Escape Velocity Calculator
How fast must an object travel to escape any celestial body's gravity?
Find out how fast an object must travel to escape a planet, moon, or star's gravitational pull forever. Enter the celestial body's mass and radius — see escape velocity in km/s, mph, and relative comparison to Earth. Assumes spherical mass distribution and no atmospheric resistance.
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How It Works
The formula, explained simply
Escape velocity is the speed a cannonball would need if fired straight up to never fall back down. Unlike orbital velocity, where an object circles the planet, escape velocity means the object overcomes gravity completely and travels to infinity. The surprising part: escape velocity is the same whether you're launching a feather or a freight train — mass of the projectile doesn't matter, only the mass and size of the celestial body.
This calculator assumes you're launching from the surface in a perfect vacuum with no air resistance. Real rockets also lose speed to atmospheric drag and to gravity during the climb, so they need more than this. The formula also assumes the body is spherically symmetric, which works well for planets, large moons and stars but not for irregular asteroids. How the density varies inside does not matter, only the total mass and the radius.
The physics becomes intuitive when you realize escape velocity is just the speed where kinetic energy equals gravitational potential energy. At exactly escape velocity, an object reaches infinite distance with zero velocity remaining — any faster and it has energy left over, any slower and gravity wins. This energy balance explains why escape velocity depends on the square root of mass divided by radius, making it much easier to escape small, low-density objects.
When To Use This
Right tool, right situation
Use this calculator for mission planning to other worlds, understanding why some moons can retain atmospheres while others cannot, or calculating the minimum energy requirements for interplanetary travel. Space agencies use escape velocity to determine fuel requirements and rocket staging for missions beyond Earth orbit.
Astrobiologists use escape velocity to predict which planets can hold onto water vapor and oxygen — if atmospheric particles reach escape velocity through thermal motion, the atmosphere slowly leaks into space. Mars's lower escape velocity (about 5 km/s) is one reason it kept less of its atmosphere than Earth.
The calculation also applies to understanding stellar evolution and planetary formation. A growing planet's escape velocity sets the minimum speed at which incoming material strikes it.
Common Mistakes
Why results sometimes look wrong
The biggest mistake is confusing escape velocity with orbital velocity. Orbital velocity keeps you circling the planet at constant altitude, while escape velocity launches you away forever. Earth's orbital velocity at sea level is 7.9 km/s, but escape velocity is 11.19 km/s, about 41% higher.
Many people think heavier objects need more speed to escape gravity, but escape velocity is mass-independent. A bowling ball and a ping pong ball need exactly the same launch speed from the same location. The confusion comes from everyday experience where heavier objects are harder to throw, but that's due to the thrower's limited force, not physics.
Another common error is ignoring atmospheric resistance in real-world calculations. The theoretical escape velocity assumes a vacuum, but Earth's atmosphere adds significant drag. Real rockets need more total velocity change (delta-v) than 11.19 km/s to escape Earth, because air resistance and gravity take a share during the climb.
The Math
Worked examples and deeper derivation
The escape velocity formula v = √(2GM/r) comes from setting kinetic energy equal to gravitational potential energy: ½mv² = GMm/r. The masses cancel out, proving escape velocity is independent of the projectile's mass. G is the gravitational constant (6.674×10⁻¹¹ m³/kg·s²), M is the celestial body's mass in kilograms, and r is the radius in meters.
For Earth: v = √(2 × 6.674×10⁻¹¹ × 5.972×10²⁴ / 6.371×10⁶) = 11,186 m/s, or 11.19 km/s. For the Moon, the same calculation yields 2.38 km/s, explaining why Apollo spacecraft needed much less fuel to leave the lunar surface than to escape Earth initially.
The formula breaks down at extreme scales. For black holes, escape velocity reaches the speed of light at the event horizon, where classical physics fails and relativity takes over. For small irregular objects, the spherical assumption fails: escape speed then differs from place to place on the surface. The square root relationship means doubling a planet's mass only increases escape velocity by 41%, and halving the radius also increases it by 41%.
Expert Unlock
The thing most explanations skip
The standard escape velocity formula assumes launch from a stationary surface, but Earth rotates at 465 m/s at the equator. Launching eastward adds this velocity for free, reducing the required rocket velocity to about 10.72 km/s. The boost shrinks with latitude, to 465 × cos(latitude) m/s: about 403 m/s at 30°, and nothing at the poles.
Why is escape velocity the same for all objects regardless of their mass?
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