Audited ·Last updated 27 Jul 2026·6 citations·Tier 1·0 uses

Escape Velocity Calculator

Calculate escape velocity for any planet or moon using mass and radius. Physics tool with step-by-step math, CODATA constants, and worked examples.

Escape Velocity Calculator

Escape velocity (m/s)
11,185.9779
Escape velocity (km/s)
11.186

Background.

The escape velocity calculator determines the minimum speed an unpowered projectile must achieve at the surface of a celestial body to break free of its gravitational pull entirely. The concept is foundational to astrodynamics, rocket engineering, and planetary science. Every interplanetary mission, every satellite launch to deep space, and every discussion of atmospheric retention on exoplanets begins with the same simple inequality: is the object's kinetic energy sufficient to climb out of the gravitational well?

The derivation traces back to Newton's law of universal gravitation and the principle of conservation of mechanical energy. If a body of mass m is launched from the surface of a much more massive body M at distance r from its center, the total mechanical energy is the sum of kinetic energy ½mv² and gravitational potential energy −GMm/r. For the projectile to reach infinity with zero residual speed, its total energy must be at least zero. Setting ½mv² − GMm/r = 0 and solving for v yields v = √(2GM/r). Remarkably, the escaping mass m cancels out, meaning a hydrogen atom and a Saturn V rocket require the same initial speed to escape Earth, though the rocket needs vastly more energy because energy scales with mass.

Earth's escape velocity of approximately 11.2 kilometers per second is one of the most referenced numbers in spaceflight. It sets the performance requirements for launch vehicles. The Atlas V, Falcon 9, and Space Launch System all must deliver payloads to speeds approaching or exceeding this threshold to reach geosynchronous transfer orbit, lunar orbit, or interplanetary trajectories. Engineers do not need to reach the full 11.2 km/s at sea level because atmospheric drag and gravity losses are offset by staging and by the Oberth effect, but the escape velocity remains the theoretical benchmark against which propulsion systems are measured.

The concept also explains why the giant planets retain hydrogen and helium atmospheres while smaller bodies like Mars and Mercury have lost most of theirs. Thermal velocities of atmospheric molecules follow the Maxwell-Boltzmann distribution; if the high-energy tail of that distribution exceeds the escape velocity, molecules leak into space over geological time. Jupiter's escape velocity of 59.5 km/s is so high that even the lightest molecules are gravitationally bound. The Moon's escape velocity of 2.38 km/s is low enough that it cannot sustain any significant atmosphere. Exoplanet hunters use estimated escape velocities together with equilibrium temperatures to assess whether a detected world could plausibly hold onto an atmosphere capable of supporting liquid water.

The calculator has value beyond professional astrodynamics. Educators use it to illustrate the interplay of mass and radius: a white dwarf with solar mass but Earth-like radius has an escape velocity approaching 6,000 km/s, a fact that underlies the conditions for Type Ia supernovae. Science fiction writers use it to check the physical plausibility of fictional worlds. Amateur astronomers use it to understand why comets on parabolic orbits have exactly zero total energy. The same formula, with M and r adjusted, applies to any spherical gravitating body, from asteroids to galaxy clusters, making it one of the most universal tools in gravitational physics.

What is escape velocity calculator?

Escape velocity is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a massive body, reaching an infinite distance with zero residual velocity. It is a scalar quantity expressed in meters per second or kilometers per second. The value depends only on the mass and radius of the attracting body, not on the mass or composition of the escaping object.

The concept applies strictly to ballistic trajectories with no thrust after the initial impulse and no non-gravitational forces such as atmospheric drag. In practice, rockets burn fuel continuously and do not need to reach the full escape velocity at launch; they can climb gradually, converting chemical energy into gravitational potential energy over time. Nevertheless, the escape velocity remains a critical design parameter because it quantifies the depth of the gravitational well that must be overcome.

Escape velocity is related to but distinct from orbital velocity. For a circular orbit just above a body's surface, the orbital velocity is v = √(GM/r), which is smaller than the escape velocity by a factor of √2. This relationship means that if a spacecraft in low orbit increases its speed by about 41.4%, it will transition to a parabolic escape trajectory.

How to use this calculator.

  1. Enter the mass of the celestial body in kilograms.
  2. Enter the radius from the body's center to the launch point in meters.
  3. The calculator computes 2GM / r using the CODATA 2018 value of G.
  4. The calculator takes the square root to obtain escape velocity in m/s.
  5. The result is also displayed in km/s for readability.
  6. Compare the output to known values (Earth 11.2 km/s, Moon 2.38 km/s) as a sanity check.

