Kepler's third law — T² ∝ a³ — derivation: Worked examples

This is a companion page to the main derivation. Here we do only one thing: we grind through every kind of problem the law $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ can throw at you, so that no exam question and no real-world scenario is unfamiliar.

Before we start, recall the three faces of the same law (each solves for a different unknown):

Here $T$ = period (time for one full lap), $a$ = semi-major axis (orbit size; for a circle just the radius), $G$ = gravitational constant $6.674 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}$ , and $M$ = mass of the central body being orbited. We will also meet $v$ = orbital speed, the distance a body covers along its path per second (in metres per second).

The scenario matrix

Every problem this law can pose falls into one of these cells. The examples below are labelled with the cell they cover. (If the table below does not render, read each Cell — distinguishing feature → Example line as one row.)

Cell A — Absolute period: given real $a$ and $M$ in SI units, find $T$ . → Ex 1
Cell B — Absolute size (inverse): given $T$ and $M$ , find $a$ . → Ex 2 (GEO)
Cell C — Ratio, same $M$ : compare two orbits, the constant cancels. → Ex 3
Cell D — Kepler's units trick: $a$ in AU, $T$ in years, Sun ⇒ $T^{2} = a^{3}$ . → Ex 4
Cell E — Ellipse, is it $a$ ?: given perihelion & aphelion, build $a$ . → Ex 5
Cell F — Solve for central mass: rearrange for $M$ ("weigh the Sun/planet"). → Ex 6
Cell G — Degenerate / limiting: $e \to 0$ (circle), $e \to 1$ (radial fall), $a \to \infty$ . → Ex 7
Cell H — Exam twist: comparable masses ⇒ use $M_{1} + M_{2}$ ; the unit trap. → Ex 8

Cell	What makes it distinct	Example
A. Absolute period	Given real $a$ and $M$ (SI units), find $T$	Ex 1
B. Absolute size (inverse)	Given $T$ and $M$ , find $a$	Ex 2 (GEO)
C. Ratio, same $M$	Compare two orbits, constant cancels	Ex 3
D. Kepler's units trick	$a$ in AU, $T$ in years, Sun → $T^{2} = a^{3}$	Ex 4
E. Ellipse — is it $a$ ?	Given perihelion & aphelion, build $a$	Ex 5
F. Solve for central mass	Rearranged for $M$ ("weigh the Sun/planet")	Ex 6
G. Degenerate / limiting	$e \to 0$ (circle), $e \to 1$ (radial fall), $a \to \infty$	Ex 7
H. Exam twist	Comparable masses → use $M_{1} + M_{2}$ ; unit trap	Ex 8

Cell A — absolute period in SI units

Worked example Ex 1 — The period of a low-Earth-orbit satellite

A satellite circles Earth at radius $r = 7.00 \times 1 0^{6}$ m (about 620 km up). Earth's $G M_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ (its standard gravitational parameter). Find the period $T$ .

Forecast: The International Space Station (ISS), a crewed satellite in low orbit, goes around in about 90 minutes. Guess before computing — will this be minutes or hours?

Step 1 — Pick the right face of the law. We know $a = r$ and $M$ ; we want $T$ . So use $T = 2 π a^{3} / GM$ . Why this step? $T$ is the isolated unknown, so we choose the form with $T$ already alone on the left.

Step 2 — Cube the radius. $r^{3} = (7.00 \times 1 0^{6})^{3} = 3.43 \times 1 0^{20} m^{3}$ . Why this step? The law depends on $a^{3}$ , not $a$ — that cube is the whole personality of Kepler's law.

Step 3 — Divide by $GM$ and take the root. $\frac{r ^{3}}{GM} = \frac{3.43 \times 1 0 ^{20}}{3.986 \times 1 0 ^{14}} = 8.605 \times 1 0^{5} s^{2}, = 927.6 s .$ Why this step? The units inside the root are $m^{3} \div (m^{3} / s^{2}) = s^{2}$ , so the square root is honestly seconds — a good sign we did not fumble the algebra.

