Kepler Third Law Orbital Period

\[T=2\pi\sqrt{\frac{a^3}{G(M+m)}}\]

Variables

Torbital period (s)

asemi-major axis of the orbit (m)

Ggravitational constant (m³/kg·s²)

Mmass of the central body (kg)

mmass of the orbiting body (kg)

Description

What is this formula?

This formula calculates the orbital period of a body using Kepler's third law in its Newtonian form. It relates the time required to complete one orbit to the semi-major axis and the masses of the two bodies.

When to use it

Use this formula for planets, moons, satellites, binary stars, or any two-body orbital system where the semi-major axis and masses are known.

Example

Calculate the orbital period of Earth around the Sun.

Given:

a=1.496×10^11 m

G=6.67430×10^-11 m³/kg·s²

M=1.989×10^30 kg

m=5.972×10^24 kg

Formula:

T=2*pi*sqrt(a^3/(G*(M+m)))

Substitution:

T=2*pi*sqrt((1.496×10^11)^3/(6.67430×10^-11*(1.989×10^30+5.972×10^24)))

Result:

T≈3.156×10^7 s

This is approximately 365.25 days.

Applications

- Planetary orbital calculations

- Satellite mission analysis

- Binary star systems

- Celestial mechanics

- Space engineering