Kepler's third law — T² ∝ a³ — derivation: Question bank

Every term used below is built in the parent note: $T$ = orbital period (time for one lap), $a$ = semi-major axis (the "average size" of the orbit; the radius $r$ for a circle), $M$ = the mass of the body being orbited, $m$ = the mass of the small body doing the orbiting, $v$ = the orbital speed (how fast the small body moves along its path, in metres per second), $G$ = gravitational constant, $e$ = eccentricity (how squashed the ellipse is; $e = 0$ is a perfect circle).

Definition Symbols and abbreviations used on this page

$r$ = the instantaneous distance from the small body to the centre of the body it orbits, at one moment (metres). For a circle $r$ never changes and equals $a$ ; for an ellipse $r$ shrinks and grows over each lap, so it is not a single fixed number.
AU = astronomical unit, the average Earth–Sun distance ( $\approx 1.496 \times 1 0^{11}$ m). Convenient for measuring planetary orbit sizes.
$M_{⊙}$ = one solar mass, the mass of the Sun ( $\approx 1.989 \times 1 0^{30}$ kg). Used as the central mass $M$ when the Sun is being orbited.
$M_{\oplus}$ = one Earth mass, the mass of the Earth ( $\approx 5.972 \times 1 0^{24}$ kg). Used as the central mass $M$ when the Earth is being orbited.
GEO = geostationary Earth orbit, the special orbit whose period equals Earth's rotation (one day), so a satellite there appears fixed over one spot on the equator.
Step 1 (from the parent derivation) = the move that sets gravity equal to the required centripetal force, $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$ — the single equation the whole law is built on.

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

True or false — justify

A heavier planet at the same distance from the Sun has a shorter year.

False. The orbiting mass

m

cancels in

GM m / r^{2} = m v^{2} / r

, so period depends only on

a

and the central mass

M

— a feather and a boulder at the same distance both take one year.

If two orbits share the same semi-major axis

a

but different eccentricities, they have different periods.

False. Period depends only on

a

. Squashing the orbit speeds the body up near periapsis and slows it near apoapsis, and these exactly cancel over one lap.

Doubling the semi-major axis doubles the period.

False.

T \propto a^{3/2}

, so doubling

a

multiplies

T

2^{1.5} \approx 2.83

— more than double, because the body both travels farther and moves slower.

T^{2} = a^{3}

(with no constant) is a universal law that works in any units.

False. That clean form only holds in the special units AU + years + solar masses, where

4 π^{2} / (G M_{⊙})

happens to equal 1. In SI units the constant is very much present.

Kepler's Third Law fails for artificial satellites because they are man-made.

False. The law knows nothing about origin — only about the gravity of the central body. Every satellite around Earth obeys

T^{2} = (4 π^{2} / G M_{\oplus}) a^{3}

, which is exactly how GEO altitude is designed.

For a planet orbiting the Sun, the correct mass to use is the planet's mass.

False.

M

is the mass being orbited (the Sun). The planet's mass

m

has already cancelled out of the derivation.

The proportionality constant

4 π^{2} / GM

is the same for a comet and a planet around the same star.

True. As long as the central mass

M

is identical, the constant is identical — that is why ratio comparisons (

T_{B}^{2} / T_{A}^{2} = (a_{B} / a_{A})^{3}

) work so cleanly.

A body in a circular orbit is not accelerating because its speed is constant.

False. Speed is constant but velocity direction changes constantly. That turning is centripetal acceleration

v^{2} / r

pointed inward — supplied entirely by gravity.

Spot the error

"Since gravity is stronger closer in, planets close to the Sun must move slower."

Error: the conclusion is backwards. Stronger gravity means a larger required centripetal force, which demands a higher speed:

v = GM / r

grows as

r

shrinks. Close planets move faster.

"For an ellipse we just plug in the current distance

r

from the Sun."

Error:

r

(the instantaneous distance) changes throughout an elliptical orbit, so it isn't a single number. The law uses the semi-major axis

a = (r_{m i n} + r_{m a x}) /2

, the average of perihelion and aphelion distances.

"When we set gravity equal to the centripetal requirement, orbiting bodies feel two forces cancelling to zero."

Error: gravity is the centripetal force here — there is only one force. The equation

GM m / r^{2} = m v^{2} / r

equates the gravitational pull to the value

m v^{2} / r

that a circle requires; they are the same force, not two opposing ones.

"The

4 π^{2}

comes from the area of a circle."

Error: it comes from squaring the circumference relation

v = 2 π r / T

. Squaring gives

v^{2} = 4 π^{2} r^{2} / T^{2}

, and that

4 π^{2}

carries straight into the final law.

