What is wrong reference point?

Zero potential energy is at infinity, not at the planet's surface or centre.

A 250 kg payload is moved from the surface of Earth to a circular orbit at altitude 500 km. Calculate the change in gravitational PE. [4 marks]

NSWPhysicsSyllabus dot point

Inquiry Question 3: How does the force of gravity determine the motion of planets and satellites?

Derive and apply the concept of gravitational potential energy in a radial gravitational field, U = -G M m / r, including the concept of escape velocity

A focused answer to the HSC Physics Module 5 dot point on gravitational potential energy in radial fields. Why U is negative, how it differs from the mgh approximation, the derivation of escape velocity, and the standard worked example using Earth.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-23

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

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What this dot point is asking
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What this dot point is asking

NESA wants you to extend the idea of gravitational potential energy from the near-Earth $U = mgh$ approximation to the full radial-field formula $U = -G M m / r$ , explain why it is negative, and apply it to problems including escape velocity. This dot point underpins every energy-based question on orbital motion in the second half of Module 5.

The answer

From $mgh$ to the radial formula

Near Earth's surface, gravitational field strength $g$ is approximately constant, and $U = mgh$ works fine. At astronomical scales, $g$ falls off as $1/r^2$ , so the potential energy must be obtained by integrating the gravitational force from infinity inward:

U(r) = -\int_{\infty}^{r} F \, dr = -\int_{\infty}^{r} \left(-\frac{G M m}{r^2}\right) dr = -\frac{G M m}{r}

The negative sign reflects two choices:

Zero potential energy at infinity (the natural reference for radial fields).
Attractive force, so moving inward releases energy.

A bound mass (closer than infinity) therefore has $U < 0$ .

Why negative U makes physical sense

Imagine releasing a stationary object from far away. Gravity does positive work pulling it inward, increasing kinetic energy. By conservation of energy, potential energy must decrease. Since we set $U = 0$ at infinity, $U$ becomes negative as the object approaches the source.

The deeper into the well, the more negative $U$ becomes. To free a mass from the well (push it to infinity), positive work must be done equal to $|U| = \frac{G M m}{r}$ .

Escape velocity

Escape velocity $v_{\text{esc}}$ is the minimum speed needed at a distance $r$ for an object to reach infinity with zero remaining kinetic energy. By conservation of energy:

\frac{1}{2} m v_{\text{esc}}^2 + \left(-\frac{G M m}{r}\right) = 0 + 0

Solving:

v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}

Key features:

Independent of the mass of the escaping object.
Depends on the mass $M$ of the source and the launch distance $r$ .
At Earth's surface: $v_{\text{esc}} \approx 11.2$ km/s.
For a black hole, $r$ shrinks until $v_{\text{esc}}$ approaches the speed of light.

Change in potential energy

When a mass moves from $r_1$ to $r_2$ :

\Delta U = U(r_2) - U(r_1) = -\frac{G M m}{r_2} - \left(-\frac{G M m}{r_1}\right) = G M m \left(\frac{1}{r_1} - \frac{1}{r_2}\right)

If $r_2 > r_1$ (moving away), $\Delta U > 0$ : potential energy increases (becomes less negative).

Work required to raise a satellite to altitude

Find the work needed to lift a $500$ kg satellite from Earth's surface ( $r_1 = R_E = 6.37 \times 10^6$ m) to an altitude of $1000$ km ( $r_2 = 7.37 \times 10^6$ m). Take $M_E = 5.97 \times 10^{24}$ kg.

Step 1: Write the change-in-potential-energy formula

Because we are dealing with a radial gravitational field rather than the near-surface approximation, the change in gravitational potential energy between two radii is:

\Delta U = G M_E m \left(\frac{1}{r_1} - \frac{1}{r_2}\right)

This form comes directly from $U(r) = -G M m / r$ : subtracting $U(r_1)$ from $U(r_2)$ and collecting terms. The result is positive when $r_2 > r_1$ because moving farther from Earth increases potential energy.

Step 2: Calculate the prefactor $G M_E m$

Multiply the three constants together first to keep the working tidy:

G M_E m = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500 = 1.99 \times 10^{17} \text{ N m}^2

Step 3: Evaluate the bracket of reciprocals

Take the reciprocal of each orbital radius and subtract:

\frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{6.37 \times 10^6} - \frac{1}{7.37 \times 10^6} = 1.570 \times 10^{-7} - 1.357 \times 10^{-7} = 2.13 \times 10^{-8} \text{ m}^{-1}

A larger $r_2$ gives a smaller reciprocal, so the bracket is positive, confirming energy must be added.

Step 4: Multiply to find $\Delta U$

\Delta U = 1.99 \times 10^{17} \times 2.13 \times 10^{-8} = 4.24 \times 10^9 \text{ J}

Final answer: About $4.2$ GJ of work is required to raise the satellite to $1000$ km altitude, ignoring the kinetic energy needed to keep it in orbit.

Try it: Escape velocity calculator - choose Earth, Moon, Mars, Jupiter or Sun (or enter custom values) and get $v_{\text{esc}}$ at the surface or any altitude.

Common traps

Forgetting the negative sign: $U = -G M m / r$ , not $+G M m / r$ . The sign matters for energy conservation.
Using $m g h$ for astronomical distances: This approximation only works for small altitude changes near a planet's surface where $g$ is roughly constant. For satellites and planetary orbits, use the radial formula.
Wrong reference point: Zero potential energy is at infinity, not at the planet's surface or centre.
Confusing escape velocity with orbital velocity: Orbital velocity at radius $r$ is $v_{\text{orb}} = \sqrt{G M / r}$ . Escape velocity is $\sqrt{2}$ times larger: $v_{\text{esc}} = \sqrt{2} \cdot v_{\text{orb}}$ .
Assuming escape velocity depends on the object's mass: It does not. A pebble and a rocket need the same escape velocity (though the rocket needs more total energy because $E = \frac{1}{2} m v^2$ ).

