Escape Velocity Explained: The √(2GM/r) Formula, the √2 Orbit Link, and the Speeds You Actually Need
Escape velocity is the slowest speed at which an object can leave the surface of a body and never fall back, ignoring air drag and any other gravity sources. The formula is short — v = √(2GM/r) — but it hides a derivation, a famous mass-cancellation, a √2 relationship to orbital velocity, and an upper limit that defines black holes. This guide walks through the maths, the worked numbers for Earth, the Moon and the Sun, and the questions people actually ask before using the escape velocity calculator.
What is escape velocity?
Escape velocity is the slowest speed an object can be flung from the surface of a planet, moon or star and never fall back, ignoring air drag and any other gravity sources. Above it, the object climbs to infinity with kinetic energy to spare. Exactly at it, the object just barely escapes, ending up at rest infinitely far away. Below it, gravity wins and the object falls back, no matter how high it gets first.
That “to infinity” framing is what makes escape velocity an unusual quantity. Almost everything else in introductory mechanics is about reaching a finite target — the top of a building, the next planet, low Earth orbit. Escape velocity is the boundary case where the target is nowhere. It is a single number that depends only on the body you are leaving, not on the rocket, the fuel, or where you are aiming. The escape velocity calculator returns it for every planet in the Solar System, the Moon, the Sun, and any custom mass and radius you care to type in.
Two important caveats. First, escape velocity is the local escape speed from a given body — it is what you need to leave that body's gravity well, not the Solar System or the galaxy. A spacecraft that reaches Earth's 11.2 km/s escape velocity is loose from Earth, but still orbits the Sun. Second, the calculation assumes a non-rotating spherical body in empty space. Real launches have to fight rotation, oblateness, atmospheric drag and the gravity of nearby bodies — the engineers' delta-v budget is always bigger than the textbook escape velocity.
The formula: v = √(2GM/r)
The derivation is one of the shortest in physics. Total energy of an object of mass m moving at speed v at distance r from the centre of a body of mass M is
E = ½mv² − GMm/r
where G = 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² is Newton's gravitational constant (CODATA 2018). The minus sign in front of the potential term sets zero potential at infinity, which is the convention that makes “escape” mean “arrive at infinity with non-negative energy”.
Set E = 0 and solve for v:
½mv² = GMm/r
v = √(2GM/r)
That is the entire derivation. The mass m of the escaping object appears on both sides and cancels — which is the famous and slightly counter-intuitive consequence that escape velocity does not depend on what is doing the escaping. A grain of dust, a satellite, a moon: same speed required at the same altitude, because gravity accelerates everything equally and the deeper potential well of a heavier object is exactly balanced by the extra kinetic energy that mass gives it. The escape velocity calculator just plugs your chosen M and r into that formula.
Worked example: Earth, the Moon and the Sun
Earth's mass is M = 5.9722 × 10²⁴ kg and its volumetric mean radius is r = 6 371 000 m (NASA Earth Fact Sheet). Substituting in:
v = √(2 × 6.6743 × 10⁻¹¹ × 5.9722 × 10²⁴ / 6 371 000)
v = √(1.2516 × 10⁸) ≈ 11 187 m/s
Or 11.19 km/s — about 25 020 mph, Mach 32.9 in sea-level air. That matches the value Apollo had to hit at trans-lunar injection. Doing the Moon the same way (M = 7.342 × 10²² kg, r = 1 737 400 m) gives 2.38 km/s, which is why the Apollo lunar module could carry a tiny ascent engine — escaping the Moon is roughly twenty-five times easier than escaping Earth in kinetic-energy terms. Punching the Sun's numbers (M = 1.989 × 10³⁰ kg, r = 6.957 × 10⁸ m) gives 617.7 km/s, fifty five times Earth's figure. The escape velocity calculator does the arithmetic for every body in the Solar System; the table below summarises the headline values for the ones people ask about most.
