About
Escape velocity is the minimum speed to leave a body's gravity without further propulsion. Earth: 11.2 km/s (25,000 mph). Moon: 2.4 km/s. Sun (from surface): 618 km/s. Black hole = c (speed of light).
Formula
v_e = √(2GM/r); G = 6.674×10⁻¹¹
What is Escape Velocity Calculator?
A Escape Velocity Calculator computes escape velocity from the inputs you provide. It applies the standard formula to the values you enter and returns the result instantly, without sending any data to a server. Free Escape Velocity Calculator. The tool.
Escape Velocity Calculator
v = √(2GM/r). Earth: 11.2 km/s. Need more for solar system escape.
Inputs
kg (Earth: 5.972e24)
meters (Earth: 6.371e6)
Escape Velocity
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Breakdown
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About escape velocity
Escape velocity is the minimum speed an unpowered object must reach at the surface of a planet, moon, or star so that it coasts away forever and never falls back, even though gravity keeps slowing it down. It is the dividing line between a ballistic arc that returns to the ground and a trajectory that breaks free of the body entirely.
The concept comes straight from Newtonian mechanics and the conservation of energy. The same idea, pushed to its limit, defines a black hole's event horizon, the radius at which escape velocity reaches the speed of light. Because the formula depends only on the central body's mass and radius, it is a clean, exact result that astronomers, mission planners, and physics students all use as a baseline before adding atmospheric drag and orbital mechanics.
How it works: the formula
Escape velocity is found by setting an object's kinetic energy equal to the gravitational potential energy binding it to the body. The launching object's own mass cancels, which is why a marble and a rocket share the same escape speed from a given surface.
v_e = sqrt( 2 G M / r ) G = 6.674 x 10^-11 N m^2 / kg^2 (gravitational constant) M = mass of the body (kg) r = distance from the body's center (m)
- Derivation: (1/2) m v_e squared = G M m / r. The object mass m divides out, giving v_e = sqrt(2GM/r).
- Surface escape uses r equal to the body's radius; from orbit, r is the orbital radius, so escape speed falls with altitude.
- Relation to orbit: v_e = sqrt(2) x v_orbit at the same radius, about 41 percent faster than a circular orbit.
- Assumptions: no atmosphere, no rotation, and a single dominant gravitating body.
Worked example
Confirm Earth's escape velocity using its mass and radius.
- Inputs: M = 5.972 x 10^24 kg, r = 6.371 x 10^6 m, G = 6.674 x 10^-11.
- Numerator: 2 G M = 2 x 6.674e-11 x 5.972e24 = 7.972 x 10^14.
- Divide by r: 7.972e14 / 6.371e6 = 1.251 x 10^8.
- Square root: sqrt(1.251e8) = 11,186 m/s.
- Convert: 11,186 m/s = 11.19 km/s = about 25,000 mph = Mach 33.
Result: Earth's surface escape velocity is 11.2 km/s, matching the textbook value. Doubling the radius (escaping from twice as far out) would cut the required speed by a factor of sqrt(2) to about 7.9 km/s.
Escape velocity reference table
Surface escape velocities for major solar-system bodies, plus two extremes. Values use each body's accepted mass and mean radius.
| Body | Escape velocity | Note |
|---|---|---|
| Moon | 2.38 km/s | Low gravity, no atmosphere |
| Mars | 5.03 km/s | Thin atmosphere |
| Earth | 11.19 km/s | Reference value, ~25,000 mph |
| Jupiter | 59.5 km/s | Largest planet |
| Sun (surface) | 617.5 km/s | From the photosphere |
| White dwarf | ~5,000 km/s | Sun-like mass, Earth-size radius |
| Black hole horizon | 299,792 km/s | Equals the speed of light c |
Common pitfalls
- Confusing escape with orbital velocity. They differ by a factor of sqrt(2). Reaching 7.9 km/s in low Earth orbit keeps you circling; you need 11.2 km/s to leave.
- Ignoring the launch altitude. Escape velocity drops with distance from the center, so escaping from orbit takes far less added speed than escaping from the ground.
- Adding the object's mass. A heavier rocket does not need more escape speed; mass cancels in the derivation. It needs more fuel to reach that speed, which is a separate propulsion question.
- Forgetting atmospheric drag. The formula assumes a vacuum. Real launches lose energy to air resistance and must also fight gravity during the climb, so practical delta-v budgets exceed the ideal escape speed.
- Mixing units. G is in SI units, so mass must be in kilograms and radius in meters. Plugging in kilometers or grams throws the answer off by powers of ten.
Related tools
Frequently asked questions
Why does escape velocity not depend on the mass of the object escaping?
Because both the kinetic energy you need and the gravitational potential energy you must overcome scale with the same mass m, so m cancels from the energy balance. Setting kinetic energy equal to the depth of the gravitational well gives (1/2)mv squared = GMm/r, and dividing through by m leaves v = sqrt(2GM/r). A pebble and a spaceship need the same 11.2 km/s to escape Earth, ignoring air resistance.
What is escape velocity from Earth, the Moon, and the Sun?
From Earth's surface it is about 11.2 km/s (roughly 25,000 mph). From the Moon it is 2.38 km/s. From the Sun's surface it is 618 km/s. From low Earth orbit, where you are already moving fast and higher up, the additional speed needed drops to roughly 3.2 km/s.
Is escape velocity the same as orbital velocity?
No. Escape velocity is exactly the square root of 2 (about 1.414) times the circular orbital velocity at the same radius. A satellite in low Earth orbit travels at about 7.9 km/s; to break free from that altitude it would need to reach 11.2 km/s. Reaching escape speed means the object follows a parabolic or hyperbolic path and never returns.
Does direction matter for escape velocity?
For the idealized airless, non-rotating case, no. Escape velocity is a scalar speed derived from energy conservation, so any direction that does not intersect the surface works. In practice rockets launch eastward to gain free speed from Earth's rotation (up to 0.46 km/s at the equator) and climb steeply to clear the dense lower atmosphere quickly.
What happens at the event horizon of a black hole?
The event horizon is the radius at which the Newtonian escape velocity equals the speed of light c. Setting sqrt(2GM/r) = c and solving gives the Schwarzschild radius r = 2GM/c squared. Inside that radius not even light can escape. The Newtonian formula gives the correct horizon radius here even though full general relativity is needed for the physics.
Sources
- CODATA (2022) recommended value of the Newtonian constant of gravitation G = 6.67430 x 10^-11 N m^2 kg^-2.
- NASA Planetary Fact Sheets, escape velocity and physical parameters for solar-system bodies.
- Halliday, Resnick and Walker, Fundamentals of Physics, chapter on gravitation and the escape-speed derivation.