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Physics · Gravitation

Escape Velocity Calculator

Find the launch speed needed to break free of any body with v = √(2GM ÷ r).

kg
m
Real bodies — tap to load
Escape velocity (v)
11.19km/s

That is 11,186 m/s. G = 6.674×10⁻¹¹ N·m²/kg²; enter M in kg and r in metres (use e-notation, e.g. 5.972e24).

Escape velocity of some bodies (km/s)

Escape velocity is the minimum speed to leave a body’s gravity without further thrust: v = √(2GM ÷ r), with G = 6.674×10⁻¹¹ N·m²/kg². For Earth (M = 5.972×10²⁴ kg, r = 6.371×10⁶ m) it works out to about 11.2 km/s — roughly 40,000 km/h.

What escape velocity means

Escape velocity is the speed an object needs at a body’s surface so that its kinetic energy exactly cancels the gravitational potential binding it — after which it coasts away and never falls back, assuming no air drag and no extra propulsion. Setting ½mv² equal to GMm ÷ r and cancelling the object’s mass m gives v = √(2GM ÷ r). It is a scalar: direction does not matter, only speed, because gravity is conservative.

v = √(2GM ÷ r)

G = 6.674×10⁻¹¹ N·m²/kg², M is the body’s mass in kg, r its radius in metres, v in m/s

Worked example

What is Earth’s escape velocity? Use M = 5.972×10²⁴ kg and r = 6.371×10⁶ m (Earth’s mean radius).

  1. 1
    Write the formula. v = √(2GM ÷ r), with G = 6.674×10⁻¹¹ N·m²/kg².
  2. 2
    Substitute the values. v = √(2 × 6.674×10⁻¹¹ × 5.972×10²⁴ ÷ 6.371×10⁶).
  3. 3
    Evaluate inside the root. 2GM ÷ r ≈ 1.251×10⁸ m²/s².
  4. 4
    Take the square root. v ≈ 11,187 m/s ≈ 11.19 km/s — the speed to escape Earth from its surface.

Escape velocity of common bodies

Computed from each body’s mass and mean radius with v = √(2GM ÷ r).

BodyEscape velocity (km/s)Escape velocity (m/s)
Moon2.382,375
Mars5.035,027
Earth11.1911,186
Jupiter59.559,500
Sun617.5617,500

Why the numbers look the way they do

It does not depend on the escaping object’s mass. Because the object’s mass m cancels out of ½mv² = GMm ÷ r, a pebble and a spacecraft need the same escape speed from the same body. Only the central body’s mass M and radius r matter.

Bigger and denser bodies demand more. Escape velocity rises with √M and falls with √r. More mass deepens the gravity well; a smaller radius puts the surface closer to the centre, where the pull is stronger. That is why the Sun — enormous mass despite its large radius — needs about 617 km/s, while the small, low-mass Moon needs only 2.4 km/s.

Does escape velocity depend on the rocket’s mass?
No. The object’s mass cancels out of the energy equation ½mv² = GMm ÷ r, so escape velocity depends only on the central body’s mass M and radius r. A marble and a spaceship need the same 11.2 km/s to escape Earth’s surface.
Why is the Sun’s escape velocity so high?
Escape velocity grows with the square root of mass, and the Sun holds about 330,000 Earth masses. Even though its radius is large, that immense mass makes its surface escape velocity roughly 617.5 km/s — over 50 times Earth’s.
What is Earth’s escape velocity?
About 11.19 km/s (≈11,186 m/s, or roughly 40,000 km/h) from the surface, using M = 5.972×10²⁴ kg and r = 6.371×10⁶ m. This is the speed needed to coast away from Earth with no further thrust.
How is escape velocity different from orbital velocity?
Orbital velocity keeps you circling a body; escape velocity lets you leave it entirely. For a circular orbit at the surface, escape velocity is exactly √2 (about 1.414) times the orbital speed, so Earth’s low-orbit speed near 7.9 km/s becomes 11.2 km/s to escape.
Does the launch direction matter?
For the speed itself, no — escape velocity is a scalar because gravity is conservative, so only the magnitude of the velocity counts. In practice, launching eastward near the equator borrows Earth’s rotation speed and lowers the fuel needed.
Why does escape velocity ignore air resistance?
The formula v = √(2GM ÷ r) comes purely from energy conservation in a vacuum. Real rockets fight atmospheric drag and rarely reach escape speed at the surface; instead they climb and accelerate gradually, but the airless value still sets the energy target.