Example 1
Launch from the Earth’s surface: M = {{u:5.972e24:mass:0}}, r = {{u:6371000:dist-m:0}}. This is the reference value for the other examples.
v₂ = 11.19 km/s
Enter the mass M of the central body and the distance r from its center; the constant G can stay empty. The calculator finds escape velocity v₂ = √(2GM/r) and also reports v₁ = v₂/√2, the parameter GM, and g = GM/r². For Earth (5.972·10²⁴ kg, 6,371 km) the result is 11.19 km/s; for the Moon it is 2.38 km/s. The model does not cover the third cosmic velocity, atmospheric drag, or continuous-thrust ascent.
Enter the mass M and the distance r. The escape velocity, the orbital speed, and the v₂(r) chart will appear here.
For the speed on a circular orbit, use the first cosmic velocity calculator; the two speeds are related by v₂ = v₁·√2. For the gravitational field, use g = GM/r², and for the force, F = GMm/r².
The escape velocity, called the second cosmic velocity in the Russian and Polish tradition, is the smallest speed a body needs at a distance r from the center of a mass M to move away to infinity with no further propulsion. It follows from energy conservation: the kinetic energy ½mv² must at least balance the potential energy −GMm/r, which gives v₂ = √(2GM/r). The mass of the escaping body cancels, so a stone and a spacecraft need the same speed. This is a textbook result of Newtonian mechanics (Halliday, Resnick, Walker; Young, Freedman), with its roots in Newton’s Principia of 1687.
The calculator works with three numbers: the gravitational constant G (empty means 6.67430·10⁻¹¹ N·m²/kg²), the mass M in kilograms, and the distance r in meters. It first evaluates GM, then 2GM/r, and finally the square root. For Earth: GM = 3.986·10¹⁴ m³/s², 2GM/r = 1.251·10⁸ m²/s², v₂ = 11,186 m/s, that is 11.19 km/s or 6.95 mi/s. For the Moon (7.35·10²² kg, 1,737 km) the result is 2.38 km/s.
What the result panel shows. Besides v₂ in km/s, the calculator reports the same speed in m/s and km/h, the first cosmic velocity v₁ = √(GM/r) = v₂/√2 (7.91 km/s for Earth), the gravitational parameter GM, the gravitational acceleration at distance r (GM/r², 9.82 m/s² at the Earth’s surface), the multiple of Earth’s escape velocity, and the entered M and r. The chart shows how v₂ falls with distance as 1/√r: twice as far from the center, √2 times less speed is needed.
Limits of the model. The formula describes a point mass or a spherically symmetric body, no atmosphere, and no rotation of the central body; it also assumes the escaping mass is negligible compared with M. The calculator does not compute the speed needed to leave the whole Solar System from Earth’s orbit (the third cosmic velocity, about 16.7 km/s), nor atmospheric losses or the boost from the planet’s rotation (about 0.47 km/s at the Earth’s equator). A rocket does not need v₂ at liftoff — it only needs to reach it after leaving the atmosphere.
The unit switch in the header changes the fields to pounds and feet and the speeds to mi/s, ft/s, and mph; the values in the text stay in SI units as a reference. Large numbers can be entered in exponential notation (5.972e24). Mass and distance must be greater than zero: without mass there is no gravitational field, and zero distance would mean infinite speed.
Examples on this page: Earth’s surface → 11.19 km/s; ISS orbit (r = 6,771 km) → 10.85 km/s; Moon → 2.38 km/s; Mars → 5.02 km/s; geostationary orbit → 4.35 km/s; the Sun from Earth’s distance → 42.1 km/s. The “Fill this example” button enters the same numbers into the fields and refreshes the result panel and the chart.
v₂ = √(2GM / r)
[v₂] = √((N·m²/kg²)·kg / m) = √(m²/s²) = m/s; v₁ = √(GM/r), v₂ = v₁·√2, g = GM/r²
Model: escape velocity from energy conservation in a Newtonian field (I. Newton, Principia 1687; total energy E = ½mv² − GMm/r = 0). The rearrangements r = 2GM/v₂² (the Schwarzschild radius for v₂ = c) and M = v₂²r/(2G) follow from the same formula, but this calculator solves only for v₂. Order of computation: GM, then 2GM/r in SI units, square root, then v₁ = v₂/√2 and g = GM/r².
