Escape velocity calculator
The speed needed to leave a planet, from its mass and radius.
Fill in the fields and the result will appear here automatically.
Compare escape speed and circular-orbit speed at the same distance r from the centre of a spherically symmetric body. In this Newtonian model without atmosphere, rotation or thrust, escape means reaching infinity with zero residual speed. The moving body has negligible mass. The mass field uses 10²⁴ kg and r is in kilometres.
How it works
Formula and logic
Escape = √(2 GM/r), orbital = √(GM/r), with G = 6.6743·10⁻¹¹.
Example
For Earth the escape velocity is 11,186 m/s — that is 40,270 km/h.
Fields and units
- Body mass — 10²⁴ kg
- Radius — km
How to use
- — Mass goes in units of 10²⁴ kilograms: Earth is 5.972, the Moon 0.07346, Mars 0.64171.
- — This is a convenient mass scale: 5.972 means 5.972·10²⁴ kg. It does not change the formula or the speed unit. Enter centre distance as r; for a starting altitude, add it to the body’s radius.
- — Orbital velocity is smaller than escape velocity by exactly √2 — you can see it in the rows.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: energy and escape speed
- Limitation
- Exterior field of a spherically symmetric body, Newtonian gravity and negligible projectile mass. Speed is relative to the centre; rotation, atmosphere, thrust and other bodies are excluded. Not applicable inside the body or at relativistic speeds.
FAQ
Does escape speed depend on the rocket's mass?
No. Only the mass of the attracting body enters the formula. A stone and a ship leave Earth at the same speed; what differs is how much fuel it takes to reach it.
Why is mass entered in units of 10²⁴ kg?
This is a convenient mass scale: 5.972 means 5.972·10²⁴ kg. It does not change the formula or the speed unit. Enter centre distance as r; for a starting altitude, add it to the body’s radius.
How does orbital velocity differ from escape velocity?
At the same r, circular speed is √(GM/r) and escape speed √(2 GM/r), a ratio of √2. A surface-skimming orbit is only a formal comparison when atmosphere or terrain is present. These speeds do not calculate rocket fuel use.
Is the atmosphere accounted for?
No. This is pure gravity. A real rocket needs margin for air resistance and for the fact that it does not accelerate instantly at the surface.