Year 12 Physics Module 5 · IQ3: Gravitational Fields 45 min Practice bank · 5 MC Lesson 16 of 18

Escape Velocity

New Horizons left Earth in 2006 at about 16.1 km/s relative to Earth. That exceeded Earth's ideal surface escape-speed threshold, but escaping Earth and escaping the Sun are different energy questions in different reference frames. This lesson builds the model carefully.

Today's hook: Earth's surface escape speed is 11.2 km/s, yet real launch vehicles do not receive one instantaneous 11.2 km/s kick. What exactly does the textbook threshold mean, and how does the ideal model differ from a powered mission?
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Define the escape model before calculating

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

Three printable resources build from foundations to mixed exam-style practice.

Build Foundations & guided practice Apply Application & data response Exam Mixed calculations & exam-style response
Think First: Escaping Earth
warm-up

Before reading, estimate: what speed would you need to escape Earth permanently, with no further propulsion after launch? Give your best estimate with reasoning. Does the mass of the projectile matter?

Learning Intentions
goals

Know, Escape Velocity Formula

  • State $v_e = \sqrt{2GM/r}$ and know it comes from setting total energy to zero
  • Recall Earth's surface escape velocity: 11.2 km/s

Understand, Energy Conservation Derivation

  • Derive $v_e$ from $\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0$
  • Explain why escape velocity is independent of projectile mass

Can Do, Apply and Analyse

  • Calculate escape velocity from any celestial body given $M$ and $r$
  • Use total-energy sign to classify bound, threshold and unbound motion
Scan these before reading
vocab
Escape velocityThe minimum initial speed for an object to escape a gravitational field permanently without further thrust.
Reference at infinityThe convention $U=0$ as $r\to\infty$, so bound states have negative total mechanical energy.
Threshold energyAt the minimum escape speed, $E_\text{total}=0$ and $v\to0$ as $r\to\infty$.
Residual speedThe non-zero speed $v_\infty$ retained by an object with positive total mechanical energy.
Cross-lesson links: L15 mapped the gravitational potential field. L16 applies it to escape, the escape velocity derivation (setting total energy to zero) is one of the most elegant derivations in HSC Physics and directly tests your understanding of the negative potential energy from L14.
Core model and reference convention

Assume no propulsion after launch, no atmosphere, a spherical isolated central mass $M$, a much smaller projectile mass $m$, and Newtonian gravity. Measure $r$ from the central mass's centre and set $U=0$ at infinity.

The formula gives a minimum speed at a specified radius, not one required direction. Direction changes the path and collision risk, but not the energy threshold at that point.

Misconceptions to fix
✗ Wrong: You need continuous thrust to maintain escape from gravity.
✓ Right: In the ideal ballistic model, escape speed is the minimum initial speed with no later thrust. At exactly the threshold, the object coasts outward and its speed approaches zero at infinity.
✗ Wrong: A heavier rocket needs a higher escape velocity than a small probe.
✓ Right: $v_e = \sqrt{2GM/r}$, the projectile mass $m$ cancels. A marble and a spaceship launched from the same point need identical initial speeds to escape.
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Derive the zero-energy threshold

Core Content

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Deriving Escape Velocity
+5 XP

Using energy conservation to find the minimum launch speed

In the stated ideal model, energy conservation produces a precise threshold. From Earth's surface it is about 11.2 km/s. A launch below the threshold has negative total energy; at the threshold the total energy is zero; above it the object retains kinetic energy at infinity.

Escape velocity diagram showing a projectile launched from a planet surface with energy conservation conditions

Energy conservation: at launch, $E_\text{total} = \frac{1}{2}mv_e^2 - \frac{GMm}{r}$. At infinity, both KE and PE are zero.

Consider a projectile of mass $m$ launched from the surface of a planet (mass $M$, radius $r$) with speed $v_e$:

  • At launch: $KE = \frac{1}{2}mv_e^2$, $U = -\frac{GMm}{r}$
  • At infinity (just barely escaping): $KE = 0$, $U = 0$

By conservation of energy, $E_\text{launch} = E_\text{infinity}$:

$$\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0$$
$$\frac{1}{2}mv_e^2 = \frac{GMm}{r} \quad \Rightarrow \quad v_e^2 = \frac{2GM}{r} \quad \Rightarrow \quad \boxed{v_e = \sqrt{\frac{2GM}{r}}}$$
The factor of 2 is mandatory: the launch kinetic energy must equal the magnitude of the negative gravitational potential energy. The projectile mass $m$ cancels, so the ideal threshold is independent of projectile mass.
HSC Trap
$v = \sqrt{GM/r}$ is orbital speed; $v_e = \sqrt{2GM/r}$ is escape speed. They differ by $\sqrt{2} \approx 1.41$. Writing $2\sqrt{GM/r}$ is also wrong, the 2 is inside the square root.
Worked Example, Escape Velocity from Earth and Jupiter

Calculate the escape velocity from the surface of Earth ($M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m) and Jupiter ($M_J = 1.90 \times 10^{27}$ kg, $R_J = 7.15 \times 10^7$ m).

