Learning Intention 1

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

Learning Intention 2

Calculate Work Done

Calculate work done moving between points in a gravitational field
Use $W = \\Delta U = GMm(1/r_1 - 1/r_2)$ correctly
Interpret positive and negative work physically

Learning Intention 3

Explain the Zero-at-Infinity Convention

Explain why GPE is defined as zero at infinity
Show that $mgh$ is an approximation for $h \\ll R$
Understand the limits of the near-surface approximation

Think First — Conceptual

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

Year 12 Physics Module 5: Advanced Mechanics 45 min Lesson 14 of 18 IQ3: Gravitational Fields

Gravitational Potential Energy

Define gravitational potential energy $U = -GMm/r$, calculate work done in gravitational fields, and understand why GPE is defined as zero at infinity.

🌍

Core Concepts

Defining Gravitational Potential Energy

Why GPE is always negative for bound objects

Gravitational Pe

Gpe Detailed

The gravitational potential energy of a mass $m$ at distance $r$ from a mass $M$ is defined as:

Gravitational Potential Energy

$U = -\\dfrac{GMm}{r}$ $U$ = gravitational potential energy (J)

$G$ = $6.67 \\times 10^{-11} \\text{ N m}^2\\text{/kg}^2$

$M$ = mass of central body (kg)

$m$ = mass of object (kg)

$r$ = centre-to-centre distance (m) = $R_{\\text{planet}} + h$

The negative sign is essential. Gravity is an attractive force, so work must be done against gravity to increase $r$ (to pull the masses apart). This means:

As $r$ decreases (object falls), $U$ becomes more negative — the system is more tightly bound
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

GPE is negative for all finite $r$ because gravity is always attractive. A bound system (planet orbiting a star, satellite orbiting Earth) has negative total energy. A positive total energy would mean the object is unbound — it would escape to infinity.

Why Zero at Infinity?

The choice of $U = 0$ at $r = \\infty$ is a convention, but a physically meaningful one. At infinite separation, there is no gravitational interaction — the masses do not influence each other. Setting $U = 0$ there 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 $U < 0$
The magnitude $|U|$ tells us the binding energy — the minimum energy needed to free the object

Worked Example — GPE of a Satellite

Find the gravitational potential energy of a 1000 kg satellite at $r = 8.0 \\times 10^6$ m from Earth's centre.

GIVEN

$m = 1000 \\text{ kg}$, $r = 8.0 \\times 10^6 \\text{ m}$, $G = 6.67 \\times 10^{-11} \\text{ N m}^2\\text{/kg}^2$, $M_E = 5.97 \\times 10^{24} \\text{ kg}$

FIND

Gravitational potential energy $U$

METHOD

Use $U = -GMm/r$ with $r$ as the centre-to-centre distance.

ANSWER

$$U = -\\frac{(6.67 \\times 10^{-11} \\text{ N m}^2\\text{/kg}^2)(5.97 \\times 10^{24} \\text{ kg})(1000 \\text{ kg})}{8.0 \\times 10^6 \\text{ m}}$$

$$U = -\\frac{3.98 \\times 10^{17} \\text{ J m}}{8.0 \\times 10^6 \\text{ m}} = -4.97 \\times 10^{10} \\text{ J}$$

$U = -4.97 \\times 10^{10} \\text{ J}$

Work Done in Gravitational Fields

Calculating energy changes when moving between orbital radii

When an object moves from one point to another in a gravitational field, the work done by the gravitational force equals the change in gravitational potential energy:

Work Done Moving Between Two Points

$W = \\Delta U = U_2 - U_1$ work done by gravity (J)

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

$\\boxed{W = \\Delta U = GMm\\left(\\dfrac{1}{r_1} - \\dfrac{1}{r_2}\\right)}$ always use this form for exact calculations

Physical Interpretation

Moving outward ($r_2 > r_1$): $(1/r_1 - 1/r_2) > 0$, so $W > 0$. Work is done against gravity (by an external agent). Gravity does negative work.
Moving inward ($r_2 < r_1$): $(1/r_1 - 1/r_2) < 0$, so $W < 0$. Gravity does positive work. The object speeds up as GPE is converted to kinetic energy.

