$, then distinguish $\Delta U$, $W_g$ and quasistatic $W_{\rm ext}$. Add one sentence explaining why $\Delta U\approx mgh$ is reliable near Earth's surface but not at satellite-scale heights.

← Previous: Circular Motion and Orbits Consolidation Next: Gravitational Potential →

Module overview · Physics · Checkpoint 3 · Module quiz