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Gravitational Force Calculator

Newton's law of universal gravitation, surface gravity and weight with step-by-step solutions
Find the gravitational force between two 1000 kg masses 10 m apart
Find the acceleration due to gravity at Earth's surface from M = 5.972e24 kg and R = 6371 km
What does a 70 kg astronaut weigh on the Moon?
Find the separation at which two 5 kg masses attract with a force of 1e-9 N

Newton's Law of Universal Gravitation

Any two masses attract each other along the line joining their centres:

F=Gm1m2r2F = G\frac{m_1 m_2}{r^2}

Symbols and SI units:

  • FF — gravitational force, newtons (N)
  • GG — gravitational constant, 6.674×10−116.674 \times 10^{-11} N·m²/kg²
  • m1,m2m_1, m_2 — the two masses, kilograms (kg)
  • rr — distance between their centres of mass, metres (m)

When it applies: to point masses, and — by a result Newton proved — to uniform spheres treated as if all their mass sat at the centre. It is the classical limit; near black holes or for the orbit of Mercury you need general relativity.

The assumption people forget: rr is measured centre to centre, not surface to surface. For an object on Earth's surface, rr is Earth's radius (6.371×1066.371 \times 10^6 m), not zero.

Surface Gravity and Weight

Set m1=Mm_1 = M (the planet) and m2=mm_2 = m (the object), and compare with F=mgF = mg:

mg=GMmR2⟹g=GMR2mg = G\frac{Mm}{R^2} \quad\Longrightarrow\quad g = \frac{GM}{R^2}

  • gg — acceleration due to gravity, m/s² (equivalently N/kg)
  • MM — mass of the planet, kg; RR — its radius, m

The object's own mass cancels, which is why everything falls at the same rate in vacuum. Weight then follows as

W=mgW = mg

in newtons. On Earth g=9.81g = 9.81 m/s²; on the Moon it is 1.621.62 m/s².

The assumption people forget: gg is not a constant of nature. It falls off as 1/r21/r^2 with altitude, so a satellite at 400400 km experiences about 8.78.7 m/s², and mass — unlike weight — does not change with location at all.

Orbits come from the same force. Setting the gravitational pull equal to the centripetal requirement, GMm/r2=mv2/rGMm/r^2 = mv^2/r, gives the circular orbital speed

v=GMrv = \sqrt{\frac{GM}{r}}

in metres per second — which is why a lower orbit is a faster orbit, not a slower one.

Common Mistakes to Avoid

  • Squaring only part of the denominator — the whole separation is squared, and doubling rr cuts the force to a quarter, not a half.
  • Confusing mass with weight — mass is in kilograms and is the same everywhere; weight is a force in newtons and depends on local gg.
  • Using surface-to-surface distance — measure between centres of mass.
  • Dropping the 10−1110^{-11} in GG — the exponent is what makes everyday gravitational forces microscopic. Two one-tonne masses ten metres apart attract with under a micronewton of force.
  • Reusing Earth's gg elsewhere — recompute g=GM/R2g = GM/R^2 for another body rather than assuming 9.819.81 m/s².
  • Mixing kilometres and metres — convert radii to metres before squaring.

Examples

Step 1: F=Gm1m2/r2F = G m_1 m_2 / r^2 with G=6.674×10−11 N\cdotpm2/kg2G = 6.674 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2
Step 2: Numerator: (6.674×10−11)(1000 kg)(1000 kg)=6.674×10−5 N\cdotpm2(6.674 \times 10^{-11})(1000\ \text{kg})(1000\ \text{kg}) = 6.674 \times 10^{-5}\ \text{N·m}^2
Step 3: Denominator: r2=(10 m)2=100 m2r^2 = (10\ \text{m})^2 = 100\ \text{m}^2
Step 4: F=(6.674×10−5 N\cdotpm2)÷(100 m2)=6.674×10−7 NF = (6.674 \times 10^{-5}\ \text{N·m}^2) \div (100\ \text{m}^2) = 6.674 \times 10^{-7}\ \text{N}
Answer: F≈6.67×10−7F \approx 6.67 \times 10^{-7} N (under a micronewton)

Step 1: Convert the radius: R=6.371×106 mR = 6.371 \times 10^6\ \text{m}
Step 2: g=GM/R2g = GM/R^2
Step 3: Numerator: (6.674×10−11)(5.972×1024)=3.986×1014 N\cdotpm2/kg(6.674 \times 10^{-11})(5.972 \times 10^{24}) = 3.986 \times 10^{14}\ \text{N·m}^2/\text{kg}
Step 4: Denominator: (6.371×106 m)2=4.059×1013 m2(6.371 \times 10^6\ \text{m})^2 = 4.059 \times 10^{13}\ \text{m}^2
Step 5: g=(3.986×1014)÷(4.059×1013)=9.82 m/s2g = (3.986 \times 10^{14}) \div (4.059 \times 10^{13}) = 9.82\ \text{m/s}^2
Answer: g≈9.82g \approx 9.82 m/s²

Step 1: gMoon=GM/R2g_{\text{Moon}} = GM/R^2 with R=1.737×106 mR = 1.737 \times 10^6\ \text{m}
Step 2: Numerator: (6.674×10−11)(7.342×1022)=4.900×1012(6.674 \times 10^{-11})(7.342 \times 10^{22}) = 4.900 \times 10^{12}
Step 3: Denominator: (1.737×106 m)2=3.017×1012 m2(1.737 \times 10^6\ \text{m})^2 = 3.017 \times 10^{12}\ \text{m}^2
Step 4: gMoon=1.62 m/s2g_{\text{Moon}} = 1.62\ \text{m/s}^2
Step 5: W=mg=(70 kg)(1.62 m/s2)=114 NW = mg = (70\ \text{kg})(1.62\ \text{m/s}^2) = 114\ \text{N}
Answer: W≈114W \approx 114 N (about one sixth of the 687687 N they weigh on Earth)

Frequently Asked Questions

F = Gm₁m₂/r², where G is 6.674 × 10⁻¹¹ N·m²/kg², the masses are in kilograms and r is the centre-to-centre separation in metres. The result is a force in newtons, directed along the line joining the two centres.

Use g = GM/R², with the planet's mass M in kilograms and its radius R in metres. The falling object's own mass cancels out, which is why all objects accelerate identically in vacuum. Earth gives 9.82 m/s² by this calculation, close to the standard 9.81 m/s².

Mass is the amount of matter, measured in kilograms, and is identical everywhere. Weight is the gravitational force on that mass, W = mg, measured in newtons, and changes with location — a 70 kg person weighs about 687 N on Earth but only 114 N on the Moon.

Because G is only 6.674 × 10⁻¹¹ in SI units. Two 1000 kg masses ten metres apart attract with about 6.7 × 10⁻⁷ N, far too small to notice. Gravity only dominates when at least one of the masses is planet-sized.

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