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Formula: $v_e = \sqrt{2GM/R}$
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Earth: $$v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}} = \sqrt{1.25 \times 10^8} = 1.12 \times 10^4 \text{ m/s}$$
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Jupiter: $$v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 1.90 \times 10^{27}}{7.15 \times 10^7}} = \sqrt{3.54 \times 10^9} = 5.95 \times 10^4 \text{ m/s}$$
Answers: Earth $v_e \approx 11.2$ km/s · Jupiter $v_e \approx 59.5$ km/s

Escape velocity derivation: set $E_\text{total} = 0$: $\tfrac{1}{2}mv_e^2 - GMm/r = 0 \Rightarrow v_e = \sqrt{2GM/r}$. Factor of 2 is INSIDE the root (not $2\sqrt{GM/r}$). Projectile mass $m$ cancels, $v_e$ depends only on central body $M$ and $r$. Earth's surface: $v_e = 11.2$ km/s. $v_e = \sqrt{2}\,v_\text{orbital}$ at same $r$.

Pause, copy the highlighted escape velocity derivation and the $\sqrt{2}$ rule into your book before moving on.

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Compare orbit and escape, then classify energy

Which formula correctly gives the escape velocity from mass $M$ at radius $r$?

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Properties of Escape Velocity and Trajectory Types
+5 XP

Key characteristics and what happens when $v$ is less than, equal to, or greater than $v_e$

We just saw that $v_e = \sqrt{2GM/r}$ is the minimum speed for zero total energy. That raises a question: what does a speed below, at or above $v_e$ tell us? This card answers it → negative energy is bound, zero energy is the escape threshold and positive energy is unbound.

Energy sign gives the robust classification. The familiar elliptical, parabolic and hyperbolic path labels apply to ideal, non-collision inverse-square two-body motion; path shape also depends on launch direction and angular momentum.

Key Characteristics

  • Independent of projectile mass$m$ cancels in the derivation
  • Depends only on $M$ and $r$ of the central body
  • At any distance $r$ from the centre (not just the surface): $v_e = \sqrt{2GM/r}$
  • $v_e = \sqrt{2} \times v_\text{orbital}$ at the same radius

Energy and trajectory classification

Applicability: the conic labels below assume an isolated two-body system under inverse-square gravity. A purely radial launch is a degenerate limiting case, and a path that intersects the body's surface ends in a collision.

Speed Trajectory Outcome
$v < v_e$ Negative energy: bound Ellipse for a non-collision orbit; otherwise returns or collides
$v = v_e$ Zero energy: threshold Parabolic escape when angular momentum is non-zero; $v\to0$ at infinity
$v > v_e$ Positive energy: unbound Hyperbolic escape when angular momentum is non-zero; $v_\infty>0$
Interactive, escape-speed calculator

Change the central body's mass and launch radius. Check that increasing $M$ raises $v_e$, while increasing $r$ lowers it.

Open fullscreen ↗

Worked Example, Escape Velocity at 2 Earth Radii

Earth's surface escape velocity is 11.2 km/s. Find the escape velocity at $r = 2R_E$ from Earth's centre.

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Formula at $r = 2R_E$: $v_e = \sqrt{2GM/(2R_E)} = \sqrt{1/2} \times \sqrt{2GM/R_E} = v_{e,\text{surface}}/\sqrt{2}$
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Calculate: $$v_e = \frac{11.2}{\sqrt{2}} = \frac{11.2}{1.414} \approx 7.92 \text{ km/s}$$
Answer: $v_e \approx 7.9$ km/s at two Earth radii, escape velocity decreases with altitude.

Energy classification: $v < v_e$ → $E<0$, bound; $v = v_e$ → $E=0$, threshold escape with $v_\infty=0$; $v > v_e$ → $E>0$, unbound with $v_\infty>0$. In ideal non-collision two-body motion, these correspond to elliptical, parabolic and hyperbolic conics. $v_e \propto 1/\sqrt{r}$.

Add the highlighted trajectory types and the $1/\sqrt{r}$ dependence to your notes before the check below.

