CLS-06

Kepler's third law

T² = 4π² a³ /(G M) for a circular (or Keplerian) orbit.

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GravityKepler 1619Newton 1687

Governing equation

T^2=\dfrac{4\pi^2 a^3}{GM}

$a \in \left[1\times 10^{8}, 1\times 10^{13}\right]$

where

M: Central mass (kg)
a: Semi-major axis (m)
T: Period (d)
T: Period (yr)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

T^2=\dfrac{4\pi^2 a^3}{GM}

$a \in \left[1\times 10^{8}, 1\times 10^{13}\right]$

$a = 1\times 10^{8}$ $T = 1.73\times 10^{-5}$

Along the curve

②

Domain of the variable — blue

a_{\min} = 1\times 10^{8},\quad a_{\max} = 1\times 10^{13}

$a \in \left[1\times 10^{8}, 1\times 10^{13}\right]$

$a = 1\times 10^{8}$ $a = 1\times 10^{8}$

Along the curve

③

Response curve — green

T^2=\dfrac{4\pi^2 a^3}{GM}

$a \in \left[1\times 10^{8}, 1\times 10^{13}\right]$

$a = 1\times 10^{8}$ $T = 1.73\times 10^{-5}$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}T}{\mathrm{d}a}

$a \in \left[1\times 10^{8}, 1\times 10^{13}\right]$

$a = 6.3\times 10^{10}$ $T^{\prime} = 6.14\times 10^{-12}$

Along the curve

Lecture brief

Historical brief

Newtonian mechanics (1687) plus energy, angular momentum, Kepler and the ballistic parabola remain the first language of motion. Every sheet here is a closed-form orbit, throw, spin or oscillator. This sheet (CLS-06 — Kepler's third law) is the form associated with Kepler 1619 · Newton 1687. Working symbols: $M$ , $a$ $\rightarrow$ $T$ , $T$ . Newton derived Kepler's T² ∝ a³ from inverse-square gravity. For a circular orbit the centripetal requirement is GM/a² = ω² a.

Purpose

Purpose: compute

T

T

from

M

a

in Classical mechanics via

T^2=\dfrac{4\pi^2 a^3}{GM}

T² = 4π² a³ /(G M) for a circular (or Keplerian) orbit. Use it when a real classical mechanics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

M = 1.989e+30\,\mathrm{kg}

a = 1.496e+11\,\mathrm{m}

, the governing relation

T^2=\dfrac{4\pi^2 a^3}{GM}

yields

T = 365.211\,\mathrm{d}

T = 0.9999\,\mathrm{yr}

. Point mass M ≫ m, circular orbit, G = 6.674×10⁻¹¹. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Central mass M1.989e+30 kgSemi-major axis a149600000000 m

Outputs

Period T365.211 d
Period T0.9999 yr

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Narration of this film

Point mass M ≫ m, circular orbit, G = 6.674×10⁻¹¹.

Newton derived Kepler's T² ∝ a³ from inverse-square gravity. For a circular orbit the centripetal requirement is GM/a² = ω² a.

Reading speed

Watch on YouTube