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 — Energy to Change Orbit

Calculate the energy required to move a satellite from $r_1 = 7.0 \\times 10^6$ m to $r_2 = 1.0 \\times 10^7$ m from Earth's centre. The satellite mass is 1000 kg.

GIVEN

$m = 1000 \\text{ kg}$, $r_1 = 7.0 \\times 10^6 \\text{ m}$, $r_2 = 1.0 \\times 10^7 \\text{ m}$, $G = 6.67 \\times 10^{-11} \\text{ N m}^2\\text{/kg}^2$, $M_E = 5.97 \\times 10^{24} \\text{ kg}$

FIND

Work done (energy required) $\\Delta U$

METHOD

Use $\\Delta U = GMm(1/r_1 - 1/r_2)$

ANSWER

$$\\Delta U = (6.67 \\times 10^{-11})(5.97 \\times 10^{24})(1000)\\left(\\frac{1}{7.0 \\times 10^6} - \\frac{1}{1.0 \\times 10^7}\\right)$$

$$\\Delta U = 3.98 \\times 10^{17} \\times (1.429 \\times 10^{-7} - 1.000 \\times 10^{-7}) \\text{ J}$$

$$\\Delta U = 3.98 \\times 10^{17} \\times 4.29 \\times 10^{-8} \\text{ J} = 8.49 \\times 10^9 \\text{ J}$$

$\\Delta U = 8.49 \\times 10^9 \\text{ J}$ (approximately 8.5 GJ)

Comparison with $mgh$

The near-surface approximation derived from the exact formula

For small heights above Earth's surface, the exact gravitational PE formula reduces to the familiar $mgh$. Here's the derivation:

Starting with the exact change in GPE when moving from Earth's surface ($R$) to height $h$ above the surface ($R + h$):

Deriving $mgh$ from the Exact Formula

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

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

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

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

The $mgh$ formula is an approximation that is valid only when the height $h$ is much smaller than Earth's radius $R$. 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 — Comparing Exact and Approximate GPE

For a 1 kg mass raised $h = 100 \\text{ m}$ above Earth's surface, compare the exact $\\Delta U$ with the $mgh$ approximation. $R_E = 6.37 \\times 10^6 \\text{ m}$.

GIVEN

$m = 1 \\text{ kg}$, $h = 100 \\text{ m}$, $R_E = 6.37 \\times 10^6 \\text{ m}$, $G = 6.67 \\times 10^{-11} \\text{ N m}^2\\text{/kg}^2$, $M_E = 5.97 \\times 10^{24} \\text{ kg}$

FIND

Exact $\\Delta U$ and approximate $mgh$

METHOD

Exact: $\\Delta U = GMm(1/R - 1/(R+h))$. Approximate: $mgh$ with $g = 9.8 \\text{ m/s}^2$.

ANSWER

Exact calculation:

$$\\Delta U = (6.67 \\times 10^{-11})(5.97 \\times 10^{24})(1)\\left(\\frac{1}{6.37 \\times 10^6} - \\frac{1}{6.3701 \\times 10^6}\\right)$$

$$\\Delta U = 3.98 \\times 10^{14} \\times (1.5700 \\times 10^{-7} - 1.5698 \\times 10^{-7}) \\text{ J}$$

$$\\Delta U = 3.98 \\times 10^{14} \\times 2.46 \\times 10^{-11} \\text{ J} = 980.3 \\text{ J}$$

Approximate calculation:

$$mgh = 1 \\times 9.8 \\times 100 = 980 \\text{ J}$$

The difference is $0.3 \\text{ J}$ (about $0.03\\%$) — negligible for everyday heights.

