Gravitational Acceleration Calculator

Enter the source mass M in kilograms and the distance r from the center in meters. You may leave G blank: the calculator then uses 6.67430×10⁻¹¹. The result is the field strength g = G·M/r², with no falling mass in the formula. Earth at r = 6371 km gives 9.82 m/s², and at 400 km altitude (r = 6771 km) it is still 8.69 m/s². The model does not compute force F = GMm/r² or local g from latitude.

Inputs

Result

Enter mass M and the distance r from the center. Leave G blank for the CODATA constant. Acceleration g, the product G·M, and the chart appear here.

For the force on a chosen mass, use the F = GMm/r² calculator. At constant g, the t = √(2h/g) calculator gives the fall time, the h = ½gt² calculator gives the distance, and the Ep = m·g·h calculator gives the potential energy. On a circular orbit, the same g is the centripetal acceleration.

How it works

How g is computed for Earth at r = 6,371,000 m
G·M3.9859×10¹⁴
r²4.0590×10¹³
g9.82 m/s²
M r g g = G·M / r²
A planet's gravitational field: acceleration g at distance r from the center of mass M.

Gravitational field strength is the acceleration that mass M gives a small body at distance r from its center. Newton's formula is g = G·M/r². The test mass cancels on the way from the force F = GMm/r² to the acceleration, so a feather and a hammer fall together in vacuum. For Earth at M = 5.972×10²⁴ kg and r = 6,371,000 m the calculator gives g = 9.82 m/s².

The same numbers, step by step. The default G = 6.67430×10⁻¹¹ multiplied by M gives the product G·M = 3.9859×10¹⁴ m³/s². The square of the radius is r² = 4.0590×10¹³ m². The quotient of those two quantities is 9.82 m/s². The textbook 9.81 m/s² appears when you enter 6.67×10⁻¹¹ in the G field: the same radius then gives 9.81 m/s².

At an altitude of 400 km the radius grows to 6,771,000 m and g falls to 8.69 m/s², which is 0.887 of standard gravity gₙ = 9.80665 m/s². Astronauts have not left the field: the station and they fall together along the orbit. At an r that is 8,849 m higher (6,379,849 m) the result is 9.79 m/s².

The result panel shows g, the product G·M, r², the ratio of g to gₙ, the mass and distance you entered, and the circular-orbit speed v = √(G·M/r). At Earth's surface that speed is 7,910 m/s. The chart compares g at fractions and multiples of the entered distance, at a fixed mass M.

The formula is for a point or a uniform sphere and measures r from the center, outside the body only. Inside a planet the acceleration decreases toward the center; this calculator does not compute that. It leaves out Earth's flattening and the centrifugal effect of rotation, so local g in a given country need not equal the quotient GM/R². The force on a chosen mass m is computed by the gravitational force calculator. Fall time at constant g is computed by the free-fall time calculator.

M and r must be greater than zero. A blank G field means 6.67430×10⁻¹¹. The result is in m/s², and after the header unit switch it is also shown in ft/s²; the values in this text stay in SI. A comma and a period in a field are the same decimal separator. The button “Fill this example” enters the numbers from the example and refreshes the panel and the chart.

Formula

g = G · M / r2

[g] = (N·m²/kg²)·kg / m² = m/s²; v = √(G·M / r)

Model: Newton's gravitational field strength (law of universal gravitation, Principia 1687). The default G is 6.67430×10⁻¹¹ (CODATA, NIST). Order of computation: g = G·M/r² in SI units, with no rounding of the product before the division.

Theory, derivation, and domain

Definition of gravitational field strength

Gravitational field strength at a point is the acceleration a small test body would have there. For a point mass, or for a sphere outside its surface, Newton wrote it as g = GM/r², where G is the gravitational constant, M is the source mass, and r is the distance from the center. The SI unit is m/s², the same unit as a kinematic acceleration.

This formula differs from the force F = GMm/r² by cancellation of the test mass m. That is why g does not depend on whether you drop a key or a feather. The distance dependence is inverse-square: twice the r gives four times less g. This is a property of the field, not of the motion that the formulas h = ½gt² and t = √(2h/g) describe later.

