Physics I: Mechanics › Oscillations & gravitation › Kepler's third law

T² = (4π²/GM) r³Say it: “T squared equals four pi squared over G M times r cubed”

Kepler's third law

Farther planets don't just take longer — they take disproportionately longer, and the cube tells you exactly how much.

Notation on this page: T is the orbital period, r the orbital radius (semi-major axis a for ellipses), M the central mass being orbited.

Before this lesson: Newton's gravitation

Where it comes from

Johannes Kepler, poring over Tycho Brahe's Mars observations, noticed in 1619 that planets obey a strict pattern: T² ∝ r³. Double a planet's distance and its year doesn't double — it grows by 2√2 ≈ 2.83×. Newton later showed this pattern is gravity itself: set the gravitational pull equal to the required centripetal force for a circular orbit:

GMm/r² = m(2πr/T)²/rgravity provides the centripetal force — the orbiter's mass m is about to cancel
Before reading on: a planet orbits twice as far out as Earth. Is its year twice as long, or longer than that? Guess the factor before reading the verdict.

Longer: (T1/T2)² = (r1/r2)³ gives T ∝ r3/2, so 23/2 = 2√2 ≈ 2.83×. Distance punishes harder than linear — the cube sees to that.

Derivation

Newton's derivation: gravity = centripetal force, then solve for T. Watch the orbiter's mass cancel in Step 3 — Kepler's pattern doesn't care what's orbiting.

GMm/r²
=
mv²/r
Step 1 — gravity = centripetal. The gravitational pull provides exactly the inward force a circular orbit demands.
=
m(2πr/T)²/r
Step 2 — orbital speed. v = circumference/period = 2πr/T. Substitute it in.
GM/r²
=
4π²r/T²
Step 3 — cancel m, tidy. The orbiter's mass divides out of both sides; (2πr/T)²/r = 4π²r/T².
T²
=
(4π²/GM) r³
Step 4 — solve for T². Cross-multiply: T² = 4π²r³/(GM). Kepler's empirical pattern, derived from F = GMm/r². ∎

The ratio form (no G, no M): (T1/T2)² = (r1/r2)³. When one orbit is known (Earth: 1 yr at 1 AU), every other orbit around the same star is a one-line ratio.

How to use it

The procedure, every time:

  1. Identify M: the central mass. Planets → the Sun's mass. Moons/satellites → the planet's mass. The orbiter's mass never appears.
  2. r is the semi-major axis. For near-circular orbits, the radius. For ellipses, half the longest diameter — not the closest approach.
  3. Prefer ratios. (T1/T2)² = (r1/r2)³ whenever a reference orbit exists — G and M cancel.
  4. Weighing stars: M = 4π²r³/(GT²) — clock an orbit, weigh the central body. This is how exoplanet host stars get weighed.
  5. Units discipline. In the full form, r in metres and T in seconds — convert years and AU before substituting.

The ratio form

(T1/T2)² = (r1/r2)³the exam-speed version — one known orbit unlocks all the othersSay it: “the period ratio squared equals the distance ratio cubed”
Common mistake: putting the orbiter's mass in for M (e.g. Earth's mass for Earth's year). M is the body being orbited — the Sun, for planets. The orbiter's mass cancelled in Step 3.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the audit: Earth's year from the formula

  1. Inventory. r = 1.496×1011 m, M = 1.989×1030 kg (the Sun).
  2. Substitute. T² = 4π²r³/(GM) = 39.48 × (1.496×1011)³/(6.674×10−11 × 1.989×1030).
  3. Evaluate. = 1.322×1035/1.327×1020 = 9.956×1014; T = √(9.956×1014) ≈ 3.155×107 s ≈ 365.2 days.
  4. Sanity check. The formula reproduces the year ✓ — Kepler's pattern, Newton's gravity, and the calendar agree.
Common mistake: using Earth's mass for M → T comes out ~18 minutes (low-Earth-orbit period!). M is the central mass — the Sun, 333,000× heavier.
Your turn — the Moon around Earth: r = 3.844×108 m, M = 5.972×1024 kg. T?

Answer: ≈ 27.5 days. T² = 39.48 × (3.844×108)³/(6.674×10−11 × 5.972×1024) = 2.242×1027/3.986×1014 = 5.626×1012; T ≈ 2.372×106 s ≈ 27.5 days — near the true sidereal month of 27.3 days (the Sun's tug perturbs the simple two-body result slightly).

Example 2 — the ratio: Mars's year (r = 2.279×1011 m)

  1. Use ratios. (TMars/TEarth)² = (rMars/rEarth)³ = (2.279/1.496)³.
  2. Evaluate. 2.279/1.496 ≈ 1.5234; cubed ≈ 3.536; square root ≈ 1.880.
  3. Scale. TMars = 365.2 × 1.880 ≈ 686.7 days ≈ 687 days ✓ — the known Martian year, no G required.
Common mistake: TMars = 365.2 × 1.5234 ≈ 556 days — applying the distance ratio linearly. The ratio must be cubed-then-square-rooted: T ∝ r3/2, not r.
Your turn — Venus: r = 1.082×1011 m. Its year?

