Newtonian Gravity Explorer

F = Gm₁m₂/r² — Newton's universal law of gravitation. Every mass attracts every other mass, and doubling the distance quarters the force.

Starting values — Newtonian Gravity Explorer

Distance (r)	3.84 × 10^8

Change any control above and the results recompute from these.

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the inverse-square law and its limits 🖖

The 1/r² dependence is not an arbitrary choice — it emerges from geometry. Gravity spreads like light from a point source: the same total flux passes through any sphere surrounding the mass, so intensity falls as 1/(4πr²). This is why all fields that spread isotropically in three dimensions share the same 1/r² profile. The law holds everywhere Newton tested it — planets, pendulums, tides — but breaks down in strong-field regimes. Mercury's perihelion advances 43 arcseconds per century more than Newtonian gravity predicts. General relativity replaces the instantaneous force with the curvature of spacetime: mass tells space how to curve, and curved space tells mass how to move. For most engineering and solar-system work, the Newtonian formula is accurate to better than one part in a million.

the pull goes both ways 🖖

The formula F = G·m₁·m₂/r² treats both masses the same, so the force Earth exerts on you is exactly the force you exert on Earth — equal and opposite, as Newton's third law demands. You don't fly toward nearby objects because acceleration is force divided by mass: Earth's enormous mass barely twitches. The constant G ≈ 6.674 × 10⁻¹¹ is so tiny that gravity only becomes noticeable once at least one mass is astronomically large.

the weakest force by a wide margin 🖖

Of nature's four fundamental forces, gravity is by far the feeblest. Compare two protons: their electric repulsion outmuscles their gravitational attraction by a factor of roughly 10³⁶. That is why a coin-sized magnet lifts a paperclip against the pull of the entire planet. Gravity only rules the cosmos because mass, unlike electric charge, never cancels out — it only ever adds up.

Problem solved in full

Checking the gravitational force against the Moon's actual orbit 5 steps
Work out the Earth–Moon gravitational force, check it against the Moon's actual orbit — and then ask whether the Earth is even the dominant pull on the Moon.
1. One formula, three numbers, and the only care needed is the square. Distances enter twice and masses once, so a 1% error in r costs 2%.
  
  $F = \frac{G m _{1} m _{2}}{r ^{2}} = \frac{6.674 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24} \times 7.34 \times 1 0 ^{22}}{( 3.844 \times 1 0 ^{8} ) ^{2}}$
2. Multiply the top, square the bottom, divide. The tool prints this figure and nothing else on the page is measured.
  
  $numerator = 2.925 \times 1 0^{37}, r^{2} = 1.478 \times 1 0^{17} ⟹ F = 1.980 \times 1 0^{20} N$
3. Force is not the check — acceleration is, because that is what an orbit shows. Divide by the Moon's mass.
  
  $a_{Moon} = \frac{F}{m _{2}} = \frac{1.980 \times 1 0 ^{20}}{7.34 \times 1 0 ^{22}} = 2.696 \times 1 0^{- 3} m/ s^{2}$
4. Now get the same acceleration from the orbit itself: the Moon covers 2πr in 27.32 days, and a circular orbit needs v²/r. The 1% gap is real and is the Earth moving too — both bodies circle their common centre.
  
  $v = \frac{2 π r}{T} = \frac{2 π ( 3.844 \times 1 0 ^{8} )}{2.361 \times 1 0 ^{6}} = 1023 m/s, \frac{v ^{2}}{r} = 2.723 \times 1 0^{- 3} m/ s^{2}$
5. Finally, run the same law with the Sun in place of the Earth. The mass ratio and the distance ratio fight each other, and the mass ratio wins.
  
  $\frac{F _{⊙ -Moon}}{F _{\oplus -Moon}} = \frac{M _{⊙}}{m _{1}} (\frac{r}{d _{⊙}})^{2} = \frac{1.989 \times 1 0 ^{30}}{5.97 \times 1 0 ^{24}} (\frac{3.844 \times 1 0 ^{8}}{1.496 \times 1 0 ^{11}})^{2} = 2.2$
Answer

1.98 × 10²⁰ N, and no: the Sun pulls the Moon 2.2 times harder. That ratio is a fact about masses and distances only — the Sun is 333,000 times heavier and 389 times further, and 333,000/389² = 2.2 — so it holds no matter what you think an orbit is. The consequence is genuinely strange: the Moon's path around the Sun is everywhere concave toward the Sun. It never loops backwards. What the Earth does is not hold the Moon captive against the Sun; it perturbs a solar orbit the two of them are already sharing, which is why the Moon's orbit is described relative to the Earth by convention rather than by necessity.

$F = \frac{G m _{1} m _{2}}{r ^{2}} = 1.980 \times 1 0^{20} N; the Sun pulls the Moon 2.2 \times harder$

References (1)

The law this tool evaluates, and the third-law pairing block 2 turns on: I. Newton, Philosophiæ Naturalis Principia Mathematica, Book III, Proposition VII. London, 1687.

By the 42Math team · Updated 2026-05-31

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Earth–Moon - Earth–Moon: F ≈ 1.98 × 10²° N — the force that locks our Moon in orbit
Earth–Sun - Sun–Earth: F ≈ 3.54 × 10²² N — holds Earth on its year-long path
You–Earth - You & Earth (70 kg): F ≈ 687.4 N — this is your weight
ISS–Earth - ISS & Earth: F ≈ 3.65 × 10⁶ N — nearly the same g as on the surface
You–Jupiter - You & Jupiter (70 kg): F ≈ 1735 N — you'd weigh 2.5× more than on Earth