History: Newton, Cavendish, and the constant G

Isaac Newton stated the law of universal gravitation in the Principia (1687): two bodies attract each other with a force proportional to the product of the masses and inversely proportional to the square of the distance. The field acceleration is that force divided by the test mass. For more than a century G stayed hidden inside the product GM, because pendulums and orbits measure GM, not G alone.

Henry Cavendish measured the attraction of lead spheres with a torsion balance in 1798 and thereby made a numerical G possible. The value recommended by CODATA and published by NIST is 6.67430×10⁻¹¹ m³/(kg·s²), with a relative uncertainty of order 2×10⁻⁵. The 2018 and 2022 adjustments give the same number, because no newer measurement changed the result. The calculator inserts it when the G field is blank.

Deriving g = GM/r² from the law of gravitation

Step 1. The law of gravitation: F = G·M·m / r². Step 2. Newton's second law for the test body: F = m·a, so a = F/m. Step 3. The mass m cancels and a = G·M/r² remains. That acceleration is written g when the source is a planet or a star. Step 4. Circular-orbit speed follows when g equals the centripetal acceleration v²/r, so v = √(GM/r) = √(g·r).

For Earth: G·M = 6.67430×10⁻¹¹ · 5.972×10²⁴ = 3.9859×10¹⁴ m³/s². At r = 6,371,000 m the square is 4.0590×10¹³ m², and the quotient is 9.82 m/s². The square root of GM/r is 7,910 m/s. None of these steps rounds the product before the division; rounding to two decimal places applies only to the g that is shown.

Assumptions: a point, a sphere, and distance from the center

The formula is exact for a point mass and, outside the surface, for a spherically symmetric sphere. The distance r is measured from the center, not from the surface. Altitude above the ground must be added to the radius: 400 km above the surface at a radius of 6371 km is r = 6771 km, not 400 km. Entering the altitude alone in the r field would produce a field that is far too strong.

Inside a uniform sphere a different expression holds: g grows linearly from zero at the center to the surface value. This calculator does not use it. It also does not compute tides, flattening, or the centrifugal effect of daily rotation. At Earth's surface those effects shift local g by a few thousandths of a meter per second squared relative to a pure GM/R², which is why 9.82 m/s² and the conventional gₙ = 9.80665 m/s² are not the same number.

Earth, the ISS, and the difference from 9.81

Standard gravity gₙ = 9.80665 m/s² was adopted by the 3rd General Conference on Weights and Measures in 1901 as a conventional value. Real g on the surface ranges from about 9.780 m/s² at the equator to about 9.832 m/s² at the poles. The quotient GM/R² for the mean radius is a different number: 9.82 m/s² with the CODATA G.

In an orbit at an altitude of 400 km, g = 8.69 m/s². That is still 0.887 gₙ, so saying there is “no gravity” on the station describes weightlessness relative to the cabin, not the disappearance of the field. A summit 8,849 m above the mean radius lowers the result only to 9.79 m/s². A mountain does not switch the field off.

The Moon, Mars, and the Sun's field on Earth's orbit

The same formula, different M and r. The Moon at 7.35×10²² kg and 1,737,000 m gives 1.63 m/s². Mars at 6.39×10²³ kg and 3,390,000 m gives 3.71 m/s². The Sun at 1.989×10³⁰ kg and a distance of 1.496×10¹¹ m gives 5.932×10⁻³ m/s². That acceleration is small, but it acts continuously and bends Earth's yearly orbit.

The calculator does not turn this g into a force, a fall time, or an escape speed as the primary result. It shows circular-orbit speed beside the result, because that speed is √(GM/r). The force between two masses has its own calculator, F = GMm/r². Escape from a given radius is covered by the second cosmic velocity calculator. This calculator reports the field itself.

Sources

The definition, the derivation, and the constant come from Newton's law of gravitation, Cavendish's measurement, and metrology recommendations: I. Newton, Philosophiæ Naturalis Principia Mathematica, 1687; H. Cavendish, Experiments to determine the density of the Earth, Philosophical Transactions of the Royal Society, 1798; CODATA 2022, Newtonian constant of gravitation, physics.nist.gov/constants (G = 6.67430×10⁻¹¹, identical to the 2018 adjustment); BIPM, The International System of Units (SI), 9th edition, and the 1901 resolution of the 3rd CGPM on gₙ; D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, the chapter on universal gravitation; R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics, Vol. I, Chapter 7; NASA National Space Science Data Center, Planetary Fact Sheet, as the source of the orders of magnitude for masses and radii in the examples, not as the formula. This page is for educational purposes.

