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Calcimator

Escape Velocity

v_esc = √(2GM/R) from mass and radius.

About this calculator

Escape velocity is the minimum speed an object needs, launched straight up with no further propulsion, to break free of a body's gravity permanently and never fall back -- it's derived by setting the object's kinetic energy equal to the gravitational potential energy binding it to the surface, so all of its kinetic energy is exactly used up as it coasts to an infinite distance with zero speed remaining. That energy balance, ½mv² = GMm/R, cancels the object's own mass m entirely and rearranges to v = √(2GM/R): escape velocity depends only on the body being escaped (its mass M and radius R) and the universal gravitational constant G, never on the mass of the escaping object itself -- a feather and a spacecraft need the identical speed to escape Earth, ignoring atmospheric drag. Mass and radius pull in opposite directions: a more massive body has stronger gravity and a higher escape velocity, while a larger radius puts the surface farther from the center of mass, weakening surface gravity and lowering escape velocity for the same mass -- which is why a body's density matters as much as its raw size.

Earth's values (5.972 x 10^24 kg, radius 6.371 x 10^6 m) give an escape velocity of about 11.2 km/s, the benchmark every rocket launch from Earth's surface has to reach in principle (real launches trade some of that against atmospheric drag and gravity losses along a curved trajectory rather than firing straight up at exactly 11.2 km/s). This calculator lets you swap in any body's mass and radius -- the Moon, Mars, a neutron star -- to compare how escape velocity scales across wildly different combinations of mass and size.

Inputs

lb
ft

Results

Escape Velocity (m/s)

11,186

Escape Velocity (km/s)11.186
How to Use This Calculator
  1. Enter the body's mass in kg (Earth = 5.972e24 kg, Moon = 7.342e22 kg).
  2. Enter the body's radius in m (Earth = 6.371e6 m).
  3. Review escape velocity in m/s and km/s.
  4. Compare to Earth's escape velocity (11.2 km/s) to understand relative gravity of other bodies.
  5. Note: Escape velocity ignores atmospheric drag -- actual launch energy requirements are higher.

How the result changes with Radius (m)

Radius (m)Escape Velocity (m/s)
3,185,50115,819.4
4,778,25112,916.5
9,556,5029,133.3
15,927,5037,074.6

What each input means

Mass (kg)
Earth ~5.972e24.
Radius (m)
Earth ~6.371e6.

How this is calculated

Formula

v_esc = √(2GM / R)

Worked example, using the default values

  1. Identify Input Parameters
    Mass = 5.972e+24 kg, Radius = 6371001 m = 2 input(s) provided
  2. Calculate Escape Velocity
    v_esc = sqrt(2GM / R)
    sqrt(2 x 6.6743e-11 x 5.972e+24 / 6371001) = 11186 m/s

Engine last updated . Checked against 2 independently-derived tests — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.

Frequently Asked Questions

Why doesn't the escaping object's own mass affect escape velocity?

Because escape velocity comes from setting kinetic energy (½mv²) equal to gravitational potential energy (GMm/R) and solving for v -- the escaping object's mass m appears in both terms and cancels out algebraically. Physically, a heavier object has more kinetic energy at a given speed, but gravity also pulls on it harder in exact proportion, so the two effects offset perfectly. This is the same mass-independence that makes all objects fall at the same rate in a vacuum (ignoring air resistance) -- it's a direct consequence of gravitational and inertial mass being equal.

Why does a larger radius lower escape velocity even if the mass stays the same?

Because surface gravity depends on how close the surface is to the center of mass, not just on total mass. Spreading the same mass over a larger radius moves the surface farther from the center, weakening the gravitational pull felt there and reducing the speed needed to escape from it. This is why radius sits in the denominator, inside a square root, in the escape velocity formula -- and why comparing two bodies of similar mass but very different density (and therefore radius) can produce meaningfully different escape velocities.

Is 11.2 km/s the actual speed a rocket needs at launch from Earth?

It's the theoretical minimum for an object launched with no further propulsion from Earth's surface, ignoring the atmosphere -- not a literal launch requirement. Real rockets burn fuel continuously along a curved ascent trajectory rather than being fired once at 11.2 km/s, and they lose energy to atmospheric drag and to fighting gravity during the slower parts of ascent (gravity losses). 11.2 km/s remains the correct physical benchmark for the total energy an escaping trajectory ultimately needs relative to Earth's surface, even though no real launch reaches it in one instantaneous burst.

How does the Moon's escape velocity compare to Earth's, and why?

The Moon's escape velocity is about 2.4 km/s, less than a quarter of Earth's 11.2 km/s, because both its mass and radius are much smaller than Earth's. Its mass is about 1.2% of Earth's, which alone would push escape velocity down sharply since it depends on the square root of mass, and its smaller radius (about 27% of Earth's) partly offsets that by putting the Moon's surface closer to its own center of mass -- but the mass difference dominates, which is the main reason lunar spacecraft need far less fuel to leave the Moon's surface than Earth's.

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