Escape Velocity Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/6/2026
Escape velocity is calculated with v = √(2 × G × M ÷ r), where M is the mass of the celestial body and r the distance to its center. For Earth, this velocity is about 11.2 km/s; for the Moon, far less massive, it is only about 2.4 km/s.
Explanation
Escape velocity is the minimum speed an object must reach, with no further propulsion afterward, to permanently break free of a celestial body's gravitational pull and never fall back, neglecting any air resistance. This velocity follows directly from energy conservation: it corresponds exactly to the speed at which the object's kinetic energy (see our kinetic energy calculator) equals the gravitational potential energy needed to theoretically bring it to an infinite distance from the body. Escape velocity depends only on the mass of the celestial body and the distance to its center, never on the mass of the object trying to escape: a feather and a rocket require exactly the same escape velocity from a given point, only the amount of energy needed to reach it differs according to their respective masses. This velocity varies enormously from one celestial body to another depending on its mass and size: about 11.2 km/s for Earth, only about 2.4 km/s for the Moon (far less massive and smaller, which explains why the lunar missions needed much less energy to leave the Moon than to leave Earth), and much higher values for very massive bodies like the Sun or a black hole, whose escape velocity at the event horizon even exceeds the speed of light.
Example: escape velocity from Earth's surface
Inputs
Earth's mass: 5.972×10²⁴ kg. Distance to the center (Earth's radius): 6,371,000 m.
Calculation
v = √(2 × 6.6743×10⁻¹¹ × 5.972×10²⁴ ÷ 6,371,000) ≈ √(125,174,000) ≈ 11,186 m/s ≈ 11.186 km/s.
Result
A speed of about 11.2 km/s must be reached to permanently escape Earth's gravitational pull.
Frequently asked questions
Why doesn't escape velocity depend on the mass of the escaping object?
Because it results from an equality between kinetic energy and gravitational potential energy, two quantities that are both proportional to the object's mass: this mass factor cancels out mathematically in the equation, exactly as in free fall, where all objects accelerate at the same rate regardless of their mass (in the absence of air resistance).
Why is the Moon's escape velocity so much lower than Earth's?
Because the Moon is both far less massive than Earth (about 1.2% of its mass) and slightly smaller in radius, two factors that both reduce escape velocity. This difference explains why the Apollo missions' lunar modules could lift off from the Moon's surface with a much less powerful engine than the one needed to leave Earth.
What does a black hole have to do with this formula?
A black hole is an object so massive and so compact that its escape velocity, calculated at a certain distance from its center (the event horizon), exceeds the speed of light itself. Since nothing can travel faster than light, no matter or radiation can escape once past this horizon — this property is what gives the black hole its name, although a rigorous description of this phenomenon actually falls under general relativity rather than this classical formula. The mass of the body used here is the same quantity that determines the attractive force in our gravitational force calculator, both formulas deriving from the same constant G.