Gravitational Field Strength Calculator
Work out gravitational field strength instantly with clear inputs, formula shown and shareable results.
About Gravitational Field Strength Calculator
Field strength is GM/r^2, the acceleration a test mass would feel. Earth's mass at its mean radius gives 9.82 m/s2, matching measured surface gravity. The same numbers give escape velocity sqrt(2GM/r) - 11.2 km/s for Earth - and the circular orbital speed, which is smaller by a factor of sqrt2 at 7.9 km/s.
Formulas
Field strength
g = G M / r^2
Escape and orbital speed
v(escape) = sqrt(2GM/r); v(orbit) = sqrt(GM/r)
Worked example
With the default values below, Gravitational Field Strength Calculator returns Gravitational Field Strength of 9.8200. Change any input above and every figure here is recalculated in your browser.
Inputs used
- Mass of the body creating the fieldminimum 0 kg
- 5,972,000,000,000,000,000,000,000 kg
- Distance from the centreminimum 0 m
- 6,371,000 m
Results
- Gravitational Field Strength
- 9.8200 m/s2
- Compared With Earth's Surface Gravity
- 1.0014
- Escape Velocity From There
- 11.1860 km/s
- Circular Orbit Speed There
- 7.9097 km/s
Step by step
- 1Values used
- 2Field strength
- 3Escape and orbital speed
- 4Gravitational Field Strength= 9.8200 m/s2
- 5Compared With Earth's Surface Gravity= 1.0014
- 6Escape Velocity From There= 11.1860 km/s
- 7Circular Orbit Speed There= 7.9097 km/s
Assumptions and limits
- The result is derived only from the 2 inputs listed above. Anything not among them is not part of the calculation.
- In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
- Figures are rounded for display to 4 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.
How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.
Frequently Asked Questions
What is Earth's escape velocity?
About 11.2 km/s from the surface, ignoring atmospheric drag and the rotation of the launch site.
Why is escape velocity sqrt2 times orbital speed?
Escaping needs enough kinetic energy to cancel the whole potential well, which is twice the energy of a circular orbit at the same radius.