■

Essential Formulae — Gravitational Potential Energy

$U = -\\dfrac{GMm}{r}$ gravitational potential energy at distance $r$ from mass $M$

$W = \\Delta U = GMm\\left(\\dfrac{1}{r_1} - \\dfrac{1}{r_2}\\right)$ work done moving from $r_1$ to $r_2$

$\\Delta U \\approx mgh$ approximation valid for $h \\ll R$

$U_{\\text{surface}} = -\\dfrac{GMm}{R}$ GPE at planet's surface

$r = R_{\\text{planet}} + h$ centre-to-centre distance (always add radius + altitude)

Common Misconceptions

✗

"GPE can be positive" — With $U = -GMm/r$, GPE is always negative for finite $r$. Zero only at infinity. A positive GPE would imply an unbound object that escapes the gravitational field.

✗

"The negative sign is just a convention with no physical meaning" — The negative sign encodes that gravity is attractive. Bound systems have negative total energy. The sign is essential for determining whether an object is bound or unbound.

✗"$mgh$ works for any height" — $mgh$ is an approximation for $h \\ll R$. For large $h$ (satellites, spacecraft), use the exact formula $\\Delta U = GMm(1/r_1 - 1/r_2)$.

Real World: Lifting a Satellite to Orbit

Energy Cost of Launching to Geostationary Orbit

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

$$\\Delta U = GMm\\left(\\frac{1}{R_E} - \\frac{1}{r}\\right)$$

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

$$\\Delta U = 3.98 \\times 10^{17} \\times (1.570 \\times 10^{-7} - 2.370 \\times 10^{-8}) \\text{ J} = 5.29 \\times 10^{10} \\text{ J}$$

This $5.29 \\times 10^{10} \\text{ J}$ (about 53 GJ) is the minimum energy needed, ignoring kinetic energy for orbit, air resistance, and other losses. In practice, rockets are only about 2% efficient due to carrying their own fuel — the vast majority of a rocket's launch mass is propellant.

Activities

Activity

Calculation Drills

Practise calculating GPE and work done in gravitational fields

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 the energy required to move a 2000 kg satellite from Earth's surface to $r = 10^7$ m.

3 At what height above Earth's surface does the $mgh$ approximation deviate by 1% from the exact $\\Delta U$?

Concept Check

The Meaning of Negative GPE

Explain the physical significance of the negative sign

A student says: "If GPE is negative, that means energy has been taken away from the object." Explain why this statement is misleading. In your answer, discuss: (a) what the zero of GPE represents, (b) what a negative GPE tells us about the binding of the object to the central mass, and (c) what would happen if an object's total energy became positive.

📝 Copy into Books

▾

Key Definitions

GPE: $U = -GMm/r$ — negative for all finite $r$
Zero of GPE: defined at $r = \\infty$ (by convention)
Binding energy: energy needed to free an object from a gravitational field

Key Formulae

$U = -GMm/r$
$\\Delta U = GMm(1/r_1 - 1/r_2)$
$\\Delta U \\approx mgh$ (for $h \\ll R$)
$U_{\\text{surface}} = -GMm/R$

Important Points

Always use $r = R + h$ (centre-to-centre)
Negative sign = attractive force, bound system
$mgh$ is an approximation only
Work done against gravity = positive $\\Delta U$

Common Errors

Using $h$ instead of $r = R + h$
Dropping the negative sign
Using $mgh$ for orbital altitudes
Confusing work done by and against gravity

Revisit Your Answer

Now that you have studied gravitational potential energy, revisit your earlier answer about why GPE is negative.

The negative sign arises because gravity is attractive. Work must be done against the field to separate the masses. Setting $U = 0$ at infinity means all finite separations have $U < 0$ — the object is bound. A bound object needs energy input to escape. If total energy were positive, the object would be unbound and could escape to infinity with kinetic energy remaining.

Has your understanding changed? Write a revised explanation:

Interactive: Gpe Graph Interactive

Inquiry Question 3: Gravitational Fields

Key Terms

Gravitational potential energy $U = -GMm/r$ — energy stored in a gravitational field, negative for bound objects

Zero-at-infinity convention GPE is defined as zero when masses are infinitely far apart

Binding energy Minimum energy required to free an object from a gravitational field ($|U|$)

Work done by gravity $W = \\Delta U = GMm(1/r_1 - 1/r_2)$

Near-surface approximation $\\Delta U \\approx mgh$, valid only when $h \\ll R$

Centre-to-centre distance $r = R_{\\text{planet}} + h$ — always measured from the centre of the central body