Escape velocity is the smallest speed a body must have at a given point of a gravitational field so that — moving freely from then on — it never returns and recedes to infinity with a speed approaching zero. In the language of energy: the body’s total mechanical energy, the sum of kinetic ½mv² and potential −GMm/r, equals zero. A body with negative energy stays bound and moves on an ellipse, a body with positive energy departs on a hyperbola, and exactly at v₂ its path is a parabola.
Three properties follow from the definition. First, v₂ does not depend on the mass of the escaping body, because m cancels in the energy equation. Second, it does not depend on the direction of launch — only the magnitude of the speed matters, as long as the path does not intersect the surface of the central body. Third, it decreases with distance as 1/√r: escape from geostationary orbit requires 4.35 km/s instead of 11.19 km/s from the surface.
The thought experiment of a cannonball fired from a high mountain comes from Newton’s De mundi systemate (published 1728): with ever greater speed the projectile lands ever farther, then circles the Earth, and finally flies away forever. In 1783 John Michell, and in 1796 Pierre-Simon Laplace, noted that for a sufficiently massive and compact body the escape velocity would exceed the speed of light — the first hint of the black hole concept; the radius at which v₂ = c is today called the Schwarzschild radius r = 2GM/c².
The names “first, second, and third cosmic velocity” took hold in Russian and Polish literature through the work of Konstantin Tsiolkovsky (1903) and astronautics textbooks; English-language texts speak of escape velocity. The first human-made object to reach escape velocity was the Soviet probe Luna 1 in January 1959; it passed the Moon and entered orbit around the Sun.
Step 1. The potential energy of a body of mass m at distance r from the center of mass M in a Newtonian field is Ep = −GMm/r, with the conventional zero at infinity. Step 2. Write energy conservation between the launch point (speed v, distance r) and infinity, where both the potential energy and — in the limiting case — the kinetic energy are zero: ½mv² − GMm/r = 0.
Step 3. Cancel m and solve for v: v² = 2GM/r, hence v₂ = √(2GM/r). Step 4. For Earth, with G = 6.67430·10⁻¹¹ N·m²/kg², M = 5.972·10²⁴ kg, and r = 6.371·10⁶ m, we get GM = 3.986·10¹⁴ m³/s², 2GM/r = 1.251·10⁸ m²/s², and v₂ = 11,186 m/s. The same formula can be written through the surface acceleration g = GM/r² as v₂ = √(2gr): for Earth √(2·9.82·6.371·10⁶) gives the same value.
The first cosmic velocity v₁ = √(GM/r) is the speed of a circular orbit just above the surface; it follows from equating gravity and the centripetal force, GMm/r² = mv²/r. Comparison with v₂ = √(2GM/r) gives the exact relation v₂ = v₁·√2 ≈ 1.414·v₁: escape needs 41 % more speed than a circular orbit, but twice the kinetic energy. For Earth v₁ = 7.91 km/s and v₂ = 11.19 km/s.
The third cosmic velocity, about 16.7 km/s relative to Earth, allows a departure from the Solar System starting in Earth orbit; it is not a simple sum but results from combining escape from Earth’s field with the surplus needed to escape the Sun while using Earth’s orbital speed (29.8 km/s). This calculator computes only v₂ for a single central body; the escape velocity from the Sun at 1 AU (42.1 km/s) appears as an example, without accounting for Earth’s motion.
The formula assumes the central body is a point mass or has a spherically symmetric mass distribution; by Newton’s shell theorem the field outside such a body is identical to that of a point of the same mass. It also assumes the escaping mass is negligible compared with M (for two comparable masses the total mass M + m would be used) and that nothing acts besides the single central body.
The model leaves out the atmosphere, the rotation of the body, and the way the speed is built up. A real rocket loses about 1.5–2 km/s to drag and gravity during ascent, but gains up to 0.465 km/s from Earth’s rotation when launched eastward from the equator. It also does not need v₂ immediately: it is enough to have that speed at engine cutoff, already outside the atmosphere at a distance r somewhat larger than the planet’s radius, where the required value is slightly lower.
