Escape Velocity
Launched on 19 January 2006, NASA's New Horizons is the fastest spacecraft ever launched at 58,000 km/h (16.1 km/s) — exceeding Earth's escape velocity of 11.2 km/s. It completed its Pluto flyby in July 2015 at 13.8 km/s and is now on a solar-system escape trajectory. Escaping the Sun entirely from Earth's distance requires 42.1 km/s — a speed no current spacecraft can reach by direct launch.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
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 $r$
- Classify trajectories as elliptical, parabolic or hyperbolic based on launch speed
Core Content
Using energy conservation to find the minimum launch speed
Throw a ball upward at 5 m/s: it rises about 1.3 m and falls back. Throw it at 50 m/s: it rises ~127 m and falls back. Launch New Horizons at 16.1 km/s (NASA, 19 January 2006): it never comes back. There is a precise speed threshold — 11.2 km/s from Earth's surface — below which any object (regardless of mass) must return, and above which any object escapes permanently with no further thrust needed. The derivation requires only energy conservation.
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, $E_\text{launch} = E_\text{infinity}$:
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).
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.
Which formula correctly gives the escape velocity from mass $M$ at radius $r$?
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 launch speed to escape. That raises a question: what trajectory does an object follow depending on whether its speed is below, at, or above $v_e$? This card answers it → elliptical (bound) below $v_e$, parabolic at $v_e$, hyperbolic above $v_e$.
The launch speed relative to $v_e$ determines the type of trajectory. This is a direct consequence of the total mechanical energy — negative means bound, zero means barely escaping, positive means escaping with residual speed.
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
Trajectory Types
| Speed | Trajectory | Outcome |
|---|---|---|
| $v < v_e$ | Elliptical or suborbital | Elliptical orbit (if tangential) or falls back to surface |
| $v = v_e$ | Parabolic | Just escapes; arrives at infinity with $v = 0$ |
| $v > v_e$ | Hyperbolic | Escapes with residual speed at infinity |
Earth's surface escape velocity is 11.2 km/s. Find the escape velocity at $r = 2R_E$ from Earth's centre.
Trajectory types: $v < v_e$ → elliptical (bound); $v = v_e$ → parabolic (just escapes, $v = 0$ at infinity); $v > v_e$ → hyperbolic (escapes with residual speed). $v_e \propto 1/\sqrt{r}$: escape velocity decreases with altitude. At $r = 2R$: $v_e = v_{e,\text{surface}}/\sqrt{2} \approx 7.9$ km/s.
Add the highlighted trajectory types and the $1/\sqrt{r}$ dependence to your notes before the check below.
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.
What happens when $v_e \geq c$?
We just saw that trajectory type depends on how $v$ compares to $v_e = \sqrt{2GM/r}$. That raises a question: what happens if we compress a mass until $v_e$ equals the speed of light? This card answers it → $v_e = c$ defines the Schwarzschild radius $R_s = 2GM/c^2$ — the event horizon of a black hole.
A profound implication of escape velocity: if we compress a mass until its escape velocity reaches or exceeds the speed of light $c$, nothing — not even light — can escape. This defines a black hole.
Setting $v_e = c$ in the escape velocity formula:
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.
Supermassive Black Holes
- M87* (imaged by the Event Horizon Telescope, 2019): $M = 6.5 \times 10^9\,M_\odot$, $R_s = 1.9 \times 10^{13}$ m $\approx 130$ AU
- Sagittarius A* (Milky Way centre): $M \approx 4 \times 10^6\,M_\odot$
Schwarzschild radius: set $v_e = c$ in escape velocity formula → $R_s = 2GM/c^2$. This is the event horizon radius — no light or matter escapes from $r < R_s$. For the Sun: $R_s \approx 3$ km. Black holes form when mass is compressed below $R_s$. Derived using Newtonian physics but result matches general relativity.
Add the highlighted Schwarzschild radius formula and derivation to your notes before the activities.
The Schwarzschild radius is the radius at which:
Earth's escape velocity is 11.2 km/s at the surface — but rockets don't launch at this speed. Instead, they ascend gradually using continuous thrust to minimise drag losses.
- 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 launched at ~16 km/s — the fastest Earth launch ever — to reach Pluto in 9 years
- 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: $\Delta E \approx mv\,\Delta v$
A planet's mass is doubled while its radius stays the same. The new escape velocity is:
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
Schwarzschild radius (enrichment): $R_s = 2GM/c^2$
Activities
Practise escape velocity calculations across different bodies
- Write the escape velocity formula $v_e = \sqrt{2GM/r}$ and explain the significance of the factor of 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)
- 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)
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?
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
- $v_e \propto \sqrt{M/r}$
Trajectory Classification
- $v < v_e$: elliptical (bound)
- $v = v_e$: parabolic (escapes, $v_\infty = 0$)
- $v > v_e$: hyperbolic (escapes, $v_\infty > 0$)
Enrichment
- Schwarzschild radius: $R_s = 2GM/c^2$
- When $v_e \geq c$: black hole
- Sun: $R_s \approx 3$ km
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?
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.
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. Explain why a satellite in a higher orbit has a lower orbital speed but a higher total mechanical energy. Use equations to support your reasoning.
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): Orbital speed $v = \sqrt{GM/r}$ decreases as $r$ increases [1 mark]. Total energy $E = -GMm/(2r)$ becomes less negative as $r$ increases — meaning the satellite has more total energy at greater altitude [1 mark]. This seems counterintuitive because moving to a higher orbit requires adding energy — the satellite gains potential energy faster than it loses kinetic energy, and the net result is a higher (less negative) total mechanical energy [1 mark].
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. $v_e = \sqrt{2} \cdot v_\text{orbital}$.
Trajectories
$v < v_e$: elliptical · $v = v_e$: parabolic ($v_\infty = 0$) · $v > v_e$: hyperbolic ($v_\infty > 0$).
Black Holes
$v_e = c$ at $R_s = 2GM/c^2$ (Schwarzschild radius). Nothing — not even light — escapes from within.
Rapid-fire questions on escape velocity derivation, trajectory classification, black holes and calculations. Destroy the incoming asteroids to bank your tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
Return to your Think First response. You should now be able to explain why NASA's New Horizons (launched 19 January 2006 at 16.1 km/s) is on a solar-system escape trajectory — and why 11.2 km/s is Earth's escape velocity.
Verify: $v_e = \sqrt{2GM/R} = \sqrt{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24} / 6.371 \times 10^6} = \sqrt{1.252 \times 10^8} = 11.2 \text{ km/s}$. New Horizons' 16.1 km/s launch speed exceeds this, so it escapes Earth. To escape the Sun from Earth's orbital distance requires 42.1 km/s — New Horizons used Jupiter's gravity (slingshot) to supplement its launch speed. The escape velocity formula is independent of the escaping object's mass because the $m$ terms cancel in the energy equation. That is why the derivation from energy conservation is one of the most elegant in HSC Physics.