Escape Velocity & Black Holes

Launch fast enough and you never come back. Squeeze a star enough and even light can't leave.

11 minUpdated 21 October 2026escape velocitybinding energyblack holeequivalence principle

Lesson at a glanceShow the key points

A body escapes if ½mu² − GMm/R ≥ 0, so the escape velocity is √(2GM/R) (11.27): about 11.2 km/s from the Earth.
Escape velocity does not depend on the mass of the escaping body or (ignoring air) on the launch direction.
Binding energy of a body at rest on the surface: GMm/R, the minimum energy to free it.
If √(2GM/R) ≥ c, nothing escapes: a black hole. Critical radius 2GM/c² (≈ 9 mm for the Earth's mass).
Inertial mass (from F = ma) and gravitational mass (from F = GMm/r²) are found experimentally to be equal; general relativity is built on this.

Throw a ball up and it comes back. Throw it harder and it goes higher. Is there a speed beyond which it never returns? Energy says yes, and it says the speed is the same whatever you throw.

The escape condition

Launch a body of mass m from the surface of a planet (mass M, radius R) at speed u. Ignoring air, energy is conserved. At height h its speed v satisfies

\tfrac12 mv^2 = \Big(\tfrac12 mu^2 - \frac{GMm}{R}\Big) + \frac{GMm}{R+h}

The last term is positive but shrinks toward zero as h grows. If the bracket is zero or positive, v² stays positive forever: the body never stops, and so never turns back. The threshold is

Escape velocity(11.27)

\tfrac12 mu^2 \ge \frac{GMm}{R} \quad\Rightarrow\quad v_{\text{esc}} = \sqrt{\frac{2GM}{R}}

about 11.2 km/s for the Earth

Interactive

Will it come back?

Fire a probe straight up (ignoring air). On the energy diagram the curve is U(r) = −GMm/r and the flat line is the total energy. The gap between them is kinetic energy. If the total energy is below zero the probe turns round where the line meets the curve; at or above zero it never returns.

escape speed from Earth11.2km/s

launch speed8.9km/s

highest point11326 km up

2GM/c^2

8.85 × 10^-3m

World

Launch speed0.80× v_esc

Worked example

Leaving Mars

Mars has mass 6.42 × 10²³ kg and radius 3390 km. Find the escape velocity from its surface.

Binding energy

A body at rest on the surface has total energy $-GMm/R$ . The minimum energy that frees it completely (brings its total to zero) is called the binding energy of the planet–body system:

E_{\text{bind}} = \frac{GMm}{R}

For a satellite already in a circular orbit of radius a, the binding energy is smaller: $GMm/2a$ .

Black holes

Imagine squeezing a star while keeping its mass fixed. $\sqrt{2GM/R}$ rises as R falls. When it reaches the speed of light c, nothing launched from the surface can escape, and since nothing travels faster than light, nothing escapes at all, not even light. Setting $\sqrt{2GM/R} = c$ :

Critical radius

R_s = \frac{2GM}{c^2}

the Schwarzschild radius (general relativity gives the same formula)

Worked example

A black-hole Earth

To what radius would the Earth (6.0 × 10²⁴ kg) have to be squeezed to become a black hole?

Astronomers have found many black holes: collapsed remnants of massive stars, and giants of millions to billions of solar masses at the centres of galaxies.

Inertial and gravitational mass

There are two independent ways to compare the masses of two bodies:

Inertial mass: apply equal forces and compare accelerations, $m_A/m_B = a_B/a_A$ . This is the m in F = ma.
Gravitational mass: compare the gravitational pulls the Earth exerts on them, $m_A/m_B = F_A/F_B$ . This is what a spring balance measures.

Nothing in Newton’s theory requires these to agree, yet every experiment finds them equal to extraordinary precision. That is why all bodies fall with the same acceleration. Einstein made this equivalence the foundation of general relativity.

What to remember

1 $v_{\text{esc}} = \sqrt{2GM/R}$ ≈ 11.2 km/s for the Earth, independent of the escaping mass.
2Binding energy at the surface: $GMm/R$ .
3Black hole: $R \le 2GM/c^2$ ; even light cannot escape.
4Inertial and gravitational mass are experimentally identical: the root of the equivalence principle.

Check yourself

How does escape velocity compare with the speed of a low circular orbit just above the surface?