Step 4 — Multiply by $2 π$ . $T = 2 π \times 927.6 = 5828 s \approx 97 min .$ Why this step? The root gave us $a^{3} / GM$ , but the formula carries a leading factor $2 π$ . That $2 π$ traces back to the fact that the orbital speed $v$ (metres travelled per second along the path) relates to period by $v = 2 π r / T$ : the body covers the whole circumference $2 π r$ in one period $T$ . Multiplying by $2 π$ turns the raw timescale $a^{3} / GM$ into the time for one complete lap.

Verify: $\approx 97$ minutes matches the ISS-class result we forecast — real low-Earth-orbit periods are 88–100 min. Units came out in seconds. ✓

Cell B — inverse problem, solve for orbit size

Worked example Ex 2 — The geostationary radius (real engineering)

A communications satellite must have period $T = 86400$ s (one day) so it hovers over one spot on Earth's equator. Find its orbital radius. ( $G M_{\oplus} = 3.986 \times 1 0^{14}$ .)

Forecast: Low Earth orbit in Ex 1 was 7000 km for 97 min. A whole day is ~15× longer. Because $T \propto a^{3/2}$ , will $a$ grow by 15× or by much less?

Step 1 — Invert the law. Solve $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ for $a$ : $a = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} .$ Why this step? Now $a$ is the unknown, so we algebraically peel it out — multiply across, then take the cube root because $a$ was cubed.

Step 2 — Build the numerator. $GM T^{2} = 3.986 \times 1 0^{14} \times (86400)^{2} = 3.986 \times 1 0^{14} \times 7.465 \times 1 0^{9} = 2.976 \times 1 0^{24} .$ Why this step? We assemble everything the law puts above the fraction bar first, because grouping the numerator before dividing keeps the powers-of-ten bookkeeping clean and error-free.

Step 3 — Divide by $4 π^{2} = 39.478$ . $\frac{2.976 \times 1 0 ^{24}}{39.478} = 7.538 \times 1 0^{22} m^{3} .$ Why this step? This is $a^{3}$ ; the units are $m^{3} / s^{2} \times s^{2} = m^{3}$ — a volume-like quantity, exactly what a cube of a length should be.

Step 4 — Cube root. $a = (7.538 \times 1 0^{22})^{1/3} = 4.224 \times 1 0^{7} m = 42, 240 km .$ Why this step? Step 3 delivered $a^{3}$ , but we want $a$ itself, so we undo the cube with a cube root — the inverse operation. Only after this does the number become a genuine length we can compare to Earth's radius.

Verify: Subtract Earth's radius $6371$ km → altitude $\approx 35, 870$ km, the textbook GEO belt. And note: period grew ~15× but radius grew only ~6× (from 7000 to 42200 km) — because the $1/3$ power tames it, exactly as forecast. ✓ See Geostationary & Geosynchronous Orbits.

Cell C — ratio form, same central mass

Worked example Ex 3 — Two of Jupiter's moons

Moon A orbits Jupiter with period $1.77$ days at $a_{A} = 4.22 \times 1 0^{5}$ km. Moon B sits at $a_{B} = 1.070 \times 1 0^{6}$ km. Find $T_{B}$ without ever using $G$ or $M_{Jupiter}$ .

Forecast: B is about 2.5× farther out. Period grows as the 3/2 power, so expect roughly $2. 5^{1.5} \approx 4$ × longer.

Step 1 — Use the ratio face. Same central body ⇒ the constant $4 π^{2} / GM$ is identical for both, so it divides out: $\frac{T _{B}^{2}}{T _{A}^{2}} = (\frac{a _{B}}{a _{A}})^{3} .$ Why this step? We are handed no $G$ and no $M$ — but we don't need them, because they're the same for both moons and cancel in the ratio. This is always the fastest route when one central mass hosts two orbits.

Step 2 — Compute the size ratio. $\frac{a _{B}}{a _{A}} = \frac{1.070 \times 1 0 ^{6}}{4.22 \times 1 0 ^{5}} = 2.536.$ Why this step? The law only ever cares about the relative sizes of the two orbits, so we collapse both distances into a single dimensionless number — how many times bigger B's orbit is. Once we have this ratio, the actual kilometres are irrelevant.

Step 3 — Raise to the 3/2 power. $T_{B} = T_{A} \times (2.536)^{3/2} = 1.77 \times 4.038 = 7.15 days .$ Why this step? Taking $(\cdot)^{3}$ folds both effects — longer track and slower speed — into a single exponent $3/2$ .

