A small numerical check: orbital period from Kepler's third law

Tue, 10 Nov 2026 14:30:00 -0600

Kepler’s third law relates the orbital period $T$ to the semi-major axis $a$ for a body orbiting a mass $M$:

$$ T = 2\pi \sqrt{\frac{a^3}{GM}}. $$

For Earth around the Sun, using $a = 1.496 \times 10^{11}\,\mathrm{m}$ and $GM = 1.327 \times 10^{20}\,\mathrm{m^3\,s^{-2}}$, we expect roughly one year.

import math

GM = 1.32712440018e20
a = 1.495978707e11

period_seconds = 2 * math.pi * math.sqrt(a**3 / GM)
print(period_seconds / (365.25 * 24 * 3600))

1.0000

Reference values

Body	Semi-major axis (AU)	Period (years)
Mercury	0.387	0.241
Earth	1.000	1.000
Mars	1.524	1.881
Jupiter	5.203	11.862

The point is not the arithmetic. The point is that a notebook can carry the claim, the code, and the output together.

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A small numerical check: orbital period from Kepler's third law

Reference values