Inverse Square Law Explorer

Force, irradiance, and flux all decay as 1/r² — because they spread over a sphere whose surface area grows as 4πr².

Worked example — Inverse Square Law Explorer

G	6.674×10⁻¹¹ Nm²kg⁻²
m₁	5.97×10²⁴ kg
m₂	1 kg
Distance r	6.37×10⁶ m
r²	4.06×10¹³ m²
Sphere surface 4πr²	5.10×10¹⁴ m²

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Voyager 1 runs on 22 W — and we can still hear it 🖖

Voyager 1 is now ~163 AU from the Sun (≈ 2.4 × 10¹³ m). Plug that into the irradiance formula: E = 3.828 × 10²⁶ / (4π × (2.4 × 10¹³)²) ≈ 0.051 W/m² — about 26,600 times fainter than sunlight on Earth, which is just 163². Solar panels would be useless. Voyager runs on radioactive decay (a radioisotope thermoelectric generator) producing roughly 250 W of usable electricity. Its radio transmitter broadcasts at 22 W — about what a dim LED strip uses. By the time the signal reaches Earth's Deep Space Network dishes (70 m across), the 1/r² law has spread those 22 W over a sphere the size of the inner solar system. The received power is around 10⁻¹⁸ W. That is why the receiving antennas need to be enormous and why NASA must chill the receivers to near absolute zero to hear the signal at all. Same law, same geometry — whether it is sunlight warming your face or the whisper of humanity's most distant artefact.

Why twice as far means a quarter as bright 🖖

The rule is pure geometry. Energy leaving a point source spreads evenly over the surface of an expanding sphere, and a sphere's area grows as 4πr². Double the distance and that same energy covers four times the area, so intensity drops to one quarter — not one half. That is why you cool off so fast when you step back from a campfire, and why this tool traces gravity, light, and electric force on the very same curve.

The 1/r² law is a fingerprint of 3D space 🖖

The exponent isn't arbitrary — it equals the number of spatial dimensions minus one. We get 1/r² because a sphere's area scales as r² in three dimensions. In a hypothetical 4D universe gravity would fall as 1/r³, and Paul Ehrenfest showed in 1917 that planetary orbits would then be unstable — planets would spiral into the star or fly away. Stable solar systems, and therefore life, may depend on space having exactly three dimensions.

Problem solved in full

The one orbit in which a satellite hangs motionless 5 steps
The Earth pulls on 1 kg at its surface with 9.82 N. Use only that to find the one orbit in which a satellite hangs motionless above a point on the equator. This is Gravity with m₁ = 5.972 × 10²⁴ kg, m₂ = 1 kg and r = 6371 km.
1. Newton's law with the numbers in it. What matters for the rest is not the 9.82 N but the product GM = 3.986 × 10¹⁴ in SI units, because the satellite's own mass is about to cancel and never come back.
  
  $r^{2} = (6.371 \times 1 0^{6})^{2} = 4.06 \times 1 0^{13} m^{2}; F = \frac{G m _{1} m _{2}}{r ^{2}} = \frac{3.986 \times 1 0 ^{14} \times 1}{4.06 \times 1 0 ^{13}} = 9.82 N$
2. A circular orbit is that same gravitational acceleration, spent as the centripetal one. Setting the two equal removes the orbiting mass and leaves a single relation between one radius and one angular rate.
  
  $\frac{GM}{r ^{2}} = ω^{2} r ⟹ r^{3} = \frac{GM}{ω ^{2}}$
3. The angular rate is imposed, not chosen: to hang over a fixed point the satellite must turn exactly with the Earth. The Earth turns once in 86,164 s rather than 86,400 s, because a solar day also has to make up the ground the Earth covered along its orbit. Using 86,400 puts the orbit 77 km too high, and a satellite at the wrong radius does not stay over its point.
  
  $ω = \frac{2 π}{86 164 s} = 7.292 \times 1 0^{- 5} rad/s, ω^{2} = 5.318 \times 1 0^{- 9} s^{- 2}$
4. Take the cube root. The 42,163 km is measured from the centre of the Earth, so subtracting the 6371 km already on the panel leaves about 35,790 km of altitude — the usual quoted 35,786 km is the same orbit measured from the equatorial radius, which is 7 km larger.
  
  $r^{3} = \frac{3.986 \times 1 0 ^{14}}{5.318 \times 1 0 ^{- 9}} = 7.495 \times 1 0^{22} m^{3} ⟹ r = 4.2163 \times 1 0^{7} m = 42163 km$
5. Gravity out there obeys the very law this page draws. The radius has grown by a factor of 6.62, so the square of that is how much weaker the pull has become.
  
  $(\frac{42 163}{6371})^{2} = 43.8 ⟹ g (r) = \frac{9.82}{43.8} = 0.2242 m/ s^{2}$
Answer

The tool prints 9.82 N on 1 kg, from a distance squared of 4.06 × 10¹³ m². What follows from it is not an altitude anyone selected: 42,163 km is the only radius whose orbital period matches the Earth's rotation, so geostationary orbit is a solution rather than a design choice. That makes the geostationary belt one circle 264,900 km around, and every satellite that must stay put — television, weather, most relays — has to sit somewhere on it. One degree of that circle is 736 km, which is why slots are allocated by international treaty instead of being claimed. And the pull out there is 43.8 times weaker than at your feet, which is this page's 1/r² curve followed out to six and a half Earth radii.

$r_{GEO} = 3 \frac{GM}{ω ^{2}} = 42163 km, 2 π r_{GEO} = 264900 km$

References (1)

Voyager transmitter power, antenna sizes and received signal levels: R. Ludwig and J. Taylor, "Voyager Telecommunications." DESCANSO Design and Performance Summary Series, Article 4. Jet Propulsion Laboratory, 2002.

By the 42Math team · Updated 2026-08-17

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

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Example problems

Earth Surface - Earth surface gravity (m2=1 kg): F ≈ 9.8 N — that is 1 kg-force
Moon Orbit - Moon orbital distance: 1 kg feels F ≈ 0.0027 N — ≈3600× weaker than at surface
Sunlight at Earth - Sunlight at Earth (1 AU): irradiance ≈ 1361 W/m² — the solar constant
Double Distance - 2× Earth distance: irradiance drops to ≈ 340 W/m² — exactly ¼ of solar constant
Protons - Proton-proton at 1 fm: Coulomb force ≈ 230 N — must be overcome by the strong force