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Separate the core model from optional enrichment

Assessability boundary

Core HSC work in this lesson: derive and apply $v_e=\sqrt{2GM/r}$, state the model assumptions, compare escape with circular-orbit speed, and classify motion using total-energy sign.

Optional enrichment: the Schwarzschild-radius discussion below. It is not assessed in this lesson's Practice or Review bank.

A projectile launched at $v = v_e$ arrives at infinity with some residual kinetic energy.

The escape velocity at twice the orbital radius is lower than at the surface.

A more massive rocket requires a higher escape velocity than a small satellite from the same location.

Optional enrichment: black holes and the Schwarzschild radius
Beyond the syllabus. Escape velocity and its energy derivation are core — keep the formula and the calculations. The Schwarzschild radius and black-hole interpretation are extension: a fascinating consequence, not exam content.
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Black Holes and the Schwarzschild Radius (Enrichment)
optional · not assessed

What happens when $v_e \geq c$?

A historical Newtonian-style heuristic sets $v_e=c$. It produces $R_s=2GM/c^2$, the same algebraic radius that appears in the Schwarzschild solution of general relativity.

Model limitation: this is not a derivation of a black hole. Newtonian mechanics does not describe light as a massive projectile or gravity as curved spacetime; the event horizon is a general-relativistic result.

Setting $v_e = c$ in the escape velocity formula:

$$c = \sqrt{\frac{2GM}{R_s}} \quad \Rightarrow \quad c^2 = \frac{2GM}{R_s} \quad \Rightarrow \quad \boxed{R_s = \frac{2GM}{c^2}}$$
The heuristic reproduces the Schwarzschild radius algebraically. For a non-rotating, uncharged mass with the Sun's mass, $R_s \approx 2.95$ km.

The Event Horizon

The event horizon at $r = R_s$ is not a physical surface but a boundary in spacetime. No information, not even light, can escape from within this boundary.

Enrichment only: setting the Newtonian escape-speed expression equal to $c$ gives $R_s=2GM/c^2$. This numerical agreement is a heuristic, not a general-relativity derivation.

Real World, Rocket Launches and Gravity Assists

Earth's 11.2 km/s value is an ideal instantaneous ballistic threshold at the surface in an Earth-centred frame. Real missions use staged powered flight and must account for atmosphere, gravity losses, Earth's rotation, launch direction and the destination's reference frame.

  • Apollo 11 reached ~10.9 km/s (translunar injection), just below Earth's escape velocity, since the Moon was the destination, not infinity
  • New Horizons departed Earth at about 16.1 km/s relative to Earth; its later heliocentric trajectory cannot be inferred by comparing that number directly with the Sun's escape speed
  • The Oberth effect: engine burns at closest approach to a planet yield maximum energy gain. The same $\Delta v$ produces a larger kinetic energy increase when orbital speed is highest: $\Delta E \approx mv\,\Delta v$
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Apply the threshold and state its limits

A planet's mass is doubled while its radius stays the same. The new escape velocity is:

Essential formulae, escape velocity

Escape velocity: $v_e = \sqrt{\dfrac{2GM}{r}}$ (factor of 2 inside root)

Relation to orbital speed: $v_e = \sqrt{2} \times v_\text{orbital}$ at same radius

Earth's value: $v_e \approx 11.2$ km/s from the surface

Model: no later thrust, no atmosphere, isolated spherical mass; $U=0$ at infinity

Activities

Activity 1, Escape Velocity Drills
ApplyBand 3

Practise escape velocity calculations across different bodies

  1. Write the escape velocity formula $v_e = \sqrt{2GM/r}$ and explain the significance of the factor of 2.
  2. Calculate the escape velocity from Mars. ($M_\text{Mars} = 6.42 \times 10^{23}$ kg, $R_\text{Mars} = 3.40 \times 10^6$ m, $G = 6.67 \times 10^{-11}$ N m²/kg²)
  3. A spacecraft is launched from Earth at 14 km/s. Show whether this exceeds Earth's escape velocity, then calculate its speed at infinity using energy conservation. ($v_e = 11.2$ km/s)
Activity 2, Independence of Projectile Mass
UnderstandBand 4

Starting from the energy conservation equation $\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0$, show algebraically that the projectile mass $m$ cancels. Hence explain why a marble and a rocket launched from the same point on Earth's surface need the same initial speed to escape. Why is this result physically reasonable?