A planet has the Earth's density but twice its radius. Its escape velocity is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchEscape velocities, circular orbits, bound and unbound orbitsWalter Lewin · MIT 8.01x Lecture 14 ReadEscape velocityHyperPhysics ReadEscape velocityWikipedia ReadSchwarzschild radiusWikipedia ReadEquivalence principleWikipedia ReadEinstein's theory of gravityOpenStax University Physics §13.7 ReadFifth forceWikipedia

PreviousSatellites, Orbits & Kepler's Laws Next chapter Simple Harmonic Motion

Escape Velocity & Black Holes

Launch fast enough and you never come back. Squeeze a star enough and even light can't leave.

11 minUpdated 21 October 2026escape velocitybinding energyblack holeequivalence principle

Lesson at a glanceShow the key points

A body escapes if ½mu² − GMm/R ≥ 0, so the escape velocity is √(2GM/R) (11.27): about 11.2 km/s from the Earth.
Escape velocity does not depend on the mass of the escaping body or (ignoring air) on the launch direction.
Binding energy of a body at rest on the surface: GMm/R, the minimum energy to free it.
If √(2GM/R) ≥ c, nothing escapes: a black hole. Critical radius 2GM/c² (≈ 9 mm for the Earth's mass).
Inertial mass (from F = ma) and gravitational mass (from F = GMm/r²) are found experimentally to be equal; general relativity is built on this.

Throw a ball up and it comes back. Throw it harder and it goes higher. Is there a speed beyond which it never returns? Energy says yes, and it says the speed is the same whatever you throw.

The escape condition

Launch a body of mass m from the surface of a planet (mass M, radius R) at speed u. Ignoring air, energy is conserved. At height h its speed v satisfies

\tfrac12 mv^2 = \Big(\tfrac12 mu^2 - \frac{GMm}{R}\Big) + \frac{GMm}{R+h}

The last term is positive but shrinks toward zero as h grows. If the bracket is zero or positive, v² stays positive forever: the body never stops, and so never turns back. The threshold is

Escape velocity(11.27)

\tfrac12 mu^2 \ge \frac{GMm}{R} \quad\Rightarrow\quad v_{\text{esc}} = \sqrt{\frac{2GM}{R}}

about 11.2 km/s for the Earth

Interactive

Will it come back?

escape speed from Earth11.2km/s

launch speed8.9km/s

highest point11326 km up

2GM/c^2

8.85 × 10^-3m

World

Launch speed0.80× v_esc

Worked example

Leaving Mars

Mars has mass 6.42 × 10²³ kg and radius 3390 km. Find the escape velocity from its surface.

Binding energy

A body at rest on the surface has total energy $-GMm/R$ . The minimum energy that frees it completely (brings its total to zero) is called the binding energy of the planet–body system:

E_{\text{bind}} = \frac{GMm}{R}

For a satellite already in a circular orbit of radius a, the binding energy is smaller: $GMm/2a$ .

Black holes

Critical radius

R_s = \frac{2GM}{c^2}

the Schwarzschild radius (general relativity gives the same formula)

Worked example

A black-hole Earth

To what radius would the Earth (6.0 × 10²⁴ kg) have to be squeezed to become a black hole?

Astronomers have found many black holes: collapsed remnants of massive stars, and giants of millions to billions of solar masses at the centres of galaxies.

Inertial and gravitational mass

There are two independent ways to compare the masses of two bodies:

Inertial mass: apply equal forces and compare accelerations, $m_A/m_B = a_B/a_A$ . This is the m in F = ma.
Gravitational mass: compare the gravitational pulls the Earth exerts on them, $m_A/m_B = F_A/F_B$ . This is what a spring balance measures.

What to remember

1 $v_{\text{esc}} = \sqrt{2GM/R}$ ≈ 11.2 km/s for the Earth, independent of the escaping mass.
2Binding energy at the surface: $GMm/R$ .
3Black hole: $R \le 2GM/c^2$ ; even light cannot escape.
4Inertial and gravitational mass are experimentally identical: the root of the equivalence principle.

Check yourself

How does escape velocity compare with the speed of a low circular orbit just above the surface?

A planet has the Earth's density but twice its radius. Its escape velocity is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousSatellites, Orbits & Kepler's Laws Next chapter Simple Harmonic Motion