Escape Velocity Converter
Calculate escape velocity for any planet, moon, or celestial body. Convert between m/s, km/s, km/h, mph, and Mach number with real gravitational physics
Escape Velocity Converter
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Formula: v_escape = sqrt(2 * G * M / R)
Worked example โ Escape velocity from Mars is about 5.03 km/s or 18,100 km/h
Formula
v_escape = sqrt(2 * G * M / R)
Escape velocity equals the square root of twice the gravitational constant (G = 6.674e-11) times the body mass (M in kg) divided by its radius (R in meters). This formula comes from setting kinetic energy equal to gravitational potential energy and solving for velocity.
Worked Examples
Example 1: Escape Velocity from Mars
Problem:Calculate the escape velocity from the surface of Mars (mass = 6.417e23 kg, radius = 3.3895e6 m).
Solution:v = sqrt(2 * G * M / R) v = sqrt(2 * 6.674e-11 * 6.417e23 / 3.3895e6) v = sqrt(2.528e7) = 5,027 m/s
Result:Escape velocity from Mars is about 5.03 km/s or 18,100 km/h
Example 2: Escape Velocity from Jupiter
Problem:Find the escape velocity from Jupiter (mass = 1.898e27 kg, radius = 6.9911e7 m).
Solution:v = sqrt(2 * G * M / R) v = sqrt(2 * 6.674e-11 * 1.898e27 / 6.9911e7) v = sqrt(3.621e9) = 60,177 m/s
Result:Escape velocity from Jupiter is about 60.2 km/s or 216,600 km/h
Frequently Asked Questions
What is escape velocity and why does it matter?
Escape velocity is the minimum speed an object must reach to break free from a celestial body's gravitational pull without any further propulsion. It is derived from the balance between kinetic energy and gravitational potential energy. For Earth, escape velocity is approximately 11.186 km/s or about 40,270 km/h. This speed determines the energy requirements for launching spacecraft into interplanetary trajectories and is fundamental to mission planning in aerospace engineering.
Does escape velocity depend on the direction of launch?
No, escape velocity is a scalar quantity that depends only on the mass and radius of the celestial body, not on the direction of travel. Whether launched straight up, horizontally, or at any angle, the same speed is needed to escape the gravitational field. However, launching eastward near the equator provides a small velocity boost from Earth's rotation of about 465 m/s, which is why most spaceports are located near the equator.
Why does the Moon have a much lower escape velocity than Earth?
The Moon's escape velocity is about 2.38 km/s compared to Earth's 11.19 km/s because escape velocity depends on the ratio of mass to radius. The Moon has roughly 1.2% of Earth's mass but about 27% of its radius. Since escape velocity scales with the square root of mass divided by radius, the Moon's much smaller mass dominates the calculation, resulting in an escape velocity about five times lower than Earth's.
How is escape velocity related to whether a body can hold an atmosphere?
A celestial body retains an atmosphere when its escape velocity significantly exceeds the average thermal velocity of gas molecules. The thermal velocity of a molecule depends on temperature and molecular mass. If gas molecules frequently reach speeds near the escape velocity, they gradually leak into space through a process called Jeans escape. This is why the Moon and Mercury have essentially no atmosphere while Earth and Venus retain thick ones.
What is the difference between escape velocity and orbital velocity?
Circular orbital velocity at radius R is sqrt(G x M / R), while escape velocity is sqrt(2 x G x M / R). Escape velocity is therefore exactly sqrt(2), about 1.414 times, the local circular orbital speed. For Earth that is roughly 7.9 km/s to stay in low orbit versus about 11.19 km/s to leave the gravity well entirely. The extra speed is what converts a bound elliptical orbit into an unbound parabolic or hyperbolic trajectory.
Why can escape velocity be written as the square root of 2 g R?
Surface gravity is g = G x M / R squared, so G x M = g x R squared. Substituting into v = sqrt(2 G M / R) gives v = sqrt(2 g R). For Earth, sqrt(2 x 9.81 x 6,371,000) is about 11,180 m/s, matching the standard 11.19 km/s. This form is handy when you know a body's surface gravity and radius but not its mass, and it shows that both a strong pull and a large radius raise the escape speed.
Why is escape velocity independent of the mass of the escaping object?
Setting kinetic energy equal to gravitational potential energy gives one half m v squared = G M m / R, where m is the escaping object's mass. That m appears on both sides and cancels, leaving v = sqrt(2 G M / R) with no trace of it. A pebble and a fully fuelled rocket need the same speed. What differs is the energy required, which scales directly with mass, which is why launch cost tracks payload mass even though the target speed does not.
How does escape velocity change if you start from altitude rather than the surface?
R in the formula is the distance from the body's centre, not from its surface, so escape velocity falls off as one over the square root of that distance. Doubling your distance from the centre lowers the required speed by a factor of sqrt(2), roughly 29 percent. This is why upper stages departing from a parking orbit need less additional speed than a hypothetical single burn from the ground, and why high-altitude staging is favoured in mission design.
How does the converter turn escape velocity into a Mach number?
It divides the speed in m/s by 343, the approximate speed of sound in dry air at sea level near 20 degrees Celsius. The result is a familiar comparison figure rather than a rigorous aerodynamic quantity. Mach number is only physically meaningful inside an atmosphere, and the local speed of sound varies with temperature and gas composition, so a Mach value for the Moon or for a vacuum-phase trajectory is illustrative only, not a flight condition.
How is escape velocity connected to the Schwarzschild radius of a black hole?
Setting the Newtonian escape velocity equal to the speed of light and solving for R gives R = 2 G M / c squared, which is precisely the Schwarzschild radius from general relativity. The agreement is a numerical coincidence, since the Newtonian derivation assumes flat space and a slow-moving projectile. Still, it gives a correct sense of scale: compress any mass inside that radius and no signal, light included, can climb out of the resulting gravity well.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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