Verify: These are Jupiter's real moons Io ( $T_{A} = 1.77$ d) and Ganymede ( $T_{B} = 7.15$ d observed). ✓ Prediction lands within rounding.

Cell D — Kepler's natural-units trick

Worked example Ex 4 — Mars's year, done in your head

Mars orbits the Sun at $a = 1.524$ AU. Find its year in Earth-years.

Forecast: Mars is a bit farther than Earth, so its year should be a bit longer than 1 — under 2 years.

Step 1 — Switch to natural units. Measure $a$ in AU and $T$ in years around the Sun. Then $\frac{4 π ^{2}}{G M _{⊙}} = 1$ by construction (because Earth's $a = 1$ , $T = 1$ define these units). The law collapses to $T^{2} = a^{3}$ . Why this step? The ugly constant is exactly the number that makes Earth fit; choosing units where Earth is "1" hides it forever. This only works for Sun-orbiting bodies with these units.

Step 2 — Apply. $T = a^{3/2} = 1.52 4^{1.5} = 1.524 \times 1.524 = 1.524 \times 1.234 = 1.881 years .$ Why this step? With the constant gone, $T^{2} = a^{3}$ rearranges to $T = a^{3/2}$ by taking the square root of both sides. We split $a^{3/2} = a \cdot a^{1/2}$ (one full factor of $a$ times its square root) so the arithmetic can be done by hand — that is the whole point of the trick.

Verify: $1.881$ yr $\times 365.25 = 687$ days — the observed Martian year. ✓

Cell E — ellipse: build $a$ from perihelion and aphelion

Worked example Ex 5 — Halley's Comet

Halley's Comet has perihelion (closest) $r_{m i n} = 0.586$ AU and aphelion (farthest) $r_{m a x} = 35.08$ AU from the Sun. Find its period.

Forecast: It swings out past Neptune; expect decades, not years.

Step 1 — The semi-major axis is the AVERAGE of the two extremes. $a = \frac{r _{m i n} + r _{m a x}}{2} = \frac{0.586 + 35.08}{2} = 17.83 AU .$ Why this step? For an ellipse, "distance from Sun" is not one number — it ranges from $r_{m i n}$ to $r_{m a x}$ . The law uses $a$ , which is exactly the midpoint of that range.

Figure — reading $a$ off the ellipse. The picture below shows the Sun sitting at one focus (not the centre) of the elongated orbit. The green dot marks perihelion ( $r_{m i n}$ ), the red dot aphelion ( $r_{m a x}$ ), and the yellow bar is the semi-major axis $a$ running from the geometric centre to a far vertex. Notice the yellow bar is literally half the full long axis, and that $a$ is the average of the green and red distances — this is the quantity Kepler's law demands, not either extreme alone.

Step 2 — Apply Kepler in AU/years (Cell D trick). $T = a^{3/2} = 17.8 3^{1.5} = 75.3 years .$ Why this step? Eccentricity does not appear — the squashing near the Sun (faster) and stretching far away (slower) trade off exactly over one lap, so only $a$ matters.

Verify: Halley's real period is ~75–76 years (last seen 1986, next 2061). ✓ The eccentricity $e = \frac{r _{m a x} - r _{m i n}}{r _{m a x} + r _{m i n}} = \frac{34.49}{35.67} = 0.967$ never entered our arithmetic — proof the law ignores shape.

Cell F — "weigh the central body"

Worked example Ex 6 — The mass of Jupiter from a moon

Io orbits Jupiter with $a = 4.22 \times 1 0^{8}$ m and $T = 1.77$ days $= 1.529 \times 1 0^{5}$ s. Find Jupiter's mass. ( $G = 6.674 \times 1 0^{- 11}$ .)

Forecast: Jupiter is the giant of the Solar System, hundreds of Earth-masses. Expect ~ $1 0^{27}$ kg.

Step 1 — Solve the law for $M$ . $T^{2} = \frac{4 π ^{2}}{GM} a^{3} \Rightarrow M = \frac{4 π ^{2} a ^{3}}{G T ^{2}} .$ Why this step? $M$ is now the unknown; every observable (the orbit's $a$ and $T$ ) sits on the right. This is literally how we weigh planets and stars we can never touch — Kepler's law is a cosmic scale.