Factors that change the answer
Mass of the body
Escape velocity scales with the square root of mass: double the mass and the answer rises by √2. The Sun is 333 000 times more massive than Earth; if it had the same radius it would have 577 times Earth's escape velocity instead of just 55, because the larger radius eats some of the difference.
Radius (distance from the centre)
Escape velocity falls as 1/√r. Doubling your altitude does not halve escape velocity — it divides it by √2 ≈ 1.414. From low Earth orbit at 400 km up, Earth's escape velocity is about 10.9 km/s rather than 11.2; from geostationary orbit at 35 800 km it is 4.3 km/s; out at the Moon's distance it is 1.4 km/s. This is why interplanetary missions wait until they are in low Earth orbit before doing the trans-injection burn — most of the escape work is already done.
Density (mass packed into a given radius)
Because v scales as √(M/r) and M for a given density scales as r³, escape velocity rises roughly linearly with radius at fixed density. That is why dense, compact bodies — neutron stars, white dwarfs, black holes — have enormous escape velocities relative to their size. A neutron star packs roughly a solar mass into a 10 km radius and has an escape velocity around 150 000 km/s, half the speed of light.
Rotation of the body
The Newtonian formula assumes a non-rotating body, but the surface of a rotating planet is already moving in the non-rotating frame. Launch eastward from Earth's equator and you start with about 465 m/s of free speed — roughly 0.46 km/s, or 4% of the escape budget. NASA and ESA both site launch pads near the equator (Cape Canaveral, Kourou) to harvest that bonus. The escape velocity calculator does not subtract rotation — it returns the local escape speed in the body's inertial frame.
Other gravity sources
Escape velocity from Earth is the speed needed to leave Earth's gravity. It is not enough to leave the Solar System. To do that you have to reach the Sun's local escape velocity at Earth's orbital distance, which is about 42.1 km/s in the Sun's rest frame. Earth itself is already moving around the Sun at 29.8 km/s, so launching prograde you only need 16.6 km/s in the Earth frame — still much more than Earth-escape alone. The formula is local; the strategy depends on which gravity well you are escaping from.
How to think about it without burning yourself
- Think in energies, not heights. “Escape” means reaching zero total energy, not reaching some particular altitude. The object will slow down as it climbs, but it never quite stops.
- The mass of the rocket really does cancel. A pebble and a planet need the same surface speed to escape Earth. What scales with rocket mass is the fuel, not the speed.
- √2 separates orbit from escape. Multiply the local circular orbital speed by √2 ≈ 1.414 and you get the local escape speed. At Earth's surface: 7.9 km/s orbit, 11.2 km/s escape.
- The answer drops fast with altitude. At infinity escape velocity goes to zero, so anything already coasting outwards through deep space is by definition escaping something.
- Direction does not change the speed. Fire at any angle other than straight into the surface and gravity does the rest of the work. Direction matters for the trajectory and the fuel budget, not the escape speed itself.
- Always check which body you are escaping. “Escape velocity” with no body specified is meaningless. Earth-escape, Solar-System-escape and galactic-escape are three different numbers.
Common mistakes
Confusing escape velocity with orbital velocity. They differ by a factor of √2 — orbital is the slower one. Mixing them up gives an answer that is wrong by 40% either way. The escape velocity calculator returns escape; for circular orbital speed at the same altitude, divide by √2.
Using diameter where the formula wants radius. v depends on the distance from the centre of the body to the surface. Plug in the diameter and the answer is too small by a factor of √2, because the formula goes as 1/√r. Always halve the diameter first.
Mixing surface escape velocity with delta-v budgets. Escape velocity is the textbook minimum for a non-rotating airless body. A real launch from Earth needs about 13–14 km/s of delta-v to reach escape — the extra 2–3 km/s pays for atmospheric drag, gravity losses during the burn, and steering. Mission planners use delta-v, not escape velocity, when sizing rockets.