The formula.

v = √(2GM ⁄ r)

The escape velocity formula is derived from the conservation of mechanical energy in a conservative gravitational field. Consider a particle of mass m at distance r from the center of a spherically symmetric mass M. The Newtonian gravitational potential energy is U = −GMm/r, where the negative sign indicates a bound state. The particle's kinetic energy is K = ½mv². The total mechanical energy E = K + U must be greater than or equal to zero for the particle to reach infinity with non-negative kinetic energy.

Setting E = 0 for the minimum escape condition gives ½mv² = GMm/r. The mass of the escaping particle cancels, leaving v² = 2GM/r, and therefore v = √(2GM/r). This result was implicit in Newton's Principia and was later refined by Lagrange and others in the context of celestial mechanics. The square-root dependence on mass means that doubling the central mass increases the escape velocity by a factor of √2 ≈ 1.414. The inverse dependence on radius means that compressing a planet to half its radius while holding mass constant would increase the escape velocity by the same factor.

The formula assumes a spherically symmetric mass distribution and ignores relativistic effects, atmospheric drag, and the gravitational influence of other bodies. For Earth, these approximations are excellent: general-relativistic corrections to the escape velocity amount to less than one part per billion. Near extremely compact objects such as neutron stars, the Newtonian formula becomes inaccurate and must be replaced by the Schwarzschild solution of general relativity, where the event horizon defines an escape velocity equal to the speed of light.

The appearance of the gravitational constant G in the formula ties escape velocity to the fundamental strength of gravity. Because G is among the least precisely known fundamental constants—CODATA 2018 lists its relative standard uncertainty as 2.2 × 10⁻⁵—planetary escape velocities carry a corresponding uncertainty in their final digits. For most pedagogical and engineering purposes, however, the standard value yields results accurate to better than one meter per second for planets and moons.

A worked example.

Example

To find Earth's escape velocity, begin with the planet's standard gravitational parameter. Using the CODATA 2018 gravitational constant G = 6.67430 × 10⁻¹¹ m³·kg⁻¹·s⁻² and Earth's mass of 5.972 × 10²⁴ kg, first compute the numerator 2GM. Multiplying 2 by 6.67430 × 10⁻¹¹ gives 1.33486 × 10⁻¹⁰; multiplying that by 5.972 × 10²⁴ gives 7.97178 × 10¹⁴ m³·s⁻². Divide this by Earth's mean radius of 6.371 × 10⁶ m to obtain 1.25126 × 10⁸ m²·s⁻². The square root of this quantity is 11,186 m/s, or approximately 11.2 km/s. This is the speed a projectile would need at sea level, ignoring air resistance, to coast indefinitely away from Earth. The Apollo lunar missions did not achieve this speed at launch; instead, the Saturn V performed a trans-lunar injection burn in low Earth orbit, leveraging the Oberth effect to reach escape energy with less total propellant than a direct ascent would require.

mass5,972,000,000,000,000,000,000,000
radius6,371,000

Frequently asked questions.

