Derive $v_e = \\sqrt{2GM/r}$ from energy conservation
Calculate escape velocity from any celestial body
Relate escape velocity to black hole formation
ESTIMATE — What speed would you need to escape Earth permanently? Give your best estimate with reasoning.
Escape Velocity
Derive escape velocity from energy conservation, calculate escape velocity from any body, and explore the concept of black holes and the Schwarzschild radius.
Deriving Escape Velocity
Escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body permanently, with no additional propulsion.
Escape Velocity
We derive it using the principle of conservation of energy. Consider a projectile of mass $m$ launched from the surface of a planet (mass $M$, radius $r$) with speed $v_e$:
- At launch (distance $r$ from centre): kinetic energy = $\\frac{1}{2}mv_e^2$, gravitational potential energy = $-\\frac{GMm}{r}$
- At infinity (just barely escaping): kinetic energy = $0$, gravitational potential energy = $0$
By conservation of energy:
Solving for $v_e$:
The factor of 2 is mandatory — it comes from needing enough kinetic energy to overcome all of the negative gravitational potential energy. The projectile must do twice as much work as would be needed merely to reach a zero-velocity state at infinity.
Properties of Escape Velocity
Key Characteristics
- Independent of projectile mass — the mass $m$ cancels in the derivation
- Depends only on $M$ and $r$ of the central body
- At distance $r$ from the centre (not necessarily the surface): $v_e = \\sqrt{2GM/r}$
Trajectory Types
| Speed | Trajectory | Outcome |
|---|---|---|
| $v < v_e$ | Elliptical or suborbital | Elliptical orbit (if tangential) or falls back |
| $v = v_e$ | Parabolic | Just escapes; arrives at infinity with $v = 0$ |
| $v > v_e$ | Hyperbolic | Escapes with excess speed at infinity |
Black Holes and the Schwarzschild Radius
A profound implication of escape velocity arises when we ask: What if the escape velocity equals or exceeds the speed of light?
If $v_e \\geq c$, then by Einstein's theory of relativity, nothing can escape — not even light. This defines a black hole.
Setting $v_e = c$ in the escape velocity formula:
This is the Schwarzschild radius — the radius of the event horizon of a black hole. It is the distance from the centre at which the escape velocity equals the speed of light.
The Event Horizon
The event horizon is the boundary at $r = R_s$. No information — not even light — can escape from within this boundary. The event horizon is not a physical surface but a boundary in spacetime.
Supermassive Black Holes
At the centres of most large galaxies lie supermassive black holes:
- M87* (imaged by the Event Horizon Telescope, 2019): $M = 6.5 \\times 10^9 \\ M_\\odot$, giving $R_s = 1.9 \\times 10^{13}$ m $\\approx 130$ AU
- Sagittarius A* (Milky Way centre): $M \\approx 4 \\times 10^6 \\ M_\\odot$
Common Misconceptions
"You need continuous thrust to escape"
Right: Once escape velocity is reached, coasting is sufficient (ignoring atmospheric drag). Rockets use continuous thrust for efficiency and trajectory control, not because it's physically required.
"Escape velocity depends on the mass of the projectile"
Right: $v_e = \\sqrt{2GM/r}$ — the projectile mass $m$ cancels out. A feather and a rocket need the same speed to escape.
"Black holes suck everything in"
Right: Objects outside the Schwarzschild radius orbit normally. Only objects that cross the event horizon are trapped. At Earth's distance, a solar-mass black hole exerts the same gravity as the Sun.
Rocket Launches
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 atmospheric lift and continuous thrust to minimise drag losses.
- Apollo 11 reached ~10.9 km/s (translunar injection) — just below Earth's escape velocity, as the Moon was the target
- New Horizons launched at ~16 km/s (the fastest launch ever) to reach Pluto in just 9 years
- The Oberth effect: burning engines at closest approach to a planet yields maximum energy change for a given $\\Delta v$
This is why deep-space probes use gravitational slingshots instead of brute-force launches — much more efficient than trying to reach $v_e$ directly from the surface.