Step 2 — Numerator. $4 π^{2} a^{3} = 39.478 \times (4.22 \times 1 0^{8})^{3} = 39.478 \times 7.515 \times 1 0^{25} = 2.967 \times 1 0^{27} .$ Why this step? We gather everything above the fraction bar first — the $4 π^{2}$ and the cubed orbit size — so that the division in Step 4 is a single clean operation rather than three tangled ones. The cube is where the orbit's size enters, exactly as in every Kepler problem.

Step 3 — Denominator. $G T^{2} = 6.674 \times 1 0^{- 11} \times (1.529 \times 1 0^{5})^{2} = 6.674 \times 1 0^{- 11} \times 2.338 \times 1 0^{10} = 1.560.$ Why this step? Now we assemble everything below the bar — the constant $G$ and the squared period. Squaring $T$ is what the law demands ( $T^{2}$ ), and doing it separately keeps the tiny $G$ and the huge $T^{2}$ from getting muddled in one line.

Step 4 — Divide. $M = \frac{2.967 \times 1 0 ^{27}}{1.560} = 1.90 \times 1 0^{27} kg .$ Why this step? With numerator and denominator each fully built, the last act is the division the formula prescribes — and only now do the units resolve to kilograms, telling us we truly have a mass.

Verify: Jupiter's accepted mass is $1.898 \times 1 0^{27}$ kg. ✓ Matches to three figures — and note $m_{Io}$ never appeared, because it cancelled in the force balance (that's cell H's warning ahead of time).

Cell G — degenerate and limiting cases

Worked example Ex 7 — What the law does at its edges

Examine three limits: (a) the circle limit $e \to 0$ ; (b) the radial-fall limit $e \to 1$ at fixed $a$ ; (c) the escape limit $a \to \infty$ .

Forecast: Guess which limit gives an infinite period.

(a) $e \to 0$ (perfect circle). Then $r_{m i n} = r_{m a x} = a$ , so $a = \frac{r + r}{2} = r$ . The general law $T = 2 π a^{3} / GM$ becomes the circular one $T = 2 π r^{3} / GM$ with no change. Why this step? It confirms the ellipse formula contains the circle as a special case — a sanity anchor, not new physics.

(b) $e \to 1$ at fixed $a$ (degenerate "orbit" = a straight plunge and return). The ellipse squashes to a line segment; the body falls straight to the centre and back. Period is still $T = 2 π a^{3} / GM$ — unchanged, because $e$ never appears. Why this step? This is the astonishing part: a near-radial comet on a razor-thin ellipse of semi-major axis $a$ takes the same time per lap as a circle of radius $a$ . Shape is irrelevant; only $a$ counts. (Note this is still a bound $e < 1$ ellipse; the moment $e$ reaches exactly $1$ the orbit stops closing — see the definition above.)

(c) $a \to \infty$ . $T = 2 π a^{3} / GM \to \infty$ . As $a$ grows without bound the ellipse's far end runs off to infinity and the orbit stops closing — it becomes the $e = 1$ parabola (escape) and beyond. An infinite period is the law's honest way of saying "no return."

Figure — three orbits, one $a$ . The picture shows a green circle, a blue moderate ellipse, and a red near-radial sliver, all sharing the same semi-major axis $a$ and the same central mass (yellow dot at a focus). Despite looking wildly different, all three complete one lap in the identical time — the visual proof that period depends on $a$ alone, never on eccentricity.

Numeric anchor for (b) vs (a): Take $GM = 1$ , $a = 1$ . Circle: $T = 2 π 1 = 6.283$ . Squashed ellipse, same $a$ : $T = 2 π 1 = 6.283$ . Identical.

Verify: Both give $T = 6.283$ in these units — equal, confirming $e$ -independence. ✓ And $a \to \infty$ giving $T \to \infty$ matches the physical boundary between bound and escape orbits — see Orbital Energy & Vis-viva Equation.

