Universal Gravitation Calculator

Calculate gravitational force between any two masses using Newton's universal law of gravitation.

Earth: 5.972e24, Moon: 7.342e22
Earth-Moon: 3.84×10⁸ m
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Universal Gravitation and the Inverse-Square Law

Newton's law of gravitation states that every pair of masses attracts with a force proportional to both masses and inversely proportional to the square of their separation. The relation F=Gm1m2/r2 works exactly for point masses and, by the shell theorem, for spherically symmetric bodies when r is measured between their centers.

The inverse square appears because a spherically spreading gravitational influence is distributed over an area proportional to 4πr2. Doubling center-to-center distance therefore reduces force to one quarter. This same dependence gives the gravitational field of a spherical body, g=GM/r2, and the weight of a smaller mass is then F=mg.

F=Gm1m2/r2,   g=GM/r2
SymbolMeaningWhy it appears / units
FMutual gravitational-force magnitudeN; each body experiences equal magnitude in the opposite direction.
GUniversal gravitational constantApproximately 6.67430×10−11N·m2/kg2.
m1, m2Interacting masseskg; force is proportional to their product.
rCenter-to-center separationm; appears squared in the denominator.
gGravitational field strengthN/kg, numerically equal to m/s2.

Near Earth's surface, using g≈9.81m/s2 is convenient because height changes are small compared with Earth's radius. At large altitude, use r=REarth+h. For irregular extended bodies or nearby nonspherical mass distributions, the mass must be integrated rather than treated as a point at its center.

Worked Examples

Example 1: Earth-Moon gravitational force
m1=5.972e24, m2=7.342e22, r=3.84e8m
Result: F=1.98×10²⁰ N
Keeps Moon in orbit
Example 2: Person 70kg on Earth surface: r=6.371e6m
F=6.674e-11×5.972e24×70/6.371e6²
Result: F=686 N = weight
Gravitational force = weight
Example 3: Two laboratory masses
m1=m2=1000kg, r=2.0m → F=G(1000)(1000)/22
Result: F≈1.67×10−5N
Even tonne-scale objects attract extremely weakly compared with everyday contact forces.
Example 4: Gravity 400 km above Earth
R=6.371×106m, h=4.00×105m → g=GM/(R+h)2
Result: g≈8.69m/s2
Orbiting astronauts are not beyond gravity; they are in continuous free fall while moving sideways fast enough to keep missing Earth.

Common Mistakes

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Using surface-to-surface distance instead of center-to-center distance

For spherical planets, moons, or stars, r is measured between centers. At altitude h above a planet, use r=R+h.

⚠️
Forgetting to square the distance

Gravity follows 1/r2, not 1/r. Increasing separation by a factor of three reduces force by a factor of nine.

⚠️
Confusing mass with weight

Mass is measured in kilograms and does not depend on local gravity. Weight is a force in newtons and changes when the gravitational field changes.

Frequently Asked Questions

G constant significance?
G=6.674×10⁻¹¹ N·m²/kg² is one of the most fundamental constants. It was first measured by Cavendish in 1798 using a torsion balance. It's still the least precisely known fundamental constant.
Inverse square law implications?
Doubling distance → force drops to ¼. Tripling → 1/9. This is why gravity weakens rapidly with distance but never reaches exactly zero. At 10 Earth radii, g = 9.81/100 = 0.0981 m/s².
Why do both objects experience the same gravitational force?
Newton's third law requires the interaction forces to be equal in magnitude and opposite in direction. Their accelerations are not equal because a=F/m, so the less massive object changes velocity more rapidly under the same force magnitude.
Why does gravity at the International Space Station remain strong?
The station is only a few hundred kilometers above a planet whose radius is about 6371 km. Because gravitational field falls as 1/r2, that modest increase in distance reduces g only moderately. Weightlessness there results from orbital free fall.
Does gravitational force ever become exactly zero?
For an isolated mass following Newton's inverse-square model, its field approaches zero as distance approaches infinity but is not exactly zero at any finite distance. In multi-body systems, vector gravitational fields from different masses can cancel at particular locations.
When is Newtonian gravity not accurate enough?
Newtonian gravity is extremely useful for weak gravitational fields and speeds far below light speed. Near black holes, for precision orbital effects such as Mercury's relativistic perihelion shift, or in cosmology, general relativity provides the more accurate description.

Formula Explorer connections

Interpretation: This relationship connects motion, force, momentum, work or energy in a mechanical system. Assumption: Choose a consistent reference direction and unit system. The model may assume constant acceleration, rigid bodies, negligible losses or an isolated system.

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