Activities
Q1. Calculate the escape velocity from the surface of Mars. $(M_{\\text{Mars}} = 6.42 \\times 10^{23}$ kg, $R_{\\text{Mars}} = 3.40 \\times 10^6$ m)
Q2. Calculate the Schwarzschild radius for Earth $(M_E = 5.97 \\times 10^{24}$ kg).
Q3. What mass would cause Earth's Schwarzschild radius to equal its actual radius ($R_E = 6.37 \\times 10^6$ m)? In other words: if Earth were a black hole, what would its mass need to be?
Copy Into Books
Escape velocity derivation (energy conservation):
$\\frac{1}{2}mv_e^2 - \\frac{GMm}{r} = 0$
$v_e = \\sqrt{\\dfrac{2GM}{r}}$
Key properties: Independent of projectile mass; depends only on $M$ and $r$; $v_e = \\sqrt{2} \\times v_{\\text{orbital}}$ at same radius.
Schwarzschild radius:
$R_s = \\dfrac{2GM}{c^2}$
Values to remember: $v_e$ (Earth) = 11.2 km/s; $R_s$ (Sun) = 3 km.
Compare your initial estimate to the actual value of 11.2 km/s. Were you close? What factors (mass of Earth, distance from centre) does escape velocity actually depend on?
Multiple Choice
1. Escape velocity depends on:
2. If a planet's mass doubles and radius stays the same, $v_e$:
3. The Schwarzschild radius is the distance where:
4. Compared to orbital velocity at the same $r$, escape velocity is:
5. A black hole with $M = 10 \\ M_\\odot$ has $R_s$ approximately:
Short Answer
Q1. 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? Calculate its excess speed at infinity.
Q2. Calculate the Schwarzschild radius for: (a) the Sun, (b) a star of 3 solar masses, (c) the supermassive black hole at the centre of the Milky Way $(4 \\times 10^6 \\ M_\\odot)$.
Q3. Assess the statement: "If the Sun collapsed to a black hole of the same mass, Earth's orbit would be unaffected." Use your knowledge of gravitational forces and the Schwarzschild radius to evaluate this claim.
SA1. $v_e = \\sqrt{\\frac{2 \\times 6.67 \\times 10^{-11} \\times 4.87 \\times 10^{24}}{6.05 \\times 10^6}} = \\sqrt{1.07 \\times 10^8} = 1.04 \\times 10^4$ m/s = 10.4 km/s.
Since 12 km/s > 10.4 km/s, the probe will escape.
Using energy conservation: $\\frac{1}{2}v^2 - \\frac{GM}{R} = \\frac{1}{2}v_\\infty^2$
$v_\\infty = \\sqrt{12^2 - 10.4^2} = \\sqrt{144 - 108.2} = \\sqrt{35.8} = \\boxed{5.99$ km/s (or ~6.06 km/s depending on rounding)
SA2. (a) $R_s = \\frac{2 \\times 6.67 \\times 10^{-11} \\times 1.99 \\times 10^{30}}{(3.00 \\times 10^8)^2} = \\boxed{2950$ m (~3 km)
(b) $R_s = 3 \\times 2950 = \\boxed{8850$ m (~8.9 km)
(c) $R_s = 4 \\times 10^6 \\times 2950 = 1.18 \\times 10^{10}$ m = $\\boxed{1.18 \\times 10^7$ km
SA3. This statement is TRUE.
The gravitational force depends only on $F = \\frac{GMm}{r^2}$. If the Sun collapsed to a black hole of the same mass $M$, the mass does not change — only its radius decreases to $R_s \\approx 3$ km.
At Earth's orbital radius (1 AU), the gravitational force would be identical to what it is now, because $r$ (Earth's orbital radius) is unchanged and $M$ is unchanged.
Earth's orbit would continue exactly as before. The difference would only be apparent for objects very close to the collapsed Sun, where tidal forces ("spaghettification") would be extreme. At 1 AU, these tidal forces are negligible.
Therefore the claim is correct — Earth's orbit would be unaffected.
Test your escape velocity knowledge in the boss battle challenge!
Boss Battle
Escape Velocity Challenge
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