How to use

  1. Leave G blank for 6.67430×10⁻¹¹, or enter the G from the problem, for example 6.67×10⁻¹¹.
  2. Enter mass M in kilograms. For Earth that is 5.972×10²⁴ kg, entered as 5.972e24.
  3. Enter r from the center in meters: 6371000 at the mean radius, 6771000 at an altitude of 400 km.
  4. Click Calculate or wait for the live result. Earth and 6371 km give 9.82 m/s².
  5. Check the panel: G·M, r², the ratio to gₙ, the mass and distance you entered, and the circular-orbit speed. Clear removes the inputs and hides the panel.

Symbol glossary

g is the field strength, G the gravitational constant, M the source mass, and r the distance from the center. For Earth at 6371 km the result is 9.82 m/s².

g
Field strength [m/s²], g = G·M/r². At M = 5.972×10²⁴ kg and r = 6,371,000 m it is 9.82.
G
Gravitational constant. A blank field means 6.67430×10⁻¹¹ m³/(kg·s²).
M
Source mass of the field [kg]. This is not the falling body's mass.
r
Distance from the center [m]. 6,371,000 m gives 9.82; 6,771,000 m gives 8.69.
G·M
Intermediate product [m³/s²]. For the default G and Earth's mass it is 3.9859×10¹⁴.
r²
Square of the distance [m²]. At 6,371,000 m it is 4.0590×10¹³.
gₙ
Standard acceleration 9.80665 m/s² (3rd CGPM, 1901). The ratio g/gₙ for the Earth example is 1.001.
v
Circular-orbit speed [m/s], √(GM/r). At Earth's surface, 7,910 m/s.

Real-life examples

Example 1

You stand at Earth's mean radius, 6371 km from the center, with a mass of 5.972×10²⁴ kg. The calculator returns Newton's field, without a correction for flattening or rotation.

g = 9.82 m/s²

Example 2

A station at 400 km altitude: the same Earth center, but r = 6771 km. The field falls with the square of distance, so the drop in g is clear and still far from zero.

g = 8.69 m/s²

Example 3

Geostationary orbit sits about 42,164 km from Earth's center. At that distance the same M produces a field much weaker than at the surface.

g = 0.224 m/s²

Example 4

The Moon at a mass of 7.35×10²² kg and a radius of 1737 km. The acceleration sits just under one sixth of the Earth field from the first example.

g = 1.63 m/s²

Example 5

Mars at a mass of 6.39×10²³ kg and a radius of 3390 km. The result is the surface value usually used in problems about a fall on Mars.

g = 3.71 m/s²

Example 6

A point at two Earth radii from the center, r = 12,742 km. Because g falls as 1/r², the field is four times weaker than at the surface.

g = 2.45 m/s²

Example 7

The Sun's field at 1 AU, with a mass of 1.989×10³⁰ kg and r = 1.496×10¹¹ m. This acceleration keeps Earth on its yearly orbit.

g = 5.932e-3 m/s²

Example 8

The same Earth model, but r is 8,849 m higher, like a summit above the mean radius. The difference from 9.82 m/s² is only a few hundredths.

g = 9.79 m/s²

Example 9

A textbook G of 6.67×10⁻¹¹ with the same Earth and r = 6371 km. That rounding of the constant, not the CODATA quotient itself, produces 9.81 m/s².

g = 9.81 m/s²

Example 10

A distance of 6800 km from Earth's center, about 429 km above the mean radius. A direct check that g is still close to the 400 km value.

g = 8.62 m/s²

Ways to use this calculator

  • You check surface g for a planet when you know its mass and the radius measured from the center.
  • You compare 9.82 m/s² at Earth's mean radius with 8.69 m/s² at an altitude of 400 km.

Common questions

What does this calculator compute, and from which formula?