Cell H — the exam twist (comparable masses)

Worked example Ex 8 — Binary stars: when

$m$ does NOT cancel cleanly Two stars orbit their common centre with a separation $a = 3.00 \times 1 0^{11}$ m and period $T = 2.00 \times 1 0^{7}$ s. Find the total mass $M_{1} + M_{2}$ . ( $G = 6.674 \times 1 0^{- 11}$ .)

Forecast: For planets we drop the small mass. Here both bodies are heavy — so which mass goes in the law?

Step 1 — Use the two-body form of the law. When both masses matter, the simple $M$ is replaced by the sum: $T^{2} = \frac{4 π ^{2}}{G ( M _{1} + M _{2} )} a^{3} \Rightarrow M_{1} + M_{2} = \frac{4 π ^{2} a ^{3}}{G T ^{2}} .$ Why this step? In the planet case we assumed the Sun sits still. When masses are comparable, both orbit a shared centre; the honest derivation (see Reduced Mass & Two-Body Problem) replaces $M$ with $M_{1} + M_{2}$ , with $a$ the separation between them. Forgetting this is the classic exam trap.

Step 2 — Numerator. $4 π^{2} a^{3} = 39.478 \times (3.00 \times 1 0^{11})^{3} = 39.478 \times 2.70 \times 1 0^{34} = 1.066 \times 1 0^{36} .$ Why this step? As in Ex 6 we build everything above the bar first — the $4 π^{2}$ times the cubed separation. Cubing the huge $3 \times 1 0^{11}$ m separation is where the orbit's size dominates the answer, so it is done carefully and on its own.

Step 3 — Denominator. $G T^{2} = 6.674 \times 1 0^{- 11} \times (2.00 \times 1 0^{7})^{2} = 6.674 \times 1 0^{- 11} \times 4.00 \times 1 0^{14} = 2.670 \times 1 0^{4} .$ Why this step? We now build everything below the bar — the constant $G$ and the squared period. Keeping the minuscule $G$ and the large $T^{2}$ in a separate line stops the exponents from colliding when we finally divide.

Step 4 — Divide. $M_{1} + M_{2} = \frac{1.066 \times 1 0 ^{36}}{2.670 \times 1 0 ^{4}} = 3.99 \times 1 0^{31} kg .$ Why this step? The formula is one big fraction; with top and bottom each assembled, the closing division yields the combined mass — and the units land in kilograms, confirming we solved for a mass and not something else.

Verify: That's about $20 M_{⊙}$ ( $M_{⊙} = 1.99 \times 1 0^{30}$ kg) split between two stars — a plausible massive binary. Units: $m^{3} / (N\cdotm^{2} / kg^{2} \cdot s^{2}) = kg$ . ✓ The trap avoided: if you had used a single planet's $M$ -formula you'd have measured only the combined system anyway — the law can never separate $M_{1}$ from $M_{2}$ alone; that needs each star's individual orbit radius.

Recall

Recall Which cell is each of these? (cover the answers)

Given $a$ (in AU) and orbiting the Sun, want $T$ in years ::: Cell D — use $T = a^{3/2}$ Given perihelion and aphelion, want the period ::: Cell E — first form $a = (r_{m i n} + r_{m a x}) /2$ Given a moon's $a$ and $T$ , want the planet's mass ::: Cell F — $M = 4 π^{2} a^{3} / (G T^{2})$ Comparing two moons of Saturn, no $G$ given ::: Cell C — ratio $(T_{2} / T_{1})^{2} = (a_{2} / a_{1})^{3}$ Two stars of similar mass ::: Cell H — use $M_{1} + M_{2}$ , $a$ = separation Does making the orbit more elliptical (fixed $a$ ) change $T$ ? ::: No — Cell G, $e$ drops out

Connections

3.2.07 Kepler's third law — T² ∝ a³ — derivation (Hinglish) — the parent derivation
Newton's Law of Universal Gravitation — supplies the $GM m / r^{2}$ that starts every derivation
Centripetal Force & Circular Motion — the $m v^{2} / r$ we set gravity equal to
Kepler's Second Law (equal areas) — why eccentricity cancels in Cells E and G
Orbital Energy & Vis-viva Equation — the $a \to \infty$ escape boundary in Cell G
Geostationary & Geosynchronous Orbits — the destination of Cell B
Reduced Mass & Two-Body Problem — justifies the $M_{1} + M_{2}$ of Cell H