Example 1
M = 5.972e+24 kg, r = 6371000 m.
Gravitational Acceleration Calculator
Enter the mass and the distance from the center. You get the field acceleration, with no falling mass in the formula. Earth at r = 6371 km is 9.82 m/s². At 400 km (r = 6771 km) you still have 8.69 m/s².
Force on a small mass: F = GMm/r². At constant g: free fall and Ep = m·g·h. In orbit compare with a = v²/r.
Result
Constant G (N·m²/kg², optional), Mass M (kg) and Distance r (m). The result shows up here.
How it works
g = G·M/r² is the acceleration that mass M gives a small body at distance r. The test mass cancels, so a feather and a hammer fall together in vacuum. At Earth’s surface, M = 5.972×10²⁴ kg and r = 6.371×10⁶ m give g = 9.82 m/s².
Same numbers: G × M = 3.986×10¹⁴, divide by r² = 4.059×10¹³, you get 9.82. School 9.81 is a rounding. At 400 km altitude r grows to 6.771×10⁶ m and g only falls to 8.69 m/s². Astronauts did not “leave” the field.
Moon: M ≈ 7.35×10²² kg, r ≈ 1.737×10⁶ m, g ≈ 1.63 m/s², about one sixth of Earth’s. Mars at M ≈ 6.39×10²³ kg and r ≈ 3.390×10⁶ m is about 3.71 m/s². You type M and r. The calculator does the rest.
The same g goes into weight F = m·g and into Ep = m·g·h while g barely changes with height. On a circular orbit gravitational g equals centripetal acceleration v²/r. That is v₁ = √(GM/r).
M and r must be positive. Empty G = 6.67430×10⁻¹¹. The result is in m/s², or ft/s² with the US switch. A comma and a period in 5.972e24 are the same.
Falling under this g lives on h = ½gt² or t = √(2h/g). Here you stay with the field itself. Force on a chosen mass is next door: F = GMm/r².
How to use
- Leave G empty for the usual 6.67430×10⁻¹¹, or type the G from the problem.
- Type mass M in kilograms, for example 5.972e24 for Earth.
- Type r from the center in meters: 6371000 at the surface, 6771000 on LEO.
- Click Calculate. Earth and 6371 km give 9.82 m/s². The same M and 6771 km give 8.69 m/s².
- Fall time from a height is on t = √(2h/g). Here you compute g itself.
Formula
g = G · M / r2
Default G = 6.67430×10−11. At Earth surface g ≈ 9.82 m/s2.
Letters in g = GM/r²
The field without a test mass: g = G·M/r². Earth at r = 6371 km is 9.82 m/s²; at 400 km (r = 6771 km) you still have 8.69 m/s².
- g
- Field acceleration in m/s². G×M = 3.986×10¹⁴ divided by r² = 4.059×10¹³ is 9.82 at the surface.
- G
- The same 6.67430×10⁻¹¹ constant, optional. There is no second mass here: a feather and a hammer fall alike.
- M
- Source mass in kilograms. Earth is 5.972×10²⁴ kg; you do not type the 70 kg falling body.
- r
- Distance from the center in meters. 6.371×10⁶ m gives 9.82; 6.771×10⁶ m at the ISS drops g to 8.69.
Real-life examples
Example 2
r ≈ 6771000 m (Earth + 400 km).
Example 3
r ≈ 42164000 m.
Example 4
M ≈ 7.35e+22 kg, r ≈ 1737000 m.
Example 5
M ≈ 6.39e+23 kg, r ≈ 3390000 m.
Example 6
r = 12742000 m.
Example 7
M_☉ ≈ 1.989e+30 kg, r ≈ 149600000000 m.
Example 8
M = solar mass, r = 10000 m.
Example 9
G = 6.67e-11, Earth surface.
Example 10
r = 6800000 m.
Ways to use this calculator
- You check surface g from a planet’s M and R.
- You compare 9.82 m/s² at the surface with 8.69 m/s² at 400 km.
Frequently asked questions
How much g for Earth at r = 6371 km?
g = 9.82 m/s². At r = 6771 km (Earth plus 400 km) you get 8.69 m/s².
How is g different from a = Δv/t?
g is the acceleration from mass M. a = Δv/t is any speed change. In a drop with no drag, a ≈ g.
Why is there “no weight” on the ISS if g ≈ 8.7?
The station and you fall together along the orbit. Relative to the cabin the fictitious force vanishes, but the field is still there.
Does g depend on the falling mass?
No. In g = GM/r² the test mass cancels. F = m g grows with m. The acceleration does not.
What g does the Moon have?
About 1.63 m/s² at M ≈ 7.35×10²² kg and r ≈ 1737 km. Mars at the surface is about 3.71 m/s².
How do I get F from g?
F = m·g, or F = GMm/r² on the gravitational-force page. 70 kg and 9.82 m/s² is 687 N.
What if the problem already gives 9.81?
This calculator computes g from M and r. Type 9.81 as g on the free-fall pages, not here.
Can I change G?
Yes. Type G in the field. Empty means 6.67430×10⁻¹¹.
How do I compute a fall at this g?
At constant g use h = ½gt² or t = √(2h/g) on the fall pages.
How much g at 2 Earth radii from the center?
r = 12,742 km, g drops by four, to about 2.45 m/s².
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.