Escape Velocity Calculator

Escape velocity from a planetary surface
Escape velocity equals the square root of 2 times G times mass divided by radius

Solution

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How It Works

Escape velocity is the minimum speed a projectile must have at a planet's surface to coast away to infinity against gravity without further propulsion. It comes from setting kinetic energy equal to gravitational potential energy: ½mvₑ² = GMm/R, which solves to vₑ = √(2GM/R). G = 6.6743 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, M is the planet's mass, and R is the planet's radius. This calculator inverts the equation to solve for the escape velocity, the central mass, or the launch radius given the other two quantities — useful for any rocky or gas body in the solar system.

Example Problem

Calculate Earth's escape velocity. Earth's mass is M = 5.972 × 10²⁴ kg and its mean radius is R = 6.371 × 10⁶ m.

  1. Write the escape velocity formula: vₑ = √(2 × G × M / R).
  2. Compute the numerator: 2 × G × M = 2 × (6.6743 × 10⁻¹¹) × (5.972 × 10²⁴) ≈ 7.969 × 10¹⁴ m³/s².
  3. Divide by R: 7.969 × 10¹⁴ / (6.371 × 10⁶) ≈ 1.251 × 10⁸ m²/s².
  4. Take the square root: vₑ = √(1.251 × 10⁸) ≈ 11,186 m/s ≈ 11.19 km/s.
  5. Convert to common units: 11.19 km/s × 3,600 s/h ≈ 40,284 km/h, or about 25,020 mph — the textbook value of Earth's escape velocity.

Key Concepts

Escape velocity is independent of the launching object's mass — a feather and a rocket both need the same vₑ to escape. It is also independent of launch direction, as long as the path doesn't re-enter the atmosphere; rocket engineers usually launch eastward to exploit Earth's rotational speed of ~465 m/s at the equator. Achieving escape velocity doesn't mean escaping immediately — at exactly vₑ the object decelerates asymptotically and reaches infinity with zero residual speed. Real rocket missions need extra delta-v to overcome atmospheric drag and to reach useful interplanetary trajectories. The deeper a body's gravitational well, the harder it is to leave: Earth needs 11.2 km/s, the Moon only 2.4 km/s, Jupiter 59.5 km/s, and the Sun a staggering 617.5 km/s.

Applications

  • Rocket and spacecraft design: setting the minimum delta-v budget for launches that must reach interplanetary or solar-escape trajectories.
  • Astrophysics: predicting which atmospheric molecules a planet can retain (the lighter the molecule, the more likely it escapes when its thermal speed nears vₑ).
  • Black hole physics: the Schwarzschild radius is where escape velocity equals the speed of light — a direct generalization of the same formula.
  • Lunar and Martian mission planning: ascent stages must reach the body's escape velocity to return to Earth's gravitational sphere of influence.
  • Comparative planetology: ranking solar system bodies by their gravitational binding energy via vₑ.

Common Mistakes

  • Confusing escape velocity with orbital velocity — orbital velocity is √(GM/R), which is exactly vₑ/√2 (about 70.7% of escape velocity). Earth's low-orbit speed is ~7.9 km/s, not 11.2 km/s.
  • Assuming escape velocity depends on the rocket's mass — it doesn't. The launching object's mass cancels out of the energy balance.
  • Using altitude above the surface as R — the formula uses the full distance from the planet's center, so for surface launches R is the planet's radius, not zero.
  • Forgetting the factor of 2 inside the square root — vₑ = √(2GM/R), not √(GM/R) (that gives orbital velocity, off by a factor of √2).
  • Treating escape velocity as a target speed that must be maintained — once an object reaches vₑ at the launch radius, gravity continuously slows it; the object decelerates but never quite stops if energy is conserved.

Frequently Asked Questions

How do you calculate escape velocity?

Apply vₑ = √(2 × G × M / R), where G = 6.6743 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, M is the central body's mass in kilograms, and R is its radius (or the launch distance from its center) in meters. Result vₑ comes out in meters per second.

What is the formula for escape velocity?

vₑ = √(2GM/R). The derivation sets the launching object's kinetic energy ½mvₑ² equal to the gravitational potential energy GMm/R needed to climb out of the gravity well to infinity. The launching mass m cancels, leaving vₑ independent of the projectile.

What is Earth's escape velocity?

About 11,186 m/s or 11.19 km/s — roughly 25,020 mph. That's the minimum speed needed at Earth's surface to coast to infinity ignoring air drag. Real rockets need more to overcome atmospheric losses and to reach useful destinations.

Does escape velocity depend on the object's mass?

No. The formula vₑ = √(2GM/R) shows escape velocity depends only on the central body's mass M and the launch radius R — not on the projectile's own mass. A feather and a rocket both need the same vₑ to escape (though rockets need fuel to reach it).

What is the escape velocity from the Moon, Mars, and Jupiter?

Moon: 2.38 km/s. Mars: 5.03 km/s. Jupiter (from its 1-bar cloud tops): 59.5 km/s. The Sun's escape velocity at its surface is about 617.5 km/s — the steepest gravitational well in our solar system.

Is escape velocity the same as orbital velocity?

No. Orbital velocity at radius r is v = √(GM/r), which equals vₑ / √2 — about 71% of escape velocity. At Earth's surface (or ISS altitude), orbital velocity is about 7.9 km/s while escape velocity is 11.2 km/s. Orbital velocity keeps you circling; escape velocity lets you leave entirely.

References:

Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th Edition.