Does escape velocity depend on the mass of the escaping object?
No. In the derivation of v = √(2GM/r), the mass of the projectile cancels out of both sides of the energy equation. A feather and a freight train require the same initial speed to escape Earth. What differs is the kinetic energy, which scales linearly with the projectile's mass. Launching a more massive object therefore requires more total energy, even though the threshold speed remains unchanged. This cancellation is a direct consequence of the equivalence of inertial and gravitational mass, which is foundational to general relativity.
Why do rockets not need to reach escape velocity at launch?
Escape velocity describes a ballistic, unpowered trajectory. Rockets are powered throughout their ascent, continuously adding energy to the system. A rocket can climb slowly out of Earth's gravity well, never reaching 11.2 km/s near the surface, and still achieve escape energy by the time its fuel is exhausted. In practice, launch vehicles reach orbital velocity first—about 7.8 km/s for low Earth orbit—and then perform additional burns to raise their energy above the escape threshold. The Oberth effect makes it most efficient to perform these escape burns at high velocity, close to the gravitating body.
What is the escape velocity of the Moon?
Using the Moon's mass of 7.348 × 10²² kg and mean radius of 1.737 × 10⁶ m, the calculator yields v = √(2 × 6.67430 × 10⁻¹¹ × 7.348 × 10²² / 1.737 × 10⁶) = 2,380 m/s, or approximately 2.38 km/s. This low value explains why the Moon has no appreciable atmosphere: thermal velocities of gas molecules at typical lunar surface temperatures exceed the escape velocity, allowing atmospheric particles to leak away over geological time scales. The Apollo lunar modules did not need powerful engines to leave the Moon; the ascent stage reached only a fraction of Earth's escape velocity.
How does escape velocity relate to black holes?
As a body is compressed, its escape velocity rises according to v = √(2GM/r). If the radius shrinks to the Schwarzschild radius r_s = 2GM/c², the Newtonian formula predicts an escape velocity equal to the speed of light c. At that point, not even light can escape, and the object becomes a black hole. The exact description requires general relativity—the Newtonian formula is only an approximation—but the Schwarzschild radius correctly identifies the event horizon for a non-rotating black hole. Stellar-mass black holes have Schwarzschild radii of a few kilometers; supermassive black holes have radii comparable to the orbit of Mercury.
Can an object with less than escape velocity still leave a planet?
Yes, if it is continuously propelled. Escape velocity is a threshold for ballistic trajectories only. A spacecraft on a sub-escape elliptical orbit can fire its engines at apogee to raise its orbital energy incrementally, eventually exceeding the escape energy. This is how ion-drive probes such as NASA's Dawn and Psyche missions operate: they use low-thrust engines over months or years to spiral out of Earth's gravity well and then the Sun's. The total energy required is the same, but the power can be applied gradually rather than in a single impulsive burn.
Is escape velocity the same everywhere on a planet?
No. Escape velocity depends on the distance r from the center of mass. At the equator of an oblate planet like Earth, the radius is slightly larger than at the poles, so the escape velocity is marginally lower. Additionally, rotation reduces the required launch speed if the launch direction is eastward: the Earth's equatorial rotation contributes about 465 m/s, meaning a rocket launched eastward from the equator needs about 10.7 km/s relative to the surface to escape, compared to 11.2 km/s from a non-rotating body. Launch sites near the equator, such as Kourou in French Guiana, exploit this advantage.
What is the escape velocity from the solar system?
From Earth's distance of 1 AU (1.496 × 10¹¹ m) from the Sun, the escape velocity relative to the Sun is v = √(2GM☉ / r) = √(2 × 1.327 × 10²⁰ / 1.496 × 10¹¹) ≈ 42.1 km/s. Because Earth already orbits the Sun at 29.8 km/s, a spacecraft launched in the direction of Earth's motion needs only about 12.3 km/s of additional speed relative to Earth to achieve solar escape. Voyager 1 and 2 used gravity assists from Jupiter and Saturn to reach these speeds without carrying prohibitive amounts of propellant.
Does atmospheric drag affect escape velocity calculations?
The theoretical escape velocity assumes no atmospheric drag. In practice, a rocket launched from Earth's surface must overcome aerodynamic drag, which dissipates kinetic energy as heat. The energy lost to drag is one reason why multi-stage rockets are used: the first stage lifts the vehicle out of the dense lower atmosphere, where drag is highest, before the upper stages accelerate to orbital or escape velocity in the thinner upper atmosphere. The calculator returns the ideal vacuum value; real mission planners add margin for drag and gravity losses.
Can escape velocity be negative?
No. Escape velocity is defined as the magnitude of the velocity vector required for escape, so it is strictly non-negative. The underlying energy condition E ≥ 0 can be negative for bound orbits, but that simply means the object is not escaping. A negative speed has no physical meaning in this context. If the calculator receives negative inputs for mass or radius, those values are unphysical and indicate an input error, since mass and distance are positive-definite quantities in Newtonian gravity.
How is escape velocity used in exoplanet research?
Astronomers estimate the escape velocities of exoplanets from their measured masses and radii, which are derived from radial-velocity and transit observations. Comparing escape velocity to the thermal speed of atmospheric gases helps assess whether a planet can retain an atmosphere over billions of years. For example, the TRAPPIST-1 planets have low escape velocities because their host star is small and the planets are relatively low-mass, making atmospheric retention a critical question in habitability studies. The James Webb Space Telescope observes transmission spectra to detect atmospheric signatures, but escape-velocity estimates inform whether those atmospheres are expected to survive.

References& sources.

  1. [1]Newton, I. (1687). Philosophiæ Naturalis Principia Mathematica. London: Royal Society. Book I, Propositions LXX and LXXI.
  2. [2]Halliday, D., Resnick, R., and Walker, J. (2021). Fundamentals of Physics, 11th ed. Hoboken, NJ: Wiley. Chapter 13: Gravitation.
  3. [3]NASA (2024). Planetary Fact Sheet — Earth. NASA Goddard Space Flight Center.
  4. [4]NIST (2024). CODATA Recommended Values of the Fundamental Physical Constants: 2018.
  5. [5]Bate, R.R., Mueller, D.D., and White, J.E. (1971). Fundamentals of Astrodynamics. New York: Dover. Chapter 2: Kepler's Equation and Orbital Energy.
  6. [6]Carroll, B.W. and Ostlie, D.A. (2017). An Introduction to Modern Astrophysics, 2nd ed. Cambridge: Cambridge University Press. Chapter 2: Interaction of Radiation and Matter.

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