Escape Velocity Calculator

Calculate the escape velocity from any planet or moon given mass and radius.

Earth = 5.972×10²⁴ kg
Earth = 6.371×10⁶ m
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Escape Velocity from Conservation of Energy

Escape velocity is the minimum initial speed that gives an unpowered object enough mechanical energy to reach infinitely far away with zero speed remaining. At distance r from the center of a spherical body, gravitational potential energy is −GMm/r. Setting the object's initial kinetic energy equal to the magnitude of that binding energy gives the familiar square-root formula.

The escaping object's mass m cancels, which is why a small probe and a massive spacecraft have the same ideal escape speed from the same location. The central body's mass M and the starting distance r are what matter. A more massive body increases the required speed, while starting farther from its center lowers it. For a launch from altitude h above a body's surface, use r = R + h rather than the surface radius R alone.

½mvesc2 − GMm/r = 0  →  vesc = √(2GM/r)
SymbolMeaningUnits / role
vescEscape velocitym/s or km/s
GGravitational constant6.674 × 10−11 N m2/kg2
MMass of central bodykg; more mass means stronger gravitational binding
rDistance from body's centerm; surface launch uses approximately the body's radius

This ideal result assumes a spherical gravitating body, no atmosphere, no rotation, and no additional propulsion after the initial speed is given. Real rockets do not need to be moving at 11.2 km/s at the launch pad; engines add energy continuously during ascent, and atmospheric drag and gravity losses also matter. At the same radius, ideal escape speed is √2 times the circular-orbit speed.

Worked Examples

Example 1: Earth escape velocity
M=5.972e24, r=6.371e6
Result: 11.19 km/s
An unpowered object needs this initial speed to coast to infinity; powered rockets can add energy during ascent
Example 2: Moon escape velocity
M=7.342e22, r=1.737e6
Result: 2.38 km/s
Why Apollo could launch from Moon with small rocket
Example 3: Escape velocity from Mars
M = 6.39 × 1023 kg, r = 3.389 × 106 m → vesc = √(2GM/r)
Result: vesc ≈ 5.02 km/s
Mars has much less mass than Earth, so its surface escape speed is less than half Earth's value even though Mars also has a smaller radius.
Example 4: Starting 400 km above Earth
M = 5.972 × 1024 kg, r = 6.371 × 106 + 4.00 × 105 = 6.771 × 106 m
Result: vesc ≈ 10.85 km/s
Escape speed decreases with altitude because the object begins farther from Earth's center and is therefore less tightly gravitationally bound.

Common Mistakes

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Using altitude alone for r

The formula needs distance from the center of the gravitating body. At altitude h, use r = R + h. Near Earth's surface, using 400 km instead of about 6771 km produces a major error.

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Including the spacecraft mass in the final formula

The spacecraft mass appears in both kinetic and gravitational potential energy and cancels. Ideal escape velocity depends on the central body's M and the starting radius r, not the payload mass.

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Confusing escape velocity with orbital velocity

A circular orbit does not escape. At the same radius, vesc = √2 vorbit in the ideal two-body model, so escape speed is about 41.4% higher than circular-orbit speed.

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Assuming a rocket must instantly reach escape speed

The textbook value applies to an object given an initial speed and then allowed to coast without propulsion. A powered spacecraft can gain the required energy over time along its trajectory.

Frequently Asked Questions

What is escape velocity?
The minimum speed to escape a body's gravity without further propulsion. Above this speed, an object has enough kinetic energy to reach infinite distance.
Does mass of the escaping object matter?
No — escape velocity is independent of the escaping object's mass. A pebble and a rocket have the same escape velocity from Earth.
Why does escape velocity decrease with altitude?
Gravitational potential energy becomes less negative as distance from the body's center increases. An object starting higher is already less tightly bound, so it needs less additional kinetic energy to reach an unbound trajectory.
What is the relationship between escape velocity and circular orbital velocity?
For the same radius around a spherical body, circular speed is √(GM/r), while escape speed is √(2GM/r). Therefore vesc = √2 times the circular orbital speed.
Does atmosphere change the theoretical escape velocity?
The gravitational escape-speed formula itself does not include atmosphere, but a real vehicle moving through an atmosphere loses energy to drag. Real launch requirements also include propulsion efficiency, gravity losses, trajectory design, and the rotation of the planet.
Can an object escape with less than the local escape speed?
An unpowered object starting at a given radius cannot coast to infinity if its initial speed is below the ideal escape speed. A powered spacecraft can begin slower and continue adding energy later, so practical missions are not limited to a single instantaneous launch speed.

Formula Explorer connections

Interpretation: This relationship connects motion, force, momentum, work or energy in a mechanical system. Assumption: Choose a consistent reference direction and unit system. The model may assume constant acceleration, rigid bodies, negligible losses or an isolated system.

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