Kepler's third law — T² ∝ a³ — derivation: Exercises

Constants you may need (keep them handy):

The full physics lives in the parent note: Kepler's Third Law derivation. Prerequisites: Newton's Law of Universal Gravitation, Centripetal Force & Circular Motion.

State whether each is true or false, and why: (a) $T^{2} \propto a^{3}$ means doubling $a$ doubles $T$ . (b) A heavier planet at the same distance orbits faster. (c) The constant $\frac{4 π ^{2}}{GM}$ depends on the central mass only.

Recall Solution

(a) False. Doubling $a$ multiplies $T$ by $2^{3/2} = 2.828$ , not $2$ . The law is on the squares and cubes, so the raw factor on $T$ is the $3/2$ power. (b) False. The orbiting mass $m$ cancels between gravity $\frac{GM m}{r ^{2}}$ and centripetal need $\frac{m v ^{2}}{r}$ . Speed $v = GM / r$ carries no $m$ . (c) True. Because $m$ cancelled, only $G$ and the central mass $M$ remain in the constant.

L1.2

In Sun-units ( $a$ in AU, $T$ in years), Jupiter has $a = 5.20$ AU. Write down (don't compute yet) the expression for its period, then evaluate.

Recall Solution

WHAT: use $T^{2} = a^{3}$ , so $T = a^{3/2}$ . Compute: $T = 5.2 0^{1.5} = 5.20 5.20 = 5.20 \times 2.280 = 11.86$ years. Real Jupiter year $\approx 11.86$ yr. ✓ WHY the trick works: in these hand-picked units $\frac{4 π ^{2}}{G M _{⊙}}$ equals exactly $1$ , so the ugly constant vanishes.

Level 2 — Application

L2.1

Mars has $a = 1.524$ AU. Find its period in years, then in days ( $1 yr = 365.25$ days).

Recall Solution

WHAT: $T = a^{3/2} = 1.52 4^{1.5}$ . $1.524 = 1.2345$ , so $T = 1.524 \times 1.2345 = 1.881$ years. In days: $1.881 \times 365.25 = 687.1$ days. ✓ (Observed Martian year $\approx 687$ days.) WHY: Mars is farther out, so both penalties apply — longer track and slower speed — pushing its year well past 1.5 yr even though it's only 1.524 AU out.

L2.2

A satellite orbits Earth in a circle of radius $r = 7.0 \times 1 0^{6}$ m. Find its orbital speed $v$ and period $T$ .

Recall Solution

Speed ( $v = GM / r$ , because gravity supplies exactly the centripetal pull): $v = \frac{3.986 \times 1 0 ^{14}}{7.0 \times 1 0 ^{6}} = 5.694 \times 1 0^{7} = 7546 m/s .$ Period ( $T = 2 π r / v$ , one circumference at that speed): $T = \frac{2 π ( 7.0 \times 1 0 ^{6} )}{7546} = \frac{4.398 \times 1 0 ^{7}}{7546} = 5828 s \approx 97.1 min .$ WHY: a low-Earth satellite laps the planet in about 90–100 minutes — this is the ISS regime.

Level 3 — Analysis

L3.1

Moon B has $4 \times$ the semi-major axis of moon A around the same planet. Without any constants, find $T_{B} / T_{A}$ .

Recall Solution

WHAT — use the ratio form. Since both orbit the same $M$ , the constant cancels: $\frac{T _{B}^{2}}{T _{A}^{2}} = (\frac{a _{B}}{a _{A}})^{3} = 4^{3} = 64 \Rightarrow \frac{T _{B}}{T _{A}} = 64 = 8.$ WHY ratios are fastest: you never need $G$ , $M$ , or units — same central body means $\frac{4 π ^{2}}{GM}$ divides out cleanly.

L3.2

Find the geostationary radius: the circular orbit whose period equals one sidereal day, $T = 86, 164$ s, around Earth.

Recall Solution

WHAT — invert the law for $r$ . From $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ : $r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} .$ Plug in: $GM T^{2} = 3.986 \times 1 0^{14} \times (86164)^{2} = 3.986 \times 1 0^{14} \times 7.424 \times 1 0^{9} = 2.959 \times 1 0^{24} .$ Divide by $4 π^{2} = 39.48$ : $7.495 \times 1 0^{22}$ . Cube root: $r = (7.495 \times 1 0^{22})^{1/3} = 4.216 \times 1 0^{7} m = 42, 160 km$ . Altitude $= r - R_{\oplus} = 42, 160 - 6, 371 = 35, 789 km$ . ✓ See Geostationary & Geosynchronous Orbits for why this exact height matters for TV dishes.

Level 4 — Synthesis

L4.1

Two planets orbit the same star. Planet X has period $8$ years; planet Y has period $27$ years. What is the ratio of their orbital speeds $v_{X} / v_{Y}$ (assume circular orbits)?