Copy Into Books, Key Formulas

Derivation

  • Set $E_\text{total} = 0$: $\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0$
  • $m$ cancels on both sides
  • $v_e = \sqrt{\dfrac{2GM}{r}}$

Key Relations

  • $v_e = \sqrt{2} \times v_\text{orbital}$ at same $r$
  • Earth: $v_e \approx 11.2$ km/s
  • $v_e \propto \sqrt{M/r}$

Energy Classification

  • $v < v_e$: $E<0$, bound
  • $v = v_e$: $E=0$, threshold
  • $v > v_e$: $E>0$, unbound

Conditions

  • No propulsion after launch
  • No atmosphere; isolated spherical mass
  • $U=0$ at infinity; state the reference frame
Revisit Your Thinking

Compare your initial estimate to the actual value of 11.2 km/s. Were you close? You now know that escape velocity depends only on $M$ and $r$ of the central body, not the projectile's mass. How did that change your thinking?

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Complete the core escape-speed assessment

01
Multiple Choice
+5 XP

A fresh set drawn from this lesson's question 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.

02
Short Answer, 10 marks
+5 XP

ApplyBand 4(3 marks) 1. Calculate the escape velocity from the surface of Venus ($M = 4.87 \times 10^{24}$ kg, $R = 6.05 \times 10^6$ m). A probe is launched at 12 km/s, will it escape? If yes, calculate its excess speed at infinity.

ApplyBand 5(4 marks) 2. Compare the escape velocities from the surface of Earth and the Moon. Calculate both values and explain which is larger and why.

Earth: $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m · Moon: $M_M = 7.34 \times 10^{22}$ kg, $R_M = 1.74 \times 10^6$ m

EvaluateBand 5–6(3 marks) 3. A probe has speed $0.90v_e$ at radius $r$ from an isolated planet. Classify its total energy and motion. Explain why calling every such path “an ellipse” needs an additional condition.

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.

Short Answer, Model Answers

SA1 (3 marks): $v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 4.87 \times 10^{24}}{6.05 \times 10^6}} = \sqrt{1.07 \times 10^8} = 1.04 \times 10^4$ m/s $\approx$ 10.4 km/s [1 mark]. Since 12 km/s > 10.4 km/s, the probe will escape [1 mark]. Speed at infinity: $v_\infty = \sqrt{v^2 - v_e^2} = \sqrt{12^2 - 10.4^2} = \sqrt{144 - 108.2} = \sqrt{35.8} \approx$ 5.99 km/s [1 mark].

SA2 (4 marks): $v_e(\text{Earth}) = \sqrt{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24} / 6.37 \times 10^6} \approx$ 11.2 km/s [1 mark]. $v_e(\text{Moon}) = \sqrt{2 \times 6.67 \times 10^{-11} \times 7.34 \times 10^{22} / 1.74 \times 10^6} \approx$ 2.38 km/s [1 mark]. Earth's escape velocity is larger because Earth has a much greater mass (roughly 81× more) while its radius is only about 3.7× larger, the combined effect of $\sqrt{M/r}$ favours Earth [2 marks: identifying mass difference + correct ratio argument].

SA3 (3 marks): Since $0.90v_e<v_e$, the probe has negative total mechanical energy [1 mark] and is gravitationally bound in the ideal model [1 mark]. A non-collision elliptical orbit requires suitable non-zero angular momentum; a radial path is a degenerate case and a trajectory intersecting the planet ends in collision [1 mark].

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Retrieve the core model and check your frame

Check what actually stuck
RAPID REVIEW
The big ideas in four tiles

Derivation

Set $E_\text{total} = 0$: $\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0$. Mass $m$ cancels, independent of projectile mass.

Formula

$v_e = \sqrt{2GM/r}$, the 2 is inside the root. Earth surface: 11.2 km/s. $v_e = \sqrt{2} \cdot v_\text{orbital}$.

Energy sign

$v < v_e$: bound · $v = v_e$: threshold ($v_\infty = 0$) · $v > v_e$: unbound ($v_\infty > 0$).

Conditions

No later thrust, no atmosphere, isolated spherical mass, Newtonian gravity and $U=0$ at infinity. State the central body and reference frame.

Asteroid Blaster, Escape Velocity
boss

Rapid-fire questions on the escape-speed derivation, energy classification, model conditions and calculations. Optional black-hole enrichment is excluded from the core pool.

How did your thinking change?

Return to your Think First response. You should now be able to explain what Earth's 11.2 km/s threshold means and why it is independent of the projectile's mass.

Verify: $v_e = \sqrt{2GM/R} = \sqrt{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24} / 6.371 \times 10^6} = 11.2 \text{ km/s}$. A speed quoted relative to Earth can be compared with Earth's threshold; a claim about escape from the Sun requires heliocentric position, velocity and energy. Do not mix those reference frames.

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