The gravitational constant G = 6.67430(15)·10⁻¹¹ N·m²/kg² (CODATA 2018) is the least precisely known fundamental constant: its relative uncertainty is 2.2·10⁻⁵. Paradoxically, the product GM for planets is known far more precisely from satellite orbits — for Earth GM = 3.986004418·10¹⁴ m³/s² to ten significant figures. That is why astrodynamics works directly with the parameter GM (denoted μ), and the calculator displays it as a separate value.
Escape velocities from the surface according to NASA data: Mercury 4.25 km/s, Venus 10.36 km/s, Earth 11.19 km/s, Moon 2.38 km/s, Mars 5.03 km/s, Jupiter 59.5 km/s (for the equatorial radius of 71,492 km; the example on this page uses the mean radius of 69,911 km and gives 60.2 km/s), Saturn 35.5 km/s, Sun 617.7 km/s. For small bodies the values are surprisingly low: from Ceres about 0.5 km/s, and from Phobos only about 11 m/s — a thrown ball would leave that moon for good.
The definition, derivation, and numerical values come from mechanics textbooks and metrology sources: I. Newton, Philosophiae Naturalis Principia Mathematica, 1687, and De mundi systemate, 1728; D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, chapter on gravitation; H. D. Young, R. A. Freedman, University Physics, chapter “Gravitation”; R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics, vol. I, chapter 7 (The Theory of Gravitation); H. D. Curtis, Orbital Mechanics for Engineering Students, Butterworth-Heinemann; CODATA 2018 (NIST), value of G; NASA Planetary Fact Sheets (planetary parameters and escape velocities); J. Michell, Philosophical Transactions 74 (1784). This page is for educational purposes.
v₂ is the escape velocity, G the gravitational constant, M the mass of the central body, r the distance from its center. For Earth the result is 11.19 km/s.
Launch from the Earth’s surface: M = {{u:5.972e24:mass:0}}, r = {{u:6371000:dist-m:0}}. This is the reference value for the other examples.
v₂ = 11.19 km/s
Escape from the altitude of the International Space Station, 400 km above Earth (r = {{u:6771000:dist-m:0}}). The ISS orbits at 7.67 km/s, so it would need another 3.2 km/s to escape.
v₂ = 10.85 km/s
The Moon’s surface: M = {{u:7.35e22:mass:0}}, r = {{u:1737000:dist-m:0}}. That is why the Apollo lunar module could lift off with a single small engine.
v₂ = 2.38 km/s
The surface of Mars: M = {{u:6.39e23:mass:0}}, r = {{u:3390000:dist-m:0}}. That is less than half the Earth value, yet still too much for a simple lander that has to return with samples.
v₂ = 5.02 km/s
Two Earth radii from the center (r = {{u:12742000:dist-m:0}}). The escape velocity drops by exactly √2 and equals Earth’s surface orbital speed.
v₂ = 7.91 km/s
Geostationary orbit, r = {{u:42164000:dist-m:0}}. A satellite there moves at 3.07 km/s; it needs only 1.27 km/s more to leave Earth.
v₂ = 4.35 km/s
Escape from the Sun at the distance of Earth’s orbit (1 AU = {{u:1.496e11:dist-m:0}}). Earth already moves at 29.8 km/s, which interstellar probes exploit.
v₂ = 42.1 km/s
Jupiter: M = {{u:1.9e27:mass:0}}, mean radius {{u:69911000:dist-m:0}}. That is more than five times the Earth value, which is why the gas giant kept its hydrogen and helium.
v₂ = 60.2 km/s
The dwarf planet Ceres: M = {{u:9.39e20:mass:0}}, r = {{u:470000:dist-m:0}}. The Dawn spacecraft entered its orbit with minimal fuel.
v₂ = 0.52 km/s
Phobos, the larger moon of Mars: M = {{u:1.06e16:mass:0}}, r ≈ {{u:11000:dist-m:0}}. A ball thrown faster than 11 m/s would fly off into space.
v₂ ≈ 11.3 m/s
The surface of the Sun: M = {{u:1.989e30:mass:0}}, r = {{u:6.957e8:dist-m:0}}. The escape velocity is about 0.2 % of the speed of light.
v₂ = 618 km/s
A school problem with the rounded constant G = 6.67·10⁻¹¹ N·m²/kg² for the Earth’s surface. The difference from the CODATA value is 4 m/s.
v₂ = 11.18 km/s
The calculator finds the escape velocity, also called the second cosmic velocity, from v₂ = √(2GM/r). It multiplies the constant G by the mass M, doubles the result, divides by the distance r, and takes the square root. For Earth (5.972·10²⁴ kg, 6,371 km) it returns 11.19 km/s.