Gravitational constant G: NIST CODATA. https://physics.nist.gov/cgi-bin/cuu/Value?bg

Worked Examples

Lunar Mission Design

What launch speed does an Apollo-style ascent stage need to escape the Moon?

The Moon has a mass of M ≈ 7.342 × 10²² kg and a surface radius of R ≈ 1.7374 × 10⁶ m. A lunar lander on its way home only has to reach escape velocity from the Moon — Earth's gravity does the rest.

  • Knowns: M = 7.342 × 10²² kg, R = 1.7374 × 10⁶ m
  • G = 6.6743 × 10⁻¹¹ N·m²/kg²
  • vₑ = √(2 × G × M / R)
  • vₑ = √(2 × 6.6743 × 10⁻¹¹ × 7.342 × 10²² / 1.7374 × 10⁶)
  • vₑ = √(5.640 × 10⁶ m²/s²)

vₑ ≈ 2,375 m/s (≈ 2.38 km/s)

Less than a quarter of Earth's 11.2 km/s. That's why a small ascent stage with a single engine can leave the lunar surface — the energy budget scales with v², so 2.4 km/s requires roughly (2.4/11.2)² ≈ 4.5% of the energy per kilogram.

Planetary Science

What is Mars' escape velocity for a return-to-Earth ascent?

Mars has M ≈ 6.4171 × 10²³ kg and R ≈ 3.3895 × 10⁶ m. A sample-return mission like Mars Sample Return needs to leave Mars at this speed (plus a margin for atmospheric drag and burn losses).

  • Knowns: M = 6.4171 × 10²³ kg, R = 3.3895 × 10⁶ m
  • G = 6.6743 × 10⁻¹¹ N·m²/kg²
  • vₑ = √(2 × G × M / R)
  • vₑ = √(2 × 6.6743 × 10⁻¹¹ × 6.4171 × 10²³ / 3.3895 × 10⁶)
  • vₑ = √(2.527 × 10⁷ m²/s²)

vₑ ≈ 5,026 m/s (≈ 5.03 km/s)

Roughly half of Earth's escape velocity but more than twice the Moon's. Mars' thin atmosphere also creates aerobraking losses, so real Mars Ascent Vehicles target ≈ 4.0 km/s burnout from the surface — orbital insertion supplies the rest.

Astronomy

What speed leaves the Solar System starting from Earth's orbit around the Sun?

Compute the heliocentric escape velocity at 1 AU. M_sun ≈ 1.989 × 10³⁰ kg and R = 1 AU = 1.496 × 10¹¹ m. This is the speed a spacecraft must reach relative to the Sun (not Earth) to leave the Solar System on a parabolic trajectory.

  • Knowns: M = 1.989 × 10³⁰ kg, R = 1.496 × 10¹¹ m
  • G = 6.6743 × 10⁻¹¹ N·m²/kg²
  • vₑ = √(2 × G × M / R)
  • vₑ = √(2 × 6.6743 × 10⁻¹¹ × 1.989 × 10³⁰ / 1.496 × 10¹¹)
  • vₑ = √(1.774 × 10⁹ m²/s²)

vₑ ≈ 42,123 m/s (≈ 42.1 km/s)

Earth already moves at ≈ 29.78 km/s in its solar orbit, so a prograde launch only needs 42.1 − 29.78 ≈ 12.3 km/s extra (plus Earth-escape and gravity-loss budget). This is exactly why Voyager and New Horizons launched in Earth's orbital direction.

Escape Velocity Formula

Escape velocity comes from setting kinetic energy equal to the gravitational binding energy at the launch radius — the minimum speed at which an unpropelled projectile can coast to infinity against a planet's gravity.

vₑ = √(2 × G × M / R)

Where:

  • vₑ — escape velocity at the launch radius (m/s)
  • G — Newton's universal gravitational constant, 6.6743 × 10⁻¹¹ N·m²/kg²
  • M — mass of the central body (kg)
  • R — distance from the body's center to the launch point (m); equal to the planet's radius for a surface launch

The formula ignores atmospheric drag and assumes a non-rotating frame. Real rockets need to add a margin for gravity losses and aerodynamic drag, and they benefit from a prograde launch that adds the planet's surface rotation speed to vₑ. Escape velocity is direction-independent and does not depend on the escaping object's mass.

Planetary Escape Velocities

Mass, mean radius, and escape velocity for the Sun, Moon, and the inner planets out to Jupiter and Saturn. Click any row to load that body's mass and radius into the calculator and solve for its escape velocity. For the full solar-system list, see the planetary data reference.

BodyMass (kg)Mean radius (km)Escape velocity (km/s)Source
Sun1.99×10³⁰696,000617.5NASA Planetary Fact Sheet
Mercury3.3×10²³2,4404.25NASA Planetary Fact Sheet
Venus4.87×10²⁴6,05210.36NASA Planetary Fact Sheet
Earth5.97×10²⁴6,37111.19NASA Planetary Fact Sheet
Moon7.34×10²²1,7372.38NASA Planetary Fact Sheet
Mars6.42×10²³3,3905.03NASA Planetary Fact Sheet
Jupiter1.9×10²⁷69,91159.5NASA Planetary Fact Sheet
Saturn5.68×10²⁶58,23235.5NASA Planetary Fact Sheet

The escape velocity shown is NASA's published value, which is referenced to each body's equatorial radius. Recomputing vₑ = √(2 × G × M / R) from the listed mass and mean radius differs by roughly 1–2% for the oblate, fast-spinning gas and ice giants (Jupiter and Saturn), whose equatorial radius is noticeably larger than their mean radius. For the small rocky bodies the two agree to within rounding.

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