"To find orbital speed I use

v = 2 π r / T

, so speed depends only on how big the orbit is."

Error: that formula is a definition (distance per lap over time), not the physics. It still contains the unknown

T

. The physics comes from

v = GM / r

, which then determines

T

"Kepler discovered

T^{2} \propto a^{3}

from Newton's gravity law."

Error: the causation is reversed. Kepler found the pattern empirically from data decades before Newton; Newton later derived it from his inverse-square gravity, confirming both.

Why questions

Why does the orbiting body's mass

m

vanish from the final formula?

Because the same

m

appears in the gravitational force (

GM m / r^{2}

) and in the inertial response (

m v^{2} / r

); dividing both sides by

m

removes it — the deep reason all masses fall alike.

Why does

T

grow faster than

a

rather than in step with it?

Two penalties stack: the path length grows like

a^{1}

, and the speed drops like

a^{- 1/2}

. Time = path/speed

\propto a^{1} / a^{- 1/2} = a^{3/2}

Why is the exponent exactly

3/2

and not, say,

2

It traces to the inverse-square law of gravity. The

1/ r^{2}

force gives

v^{2} \propto 1/ r

, and combining with the circumference relation forces the

3/2

power. A different force law would give a different exponent.

Why can we derive the law using a circle when real orbits are ellipses?

The full elliptical derivation (via Kepler's equal-area law) collapses to the identical formula with

r \to a

; eccentricity cancels out. The circle gives the right answer with far less algebra.

Why does the same law apply to the Moon around Earth and Earth around the Sun with different constants?

The form

T^{2} \propto a^{3}

is universal, but the constant

4 π^{2} / GM

scales with the central mass

M

— the Sun and Earth have vastly different

M

, hence different constants.

Why must the inward pull exactly equal

m v^{2} / r

, not more or less, for a circular orbit?

If gravity exceeded

m v^{2} / r

the body would spiral inward; if it were less, it would drift outward. Perfect equality is what keeps the radius constant — a true circle.

Edge cases

What happens to the period as

a \to \infty

(orbit infinitely large)?

T \propto a^{3/2} \to \infty

: the period grows without bound. Infinitely far out, gravity is vanishingly weak and the path is infinite, so a lap would take forever.

What is the "orbit" when eccentricity

e = 1

The ellipse degenerates into a parabola — an unbound, one-way escape trajectory. There is no repeating period, so Kepler's Third Law (which assumes a closed orbit) no longer applies.

What about eccentricity

e > 1

That is a hyperbolic trajectory — even more strongly unbound than the parabola: the body flies past once and never returns. Like the parabola it has no period, and there is no semi-major axis in the usual bound sense, so the law simply does not apply.

Does the law hold if the two masses are comparable, like a binary star?

Only with a fix: replace

M

by the total

M_{1} + M_{2}

and measure

a

as the separation. Both stars then orbit their common centre of mass with this shared period.

Does the period change if the orbit is tilted (different inclination) or run backwards (retrograde vs prograde)?

No. Gravity is a central force — it depends only on the distance

r

, not on the orbit's tilt or direction of travel. Two orbits with the same

a

around the same

M

share the same period regardless of orientation.

What does the formula predict as

r \to 0

(grazing the surface)?

T = 2 π r^{3} / GM \to 0

: an infinitely low orbit has a vanishing period. In reality the surface (and atmosphere) stops you long before, but the trend — lower orbits are faster — is real.

Is there a distance at which a satellite's period equals Earth's rotation?

Yes — that specific

a \approx 42, 200

km makes

T = 86400

s, the geostationary radius. See Geostationary & Geosynchronous Orbits; it is a direct solve of the law for

r

given

T

Can a satellite "hover" motionless over a point on the equator by balancing gravity with nothing?

No. It must orbit at geostationary speed; there is no force to cancel gravity while stationary. It only appears fixed because its orbital period matches Earth's spin.

Connections

Newton's Law of Universal Gravitation — the inverse-square force that forces the $3/2$ exponent.
Centripetal Force & Circular Motion — the $m v^{2} / r$ requirement equated in Step 1.
Kepler's Second Law (equal areas) — why eccentricity cancels for the elliptical case.
Orbital Energy & Vis-viva Equation — the speed–distance relationship generalised.
Reduced Mass & Two-Body Problem — the $M \to M_{1} + M_{2}$ fix for comparable masses.
Geostationary & Geosynchronous Orbits — the law inverted to design real orbits.