By default the mass is in kilograms, the distance in meters, and G in N·m²/kg²; the result is in km/s as well as m/s and km/h. After switching the header to US units the mass is in pounds, the distance in feet, and the speeds in mi/s, ft/s, and mph. Exponential notation, such as 5.972e24, is accepted.
No. An empty field means the CODATA 2018 value: 6.67430·10⁻¹¹ N·m²/kg². Enter your own value only when a problem specifies a rounded constant, for example 6.67·10⁻¹¹; for Earth that changes the result by about 4 m/s.
The first, v₁ = √(GM/r), keeps a body on a circular orbit just above the surface; the second, v₂ = √(2GM/r), lets it recede to infinity. The relation is exact: v₂ = v₁·√2. For Earth v₁ = 7.91 km/s and v₂ = 11.19 km/s; the result panel reports both values.
No. In the energy equation ½mv² = GMm/r the mass m cancels, so a stone, a satellite, and a crewed spacecraft need the same speed. The energy and the amount of propellant do depend on the mass, because kinetic energy grows in proportion to m.
In the ideal model no: only the magnitude of the speed matters, because the potential energy depends solely on distance. In practice the direction must avoid the planet’s surface, and an eastward launch adds the speed of Earth’s rotation — 0.465 km/s at the equator.
It is the product of the gravitational constant and the mass of the body, denoted μ in astrodynamics. For Earth it equals 3.986·10¹⁴ m³/s². It is known from satellite observations far more precisely than G or the mass alone, which is why engineers compute orbits directly from GM.
Because r in the formula is the distance from the center of mass. For launch from the surface r equals the radius of the body; for an orbit the altitude must be added to the radius. The ISS orbits 400 km above Earth, so r = 6,371 + 400 = 6,771 km and v₂ = 10.85 km/s.
No. Escape velocity applies to a body moving freely, without propulsion. A rocket accelerates gradually, and it is enough to reach the v₂ appropriate for the distance at which the engine shuts down — usually already outside the atmosphere, where the required value is slightly lower than at the surface.
This calculator does not. In practice the losses to drag and gravity during ascent total about 1.5–2 km/s, which a rocket must add to the theoretical value. For bodies without an atmosphere, such as the Moon, the calculator’s result is close to reality.
It is the ratio of the computed escape velocity to the value 11.19 km/s for the Earth’s surface. It makes comparisons easy: the Moon has 0.21, Mars 0.45, Jupiter about 5.4, and the Sun 55. A value of 1.000 means exactly Earth conditions.
The Newtonian formula with v₂ = c gives the radius r = 2GM/c², which formally coincides with the Schwarzschild radius of general relativity; that was the argument of Michell and Laplace in the 18th century. A correct description of motion near such an object, however, requires relativity, not this calculator.
Use v₂ = √(2GM/r). For Earth with M = 5.972·10²⁴ kg and r = 6.371·10⁶ m you get ≈ 11.19 km/s. The panel also shows v₁ = v₂/√2 ≈ 7.91 km/s.
About 11.19 km/s from the surface (r = 6,371 km). From ISS altitude (r = 6,771 km) it falls to 10.85 km/s.
About 2.38 km/s at M = 7.35·10²² kg and r = 1,737 km. That is why the Apollo lunar module could lift off with one small engine.
No. The third is escape from the Solar System from Earth’s orbit — about 16.7 km/s relative to Earth. This calculator returns only v₂ relative to one body M.
Yes, when g = GM/r² at the same point: √(2gr) = √(2GM/r). For Earth √(2·9.82·6.371·10⁶) matches the full G·M formula.
The definition, unit, history, and derivation below come from recognized physics and metrology sources.
Page updated November 3, 2026.