Answer: ≈ 225 days. Ratio 1.082/1.496 ≈ 0.72326; cubed ≈ 0.37835; square root ≈ 0.61510; T = 365.2 × 0.61510 ≈ 224.6 days — the known 224.7-day Venusian year.

Example 3 — parking a satellite: geostationary orbit

  1. Target: T = 24 h = 86,400 s (one spin of Earth). Unknown: r.
  2. Rearrange. r³ = GMT²/(4π²) = 3.986×1014 × (86,400)²/39.48.
  3. Evaluate. = 2.975×1024/39.48 = 7.537×1022; r = ∛(7.537×1022) ≈ 4.224×107 m.
  4. Altitude. 42,241 km − 6,371 km = 35,870 km above the equator ≈ the real Clarke belt (35,786 km with the sidereal day) ✓
Common mistake: reporting r = 42,241 km as the altitude. r is measured from Earth's centre — subtract R = 6,371 km to get height above ground.
Your turn — the ISS: r = 6.771×106 m. Its period?

Answer: ≈ 92.4 min. T² = 39.48 × (6.771×106)³/3.986×1014 = 1.225×1022/3.986×1014 = 3.074×107; T ≈ 5545 s ≈ 92.4 min — the familiar ~92-minute low orbit.

Before reading on: you know Earth's orbit (r, T) and G. Can you weigh the Sun without ever visiting it? Which rearrangement gives M?

Example 4 — judgment call: weighing the Sun

  1. Rearrange for M. T² = 4π²r³/(GM) → M = 4π²r³/(GT²).
  2. Substitute Earth's orbit. M = 39.48 × (1.496×1011)³/(6.674×10−11 × (3.155×107)²).
  3. Evaluate. = 1.322×1035/6.645×104 ≈ 1.99×1030 kg — the Sun's mass, weighed from 150 million km away ✓
Common mistake: “you can't weigh something without a scale.” Orbits are scales: the central mass is encoded in every period. This trick weighs exoplanet host stars and black holes today.
Your turn — weigh Jupiter from its moon Io: r = 4.217×108 m, T = 1.769 days = 152,842 s. M?

Answer: ≈ 1.90×1027 kg. T = 1.769 × 86,400 = 152,842 s. M = 39.48 × (4.217×108)³/(6.674×10−11 × 152842²) = 2.961×1027/1.559 ≈ 1.90×1027 kg — Jupiter's known mass, weighed from Io's orbit.

Memorization tips

  • Chant it: “T squared equals four pi squared over G M, r cubed.” Periods squared, distances cubed.
  • The ratio form is the workhorse: (T1/T2)² = (r1/r2)³. Earth (1 yr, 1 AU) is the universal reference — no G needed.
  • T ∝ r3/2: twice the distance → 2.83× the period. “Farther punishes harder than linear.”
  • M is the centre. The orbiter's mass cancelled — the most-tested trap. Planets → Sun's M; satellites → Earth's M.
  • Weigh anything: M = 4π²r³/(GT²). Clock an orbit, weigh the star — the formula runs backwards.
  • Derive in 30 seconds: GMm/r² = m(2πr/T)²/r, cancel m, solve for T². Gravity = centripetal is the whole proof.

Final challenge

Five mixed questions — ratios, weighing, and the traps, all in one. Score 5/5 and Kepler's third law is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Does Kepler's third law work for elliptical orbits?

Yes — use the semi-major axis a (half the longest diameter) in place of r: T² = (4π²/GM)a³. Our circular derivation is the special case a = r. Kepler himself discovered it for ellipses.

Which mass goes in the formula — the planet's or the Sun's?

The central mass M — the body being orbited. For planets it's the Sun's mass; for the Moon or satellites it's Earth's. The orbiter's mass cancels out entirely.

Why is it T squared and r cubed?

The exponents come from combining circular motion (which brings T²) with the inverse-square law (which brings r³ after clearing denominators). Kepler found the pattern empirically in 1619; Newton derived the exponents from F = GMm/r².

Who was Kepler and how did he find the law?

Johannes Kepler (1571–1630) inherited Tycho Brahe's exquisite naked-eye observations of Mars and spent years fitting them. The third law — T² proportional to a³ — was published in 1619 in Harmonices Mundi, decades before Newton explained it.

Can I use Kepler's third law for moons of Jupiter?

Yes — with Jupiter's mass as M. The law is universal: any central mass works. That's how astronomers weigh exoplanet host stars and black holes today.

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