Escape VelocityFree online science calculator — instant results, no sign-up
Calculate the escape velocity of a planet from its mass and radius.
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About the Escape Velocity
The Escape Velocity calculator estimates the minimum speed needed to escape a planet's gravity from its surface using the planet's mass and radius. It reports the result in meters per second and kilometers per second, assuming no atmospheric drag or further propulsion. The default inputs, 5.972 × 10²⁴ kg and 6.371 × 10⁶ m, represent Earth.
- Planet mass (×10²⁴ kg)
- The planet's total mass in units of 10²⁴ kilograms, so the default entry of 5.972 means 5.972 × 10²⁴ kg.
- Planet radius (×10⁶ m)
- The distance from the planet's center to its surface in units of 10⁶ meters, so the default entry of 6.371 means 6.371 × 10⁶ m.
Formula and method
M = m × 10²⁴ kg; R = r × 10⁶ m; v_esc = √(2GM/R); v_km/s = v_esc/1000. With the defaults: v_esc = √[2 × (6.67430 × 10⁻¹¹) × (5.972 × 10²⁴) / (6.371 × 10⁶)] ≈ 11,186 m/s = 11.186 km/s.
- mThe numerical entry in the planet mass field; the default is 5.972.
- rThe numerical entry in the planet radius field; the default is 6.371.
- MPlanet mass converted to kilograms.
- RPlanet radius converted to meters.
- GNewtonian gravitational constant, approximately 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻².
- v_escEscape velocity at the planet's surface in meters per second.
- v_km/sThe same escape velocity expressed in kilometers per second.
Convert the entered mass and radius to kilograms and meters, calculate 2GM/R, and take its square root. Divide the speed in meters per second by 1,000 to express it in kilometers per second. The formula comes from setting the initial kinetic energy equal to the energy needed to overcome the planet's gravitational attraction. With the default inputs, the displayed results are 11,186 m/s and 11.186 km/s after rounding.
Worked example
These are the calculator's starting values and the results it actually produces for them.
Inputs
- Planet mass5.972 ×10²⁴ kg
- Planet radius6.371 ×10⁶ m
Results
- Escape velocity11,186 m/s
- In km/s11.186
- NoteDefaults are Earth's values
How to use the Escape Velocity
- 1Enter the planet's mass in units of 10²⁴ kg; leave 5.972 for the Earth example.
- 2Enter the planet's radius in units of 10⁶ m; leave 6.371 for the Earth example.
- 3Check that both entries are positive and that you entered the scaled values rather than the full values in kilograms and meters.
- 4Run the calculation to obtain escape velocity in meters per second and kilometers per second.
- 5For the defaults, confirm Escape velocity = 11,186 m/s, In km/s = 11.186, and Note = Defaults are Earth's values.
Reading your result
The 'Escape velocity' row gives the minimum initial speed in m/s needed to escape the idealized planet without additional propulsion, reaching infinitely far away with zero remaining speed. The 'In km/s' row expresses the same speed in km/s, not a separate calculation of a different physical quantity. For the default inputs, these rows read 11,186 m/s and 11.186. The note 'Defaults are Earth's values' identifies the preset mass and radius; it does not mean that custom entries necessarily describe Earth.
Limitations
The calculation assumes an isolated, spherical planet and Newtonian gravity. It ignores atmospheric drag, planetary rotation, terrain, other celestial bodies, and relativistic effects. Both mass and radius must be positive. The result is a speed threshold, not a launch trajectory, fuel requirement, or rocket delta-v budget; a powered spacecraft need not acquire this entire speed at the surface in one instant. Input precision and output rounding limit the precision of the result.
Common questions
Why do the default inputs produce an escape velocity of 11.186 km/s?+
The defaults use Earth's approximate mass, 5.972 × 10²⁴ kg, and radius, 6.371 × 10⁶ m. Substituting them into √(2GM/R) gives approximately 11,186 m/s, displayed as 11.186 km/s.
How do mass and radius affect a planet's escape velocity?+
Escape velocity is proportional to the square root of mass and inversely proportional to the square root of radius. Increasing mass while holding radius fixed raises escape velocity; increasing radius while holding mass fixed lowers it.
Is escape velocity the same as the speed needed for a circular orbit?+
No. At the same distance from the center of a spherical planet, circular orbital speed is √(GM/R), while escape velocity is √(2GM/R). Escape velocity is therefore √2 times the circular orbital speed.
Can I estimate escape velocity from above the planet's surface?+
Yes, under the same idealized assumptions: use the distance from the planet's center to the starting point in place of the surface radius. Add altitude to the planet's radius, then express that total distance in units of 10⁶ m before entering it.
Does the calculator save my information?+
Calculations run in your browser. While signed in, inputs and results are automatically recorded in account history; saving a result creates a separate saved record.
Further reading
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Results are estimates for informational purposes and are not financial, medical or legal advice.