Forgetting that escape velocity drops with altitude. From a hovering platform 100 000 km above Earth, escape velocity is about 2.2 km/s, not 11.2 — most of the work is already done by being up there. This is why orbit-then-burn is the standard interplanetary strategy: parking in low Earth orbit, then waiting and burning, is enormously cheaper than burning straight outward from the launchpad.
Where escape velocity actually matters
Spaceflight is the obvious place. Every mission to the Moon, to Mars, or out of the Solar System has to clear Earth's 11.2 km/s, and the rocket equation makes that one of the dominant cost drivers in mission design. For human exploration of the Moon, the much smaller 2.4 km/s lunar escape speed is what makes the ascent stage a tractable engineering problem in the first place.
Astrophysics is the other big one. Stars hold on to their atmospheres because thermal molecular speeds are below escape velocity; the Moon and Mercury have lost most of their atmospheres because they are not. Jupiter holds hydrogen and helium that Earth cannot because Jupiter's escape velocity is 59.5 km/s versus Earth's 11.2 — more than five times higher, more than twenty-five times the kinetic energy at a given molecular mass. The escape velocity calculator gives the values for every Solar System body, which makes atmospheric retention back-of-the-envelope an easy calculation.
Cosmology pushes the formula to its limit. The Schwarzschild radius of a black hole — the boundary inside which escape is impossible because the required speed would exceed c — is exactly the radius at which v = √(2GM/r) equals the speed of light. For one solar mass that is 2.95 km. For Sagittarius A*, the 4.1-million-solar-mass black hole at the centre of the Milky Way, it is 12 million km. The Newtonian formula gives the right answer in this limit by a happy mathematical coincidence; general relativity is what you would normally need for the proper derivation.
When to stop and ask a physicist
The escape velocity formula is exact for a non-rotating spherical body in empty space. It is a good approximation for: every planet, moon and star in the Solar System; first-pass mission design; atmospheric retention arguments; rough sizing of black-hole event horizons.
Reach for something more sophisticated when: the body is spinning fast enough to deform (close binary systems, millisecond pulsars); the gravity field is non-spherical (asteroids, comets, Phobos); the escape speed is a significant fraction of c (neutron stars, black holes — switch to general relativity); the trajectory passes close to another massive body (use the patched-conic or full-N-body method). For any of these, the escape velocity calculator gives you a sensible first estimate, but a textbook in orbital mechanics or general relativity is the real tool.
Frequently asked questions
What is escape velocity in plain English?
The slowest speed at which something can be flung from the surface of a planet, moon or star and never fall back. Above it the object escapes; below it, gravity eventually pulls it back, no matter how high it climbs first.
What is Earth's escape velocity?
About 11.186 km/s from sea level — call it 11.2 km/s, or 25 020 mph, or Mach 33 in sea-level air. It falls to roughly 10.9 km/s at low-Earth-orbit altitude and varies slightly with latitude.
Why does the rocket's mass not appear in the formula?
Because gravity accelerates everything equally. The extra kinetic energy a heavier object has at a given speed is exactly balanced by the deeper potential well it sits in, so the speed required to escape is the same for any mass. Fuel scales with rocket mass; escape speed does not.
How does escape velocity relate to orbital velocity?
Escape velocity is √2 times the circular orbital velocity at the same radius. At Earth's surface that is 11.2 km/s versus 7.9 km/s — the famous factor-of-1.4 gap between being in orbit and being on your way out.
What is the escape velocity of a black hole?
At the event horizon, exactly the speed of light. That is the definition of a black hole. Setting v = c in the formula gives the Schwarzschild radius r = 2GM/c², about 3 km per solar mass.
Does this calculator account for atmospheric drag?
No. It returns the idealised non-rotating, non-atmospheric escape velocity from the body's surface. Real launches need a delta-v budget that is roughly 20–30% larger than the textbook escape velocity to cover drag, gravity losses and steering.
Related calculators
- Escape Velocity Calculator — the parent tool, every planet plus a custom-body mode.