Multiple Choice

1. Gravitational potential energy is defined as zero at:

A Earth's surface

B Infinity

C The centre of Earth

D Low Earth orbit

2. The gravitational PE of a mass $m$ at distance $r$ from mass $M$ is:

A $GMm/r$

B $-GMm/r$

C $GMm/r^2$

D $-GMm/r^2$

3. Moving a satellite to a higher orbit:

A Decreases its GPE (more negative)

B Increases its GPE (less negative)

C Doesn't change GPE

D Makes GPE positive

4. Work done moving from $r_1$ to $r_2$ in a gravitational field:

A $GMm(1/r_2 - 1/r_1)$

B $GMm(1/r_1 - 1/r_2)$

C $GMm/r_1 - GMm/r_2$

D $mg(r_2 - r_1)$

5. The $mgh$ approximation is valid when:

A $h$ is very large

B $h$ is much less than Earth's radius

C $h$ equals Earth's radius

D Always

Short Answer

Apply Band 4 3 marks

Calculate the gravitational potential energy of a 1500 kg spacecraft at altitude 400 km above Earth. Calculate the energy required to move it to an altitude of 1000 km.

Analyse Band 5 3 marks

Show that for small heights $h$ above Earth's surface, $\\Delta U \\approx mgh$. Start from the exact expression and use the binomial approximation.

Evaluate Band 6 4 marks

Evaluate the $mgh$ approximation by calculating the percentage error when using it to find the potential energy change of 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.

Model Answers

▾

Question 1 (3 marks)

$r_1 = R_E + h_1 = 6.37 \\times 10^6 + 400 \\times 10^3 = 6.77 \\times 10^6 \\text{ m}$ (1 mark)

$r_2 = R_E + h_2 = 6.37 \\times 10^6 + 1000 \\times 10^3 = 7.37 \\times 10^6 \\text{ m}$

$U_1 = -\\dfrac{(6.67 \\times 10^{-11})(5.97 \\times 10^{24})(1500)}{6.77 \\times 10^6} = -8.83 \\times 10^{10} \\text{ J}$

$U_2 = -\\dfrac{(6.67 \\times 10^{-11})(5.97 \\times 10^{24})(1500)}{7.37 \\times 10^6} = -8.11 \\times 10^{10} \\text{ J}$

$\\Delta U = U_2 - U_1 = (-8.11 \\times 10^{10}) - (-8.83 \\times 10^{10}) = 7.2 \\times 10^9 \\text{ J}$ (2 marks)

Question 2 (3 marks)

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

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

For $h \\ll R$: $R+h \\approx R$, so $\\Delta U \\approx \\dfrac{GMmh}{R^2} = mgh$ since $g = GM/R^2$ (1 mark)

Question 3 (4 marks)

(a) $h = 100 \\text{ m}$: Exact $\\Delta U = 980.3 \\text{ J}$, $mgh = 980 \\text{ J}$, error $= 0.03\\%$. The approximation is excellent for everyday heights. (1 mark)

(b) $h = 1000 \\text{ km}$: Exact $\\Delta U = 8.49 \\times 10^6 \\text{ J}$, $mgh = 9.8 \\times 10^6 \\text{ J}$, error $= 15.4\\%$. The approximation is poor at orbital altitudes. (1 mark)

(c) $h = 36{,}000 \\text{ km}$: Exact $\\Delta U = 5.29 \\times 10^7 \\text{ J}$, $mgh = 3.53 \\times 10^8 \\text{ J}$, error $= 567\\%$. The approximation is completely invalid for GEO. (1 mark)

The $mgh$ approximation is excellent for everyday heights ($h < 1 \\text{ km}$), acceptable for low Earth orbit only as an order-of-magnitude estimate, and completely invalid for geostationary altitude. The exact formula must be used for any significant altitude. (1 mark for discussion)

Boss Battle

✓

Mark Lesson 14 Complete

Tick this box when you have finished all sections of this lesson.

← Previous: Lesson 13 Next: Lesson 15 →

Module Overview Physics Subject Page Checkpoint 3