Examples in context

Example 1. Escape velocity for an ANU Mt Stromlo deep-space probe. A research probe is launched from Earth's surface ( $M_E = 5.97 \times 10^{24} \text{ kg}$ , $r_E = 6.37 \times 10^6 \text{ m}$ ). The escape speed is $v_{\text{esc}} = \sqrt{2 G M_E / r_E} = \sqrt{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24} / 6.37 \times 10^6} = 1.12 \times 10^4 \text{ m/s}$ (about $11.2 \text{ km/s}$ ). At the standard rocket exhaust speed of $4.4 \text{ km/s}$ , the Tsiolkovsky equation requires a mass ratio $e^{11.2/4.4} = 12.7$ , so a $1000 \text{ kg}$ probe needs $\sim 12.7$ tonnes of propellant. This is why Australian Mt Stromlo deep-space work relies on overseas launchers with multi-stage Saturn-class capacity.

Example 2. Lifting water in Snowy Hydro 2.0 pumped storage. Snowy 2.0 lifts water from Talbingo reservoir to Tantangara reservoir through a vertical head of $h = 700 \text{ m}$ . For each $1.0 \text{ kg}$ of water raised, the gain in gravitational PE is $\Delta U = m g h = 1.0 \times 9.8 \times 700 = 6.86 \times 10^3 \text{ J}$ . At a flow rate of $300 \text{ m}^3/\text{s}$ (so $3.0 \times 10^5 \text{ kg/s}$ ), the pumping power required is $P = 3.0 \times 10^5 \times 6.86 \times 10^3 = 2.06 \times 10^9 \text{ W} = 2.06 \text{ GW}$ , matching the station's nameplate generating capacity once the energy is later released through the Francis turbines.

Try this

Q1. Write the formula for gravitational potential energy of a mass $m$ at distance $r$ from a body of mass $M$ and explain the choice of zero. [2 marks]

Cue. $U = -G M m / r$ ; zero taken at $r = \infty$ , so all finite- $r$ values are negative.

Q2. A $250 \text{ kg}$ payload is moved from the surface of Earth to a circular orbit at altitude $500 \text{ km}$ . Calculate the change in gravitational PE. [4 marks]

Cue. $\Delta U = -G M_E m (1/r_2 - 1/r_1)$ with $r_1 = 6.37 \times 10^6$ , $r_2 = 6.87 \times 10^6$ . Compute $\Delta U \approx 1.14 \times 10^9 \text{ J}$ .

Q3. A $1500 \text{ kg}$ spacecraft is fired straight up at $9.0 \text{ km/s}$ from Earth's surface. (a) Calculate its total mechanical energy. (b) Find the maximum height above Earth's centre reached. (c) Determine whether it escapes Earth's gravity. [2+3+2 marks]

Cue. (a) $E = \tfrac{1}{2} m v^2 - G M_E m / r_E = 6.08 \times 10^{10} - 9.38 \times 10^{10} = -3.30 \times 10^{10} \text{ J}$ . (b) $r_{\max} = -G M_E m / E = 1.81 \times 10^7 \text{ m}$ . (c) $E < 0$ , does not escape.

Exam-style practice questions

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

2021 HSC4 marksCalculate the escape velocity from the surface of the Earth. (Mass of Earth = 5.97 x 10^24 kg, radius of Earth = 6.37 x 10^6 m, G = 6.67 x 10^-11 N m^2/kg^2.)

Show worked answer →

Escape velocity is the minimum speed an object needs at a distance $r$ from the centre of a mass $M$ to just barely escape to infinity with zero kinetic energy remaining.

Conservation of energy from the surface to infinity (where both $U$ and $K$ are zero):

$\frac{1}{2} m v_{\text{esc}}^2 - \frac{G M m}{r} = 0$ .

Solving for $v_{\text{esc}}$ :

$v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}$ .

Substituting:

$v_{\text{esc}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}}$
$v_{\text{esc}} = \sqrt{\frac{7.96 \times 10^{14}}{6.37 \times 10^6}}$
$v_{\text{esc}} = \sqrt{1.25 \times 10^8}$
$v_{\text{esc}} = 1.12 \times 10^4$ m/s, or about $11.2$ km/s.

Markers reward the derivation from conservation of energy, the correct identification of the boundary conditions (zero energy at infinity), and the final answer with units. Note that escape velocity does not depend on the mass of the escaping object.

2019 HSC3 marksExplain why gravitational potential energy in a radial field is taken to be negative, and why this is physically reasonable.

Show worked answer →

Gravitational potential energy in a radial field is defined as:

$U = -\frac{G M m}{r}$ .

The negative sign arises from the choice of zero potential energy at infinity (the natural reference point for radial fields). Because gravity is always attractive, the gravitational force does positive work on a mass as it moves from infinity inward, reducing its potential energy below zero.

Physically, this means a mass that is bound to a gravitational source (such as a planet in orbit) has less energy than a free mass at infinity. To escape the field, work must be done against gravity equal to $\frac{G M m}{r}$ , which raises the potential energy to zero at infinity.

The familiar formula $U = m g h$ is the linear approximation valid near a planet's surface where $g$ is approximately constant. At astronomical scales, $g$ varies with $r$ and the full radial formula must be used.

Markers reward the explicit reference point at infinity, the link between attractive force and negative potential energy, and the comparison with $mgh$ .