- Force Calculator (F = m·a) — Newton's second law, the other side of the gravity story.
- Speed Converter — convert escape velocities between km/s, mph, km/h and knots.
- Age on Other Planets — same planetary cast, different physics question.
- Distance Calculator — distance, speed and time relationships for orbital back-of-envelope work.
Frequently asked questions
What is escape velocity in plain English?
It is the slowest speed at which an object can be flung straight up from the surface of a planet, moon or star and never fall back, ignoring air resistance and any other gravity sources. Above that speed, the kinetic energy is greater than the well of gravitational potential energy the object sits in, so it can climb to infinity with some speed to spare. Exactly at that speed, it just barely escapes, ending up at rest infinitely far away.
What is the escape velocity of Earth?
About 11.186 km/s from the surface — roughly 11.2 km/s, or 25 020 mph, or Mach 33 in sea-level air. NASA quotes 11.186 km/s as the canonical value. It varies slightly with altitude: from the surface it is 11.19 km/s, from low Earth orbit at 400 km it drops to about 10.9 km/s. Rotation helps too — launching east from the equator gives you about 0.46 km/s for free from Earth's spin.
Why does escape velocity not depend on the rocket's mass?
In the energy equation ½mv² = GMm/r, the mass m of the escaping object appears on both sides and cancels, leaving v = √(2GM/r). The same physics is why a heavy rock and a feather fall at the same rate in a vacuum: gravity accelerates everything equally, so a heavier object has both more kinetic energy and a deeper potential well to climb out of. What does scale with rocket mass is the fuel needed to reach that speed, not the speed itself.
Is escape velocity the same as orbital velocity?
No, but they are linked by a factor of √2. Circular orbital velocity at radius r is v_orb = √(GM/r); escape velocity is v_esc = √(2GM/r). So escape velocity is exactly √2 ≈ 1.414 times the local circular orbital velocity. At Earth's surface that gives 7.9 km/s for orbit and 11.2 km/s for escape — the well-known factor-of-1.4 gap that separates being "in orbit" from being "on your way out".
What is the escape velocity of a black hole?
A black hole is defined by the surface — the event horizon — where escape velocity equals the speed of light, c ≈ 299 792 458 m/s. Setting v = c in v = √(2GM/r) and solving for r gives the Schwarzschild radius r_s = 2GM/c², about 3 km for one solar mass and about 12 million km for Sagittarius A* at the centre of the Milky Way. The Newtonian formula gives the right answer here by coincidence — general relativity reproduces the same number for a non-rotating black hole.
Why is the Sun's escape velocity so much bigger than Earth's?
Mass dominates the formula and the Sun is roughly 333 000 times more massive than Earth. Its surface radius is also about 109 times bigger, which softens the answer because v scales as √(M/r), but the mass term still wins. The result is about 617.7 km/s at the surface — fifty-five times Earth's escape velocity. This is why interplanetary probes use Jupiter gravity assists and the Oberth effect rather than trying to burn straight outward from low Earth orbit.
Where do the planetary masses and radii come from?
From the NASA Planetary Fact Sheet at nssdc.gsfc.nasa.gov/planetary/factsheet/, the standard public reference for Solar System parameters. Solid bodies use the volumetric mean radius (the radius of a sphere with the same volume), gas giants use the 1-bar equatorial radius (the level where atmospheric pressure equals one Earth atmosphere). That convention matches the canonical escape-velocity figures NASA publishes alongside the data.
Does the direction of launch matter for escape velocity?
For the speed, no — escape velocity is a scalar. Fire it in any direction other than straight into the body and gravity does the rest of the work of slowing it as it climbs. For the fuel and the trajectory, direction matters a lot: a launch eastward at low latitude exploits the body's rotation, a polar launch does not. Mission planners separate these into the local escape speed (this calculator) and the delta-v budget (which adds rotation, atmosphere and gravity losses).
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