Recall Solution

Step 1 — get the size ratio from the period ratio. $T^{2} \propto a^{3} \Rightarrow a \propto T^{2/3}$ : $\frac{a _{X}}{a _{Y}} = (\frac{T _{X}}{T _{Y}})^{2/3} = (\frac{8}{27})^{2/3} = (\frac{2}{3})^{2} = \frac{4}{9} .$ Step 2 — turn size into speed. $v = GM / r \Rightarrow v \propto a^{- 1/2}$ : $\frac{v _{X}}{v _{Y}} = (\frac{a _{X}}{a _{Y}})^{- 1/2} = (\frac{4}{9})^{- 1/2} = \frac{3}{2} = 1.5.$ WHY: X is the inner planet ( $a_{X} < a_{Y}$ ), and inner planets move faster — its speed is $1.5 \times$ Y's. Two proportionalities chained, no constants ever needed. (This speed scaling also underlies Orbital Energy & Vis-viva Equation.)

L4.2

An asteroid has perihelion (closest) distance $r_{m i n} = 0.5$ AU and aphelion (farthest) $r_{m a x} = 3.5$ AU. Find its orbital period.

Recall Solution

Step 1 — the shape doesn't matter, only $a$ . The law needs the semi-major axis, the average of the two extreme distances: $a = \frac{r _{m i n} + r _{m a x}}{2} = \frac{0.5 + 3.5}{2} = 2.0 AU .$ Step 2 — apply Sun-units. $T = a^{3/2} = 2. 0^{1.5} = 22 = 2.828 years .$ WHY only $a$ ? Squashing the orbit speeds the asteroid near perihelion and slows it near aphelion; by equal areas these exactly trade off over one lap. Eccentricity never enters the period.

Level 5 — Mastery

L5.1

Derive the mass of the Sun from Earth's orbit alone: $a = 1.496 \times 1 0^{11}$ m, $T = 3.156 \times 1 0^{7}$ s. Compare to the tabulated $M_{⊙} = 1.989 \times 1 0^{30}$ kg.

Recall Solution

WHAT — rearrange the law to isolate $M$ . From $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ : $M = \frac{4 π ^{2} a ^{3}}{G T ^{2}} .$ WHY this is powerful: we weigh the Sun using only a planet's distance and year — no scale needed. This is exactly how astronomers mass distant stars. Compute numerator: $a^{3} = (1.496 \times 1 0^{11})^{3} = 3.348 \times 1 0^{33}$ . Times $4 π^{2} = 39.48$ : $1.322 \times 1 0^{35}$ . Denominator: $G T^{2} = 6.674 \times 1 0^{- 11} \times (3.156 \times 1 0^{7})^{2} = 6.674 \times 1 0^{- 11} \times 9.960 \times 1 0^{14} = 6.647 \times 1 0^{4}$ . Divide: $M = \frac{1.322 \times 1 0 ^{35}}{6.647 \times 1 0 ^{4}} = 1.989 \times 1 0^{30} kg$ . ✓ Matches the tabulated Sun mass to three figures.

L5.2

For comparable masses the law becomes $T^{2} = \frac{4 π ^{2}}{G ( M _{1} + M _{2} )} a^{3}$ , where $a$ is the semi-major axis of the relative orbit. A binary star system has two equal stars ( $M_{1} = M_{2} = M$ ) orbiting with relative-orbit semi-major axis $a = 2.0 AU$ and period $T = 1.0$ year. Find $M$ in solar masses.

Recall Solution

Step 1 — write the two-body law in Sun-units. With $a$ in AU, $T$ in years, and total mass in $M_{⊙}$ : $T^{2} = \frac{a ^{3}}{M _{1} + M _{2}} (Sun-units) \Rightarrow M_{1} + M_{2} = \frac{a ^{3}}{T ^{2}} .$ Step 2 — plug in. $M_{1} + M_{2} = \frac{2. 0 ^{3}}{1. 0 ^{2}} = \frac{8.0}{1} = 8.0 M_{⊙} .$ Step 3 — split equally. Since $M_{1} = M_{2} = M$ : $2 M = 8.0 \Rightarrow M = 4.0 M_{⊙}$ . WHY the $M_{1} + M_{2}$ ? When masses are comparable, both stars orbit their shared centre of mass; the effective central mass is the total. See Reduced Mass & Two-Body Problem for the full reduction. For a planet round the Sun, $m ≪ M$ so $M_{1} + M_{2} \approx M_{⊙}$ and we recover the simple law.

Self-check summary

Recall Which technique for which problem?

Given $a$ , want $T$ (or reverse) ::: plug into $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ , one variable at a time Two orbits, same central mass ::: use the ratio form $(T_{1} / T_{2})^{2} = (a_{1} / a_{2})^{3}$ Given $T$ , want $r$ ::: invert: $r = (GM T^{2} /4 π^{2})^{1/3}$ Want central mass $M$ ::: rearrange: $M = 4 π^{2} a^{3} / (G T^{2})$ Elliptical orbit ::: use $a = (r_{m i n} + r_{m a x}) /2$ ; eccentricity drops out Comparable masses (binary) ::: replace $M$ with $M_{1} + M_{2}$

Exercises — Kepler's third law — T² ∝ a³ — derivation

Level 1 — Recognition

L1.1

L1.2

Level 2 — Application

L2.1

L2.2

Level 3 — Analysis

L3.1

L3.2

Level 4 — Synthesis

L4.1

L4.2

Level 5 — Mastery

L5.1

L5.2

Self-check summary