The calculator finds the gravitational field strength g = G·M/r². It multiplies the constant G by the source mass M and divides that product by the square of the distance from the center. For M = 5.972×10²⁴ kg and r = 6,371,000 m the result is 9.82 m/s².

Which units should I enter for mass, distance, and G?

Mass is in kilograms, distance r is in meters, and the result is in m/s². G stays in SI even after the header is switched to US units. A comma and a period in a field are the same decimal separator.

What happens if I leave G blank?

A blank field means G = 6.67430×10⁻¹¹ m³/(kg·s²), the CODATA value published by NIST (the same in the 2018 and 2022 adjustments). Enter another number only when the problem gives its own G.

Does g depend on the mass of the falling body?

No. In g = GM/r² the test mass cancels, so a feather and a hammer fall together in vacuum. The force F = m·g grows with the body's mass. The acceleration does not. 70 kg at 9.82 m/s² is 687.4 N.

How is this result different from an acceleration a = Δv/t?

g is the field acceleration of mass M at distance r. a = Δv/t describes any change of speed, including one from an engine or from drag. In a drag-free drop near the surface, a is approximately this g.

Why does weight seem to vanish in orbit if the field is still there?

The station and the person fall together along the orbit, so the apparent force relative to the cabin vanishes. The field does not: at r = 6,771,000 m the calculator gives 8.69 m/s², which is 0.887 of standard gₙ.

Can the formula be used inside a planet?

No. g = GM/r² describes a point or a uniform sphere outside the body, with r measured from the center. Inside a planet the acceleration decreases toward the center. This calculator always uses the exterior formula.

Why is the result at Earth's surface not equal to 9.81?

The quotient for M = 5.972×10²⁴ kg and r = 6,371,000 m is 9.82 m/s². The textbook 9.81 m/s² appears after you enter G = 6.67×10⁻¹¹. Standard gravity gₙ = 9.80665 m/s² is the 1901 convention. Local g also depends on latitude and flattening, which this formula does not compute.

How do I get a force, a fall time, or an orbit speed from this g?

The force on a chosen mass is computed by the F = GMm/r² calculator. Fall time at constant g is t = √(2h/g), and distance is h = ½gt². Circular-orbit speed is on the result panel as √(GM/r); at Earth's surface it is 7,910 m/s.

What does the result panel show?

The panel shows g, the product G·M, r², the ratio of g to gₙ = 9.80665 m/s², the mass and distance you entered, and the circular-orbit speed. The chart compares g at fractions and multiples of the entered distance, at the same mass M.

Can the calculator find g for the Moon, Mars, and the Sun?

Yes, once you enter the mass and the distance from the center. The Moon at 7.35×10²² kg and 1737 km gives 1.63 m/s², Mars at 6.39×10²³ kg and 3390 km gives 3.71 m/s², and the Sun at 1 AU gives 5.932×10⁻³ m/s².

Why is there no result when mass or distance is zero?

The source mass and the distance from the center must be greater than zero. Zero in the denominator r² is meaningless, and a zero mass produces no field. While either value is out of range, the calculator shows a message instead of a number.

How do I calculate gravitational acceleration with g = GM/r²?

Multiply G by the source mass M and divide by r². At M = 5.972×10²⁴ kg and r = 6,371,000 m you get about 9.82 m/s².

What is g at 400 km above Earth?

About 8.69 m/s². Add 400 km to the mean radius 6371 km, so r = 6771 km — not 400 km in the distance field.

Is r altitude above the surface or distance from the center?

Distance from the center of mass. Add altitude h to the body radius: r = R + h. Entering h alone in the r field overstates the field by many orders of magnitude.

How is this different from the gravitational force calculator?

Here the result is acceleration g = GM/r² with no second mass. Force F = GMm/r² needs a test mass m and belongs on the gravitational-force page.

What is g on the Moon at 7.35×10²² kg and r = 1737 km?

About 1.63 m/s². That is roughly one sixth of the Earth surface value from the first example under the same Newton model.

The result is the quotient g = G·M/r²: for 5.972×10²⁴ kg and r = 6,371,000 m it is 9.82 m/s². The force on a chosen mass is computed by the gravitational force calculator.

Sources

The definition, the constant, and the history below come from the law of gravitation, CODATA/NIST, and SI metrology.

Page updated November 3, 2026.