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

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

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

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