Escape Velocity Calculator
Calculate escape velocity for any planet, moon, or custom body. Compare escape speeds across the solar system with orbital velocity, Mach number.
Escape Velocity Calculator
Calculator
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Formula: v_escape = √(2GM / r)
Worked example — 11.19 km/s (40,270 km/h or 25,020 mph) — about Mach 32.6
Formula
v_escape = √(2GM / r)
Escape velocity equals the square root of (2 × gravitational constant × body mass ÷ body radius). G = 6.674 × 10⁻¹¹ N·m²/kg². This is derived from setting kinetic energy equal to gravitational potential energy. Orbital velocity is v_orbit = √(GM/r), which is exactly v_escape / √2.
Worked Examples
Example 1: Earth's Escape Velocity
Problem:Calculate the escape velocity from Earth's surface.
Solution:v_e = sqrt(2GM/r) = sqrt(2 × 6.674×10⁻¹¹ × 5.972×10²⁴ / 6.371×10⁶) = sqrt(1.249×10⁸) = 11,186 m/s = 11.19 km/s
Result:11.19 km/s (40,270 km/h or 25,020 mph) — about Mach 32.6
Example 2: Mars Escape Velocity
Problem:Calculate the escape velocity from Mars for mission planning.
Solution:v_e = sqrt(2 × 6.674×10⁻¹¹ × 6.417×10²³ / 3.3895×10⁶) = sqrt(2.527×10⁷) = 5,027 m/s = 5.03 km/s
Result:5.03 km/s — about 45% of Earth's, making Mars launches easier
Frequently Asked Questions
What is escape velocity?
Escape velocity is the minimum speed an object must reach to break free from a celestial body's gravitational pull without further propulsion. It is calculated using v_e = sqrt(2GM/r), where G is the gravitational constant, M is the body's mass, and r is its radius. For Earth, escape velocity is approximately 11.19 km/s (about 40,270 km/h or 25,020 mph). Note that rockets don't actually need to reach escape velocity all at once — they can escape with continuous thrust at lower speeds.
Why doesn't escape velocity depend on the object's mass?
Escape velocity depends only on the celestial body's mass and radius, not on the escaping object's mass. This comes from energy conservation: kinetic energy (½mv²) must equal gravitational potential energy (GMm/r). The object's mass m cancels on both sides, giving v = sqrt(2GM/r). A feather and a rocket have the same escape velocity — though practically, heavier objects need more energy (force × distance) to reach that speed.
What is the relationship between escape velocity and orbital velocity?
Escape velocity is exactly sqrt(2) ≈ 1.414 times the circular orbital velocity at the same altitude. Orbital velocity keeps an object in a circular orbit (v_o = sqrt(GM/r)), while escape velocity provides enough energy to reach infinity. To go from orbit to escape, you only need to increase speed by about 41.4%. This relationship holds at any altitude.
What happens if you exceed escape velocity?
If an object exceeds escape velocity, it follows a hyperbolic trajectory and will never return to the body. The excess speed above escape velocity determines the 'hyperbolic excess velocity' — the speed the object retains at infinite distance. This is crucial for interplanetary missions: spacecraft must leave Earth with enough hyperbolic excess to reach their target planet on a transfer orbit.
Does the direction of launch change the escape speed required?
No, because the condition is an energy balance rather than a vector one: any direction that clears the surface needs the same speed, since gravitational potential energy depends only on distance from the centre. Launch direction still matters practically. A body's rotation already supplies part of the speed, about 465 m/s eastward at Earth's equator, which is why launch sites near the equator are favoured and why eastward launches are the norm.
Why can a planet have modest surface gravity yet a very high escape velocity?
Substituting g = GM/r squared into the formula gives v_e = sqrt(2gr), so escape speed depends on surface gravity and radius together. Uranus illustrates the point: its surface gravity is close to Earth's, near 9 m/s squared, but its radius is roughly four times larger, so its escape velocity is about 21 km/s against Earth's 11.19 km/s. Gravity measures how hard you are pulled; escape velocity measures how far the well extends.
How does escape velocity determine which gases a planet keeps?
Gas molecules have a distribution of thermal speeds that rises with temperature and falls with molecular mass, so light molecules move fastest. A common rule of thumb is that a body retains a gas over geological time only if its escape velocity is several times the typical molecular speed. Hydrogen and helium therefore leak away from Earth while nitrogen, oxygen and carbon dioxide are retained, and the Moon, at 2.38 km/s, holds essentially nothing.
What is the connection between this formula and the Schwarzschild radius?
Set the escape speed equal to the speed of light and solve for radius: r = 2GM/c squared, the Schwarzschild radius. Compressing Earth's mass inside about 8.87mm would produce that condition. The agreement with general relativity is partly coincidental, since the Newtonian derivation treats light as an ordinary projectile, but it explains why John Michell and Laplace could reason about invisible dark stars more than a century before the field equations existed.
How much extra speed does a spacecraft already in orbit need to escape?
Escape velocity falls off as one over the square root of distance from the centre, so from a 400 km parking orbit the requirement drops to roughly 10.85 km/s while the circular orbital speed there is about 7.67 km/s. The difference of roughly 3.2 km/s is the classic delta-v cost of an Earth escape burn from low orbit, and it is precisely the square root of two relationship applied at orbital radius rather than at the surface.
What speed is needed to escape the Sun from Earth's orbit?
Escaping Earth only leaves a spacecraft orbiting the Sun near 1 AU, where the circular speed is about 29.8 km/s and the solar escape velocity is sqrt(2) times that, roughly 42.1 km/s. Earth's own motion already supplies the 29.8 km/s, so only about 12.3 km/s of extra speed is required, and only if the departure is aligned with Earth's orbital direction. Gravity assists are used to build that excess instead of propellant.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer · Editorial policy
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