Kepler’s third law connects the size of an orbit with its orbital period. In its simplest form,
for objects orbiting the same central body,
\[
T^2 \propto a^3.
\]
This means larger orbits take longer. More precisely, period grows as
\[
T\propto a^{3/2}.
\]
1. Newtonian form
Newton’s version of Kepler’s third law is
\[
\boxed{
T^2=\frac{4\pi^2a^3}{G(M+m)}
}.
\]
Here:
- \(T\) is the orbital period,
- \(a\) is the semi-major axis,
- \(G\) is the gravitational constant,
- \(M\) is the central mass,
- \(m\) is the orbiting body mass.
For planets, moons, and artificial satellites, the orbiting mass is often tiny compared with the central mass.
In that case,
\[
M+m\approx M.
\]
So the formula becomes
\[
\boxed{
T^2=\frac{4\pi^2a^3}{GM}
}.
\]
2. Solving for period
Starting from
\[
T^2=\frac{4\pi^2a^3}{G(M+m)},
\]
take the square root:
\[
\boxed{
T=2\pi\sqrt{\frac{a^3}{G(M+m)}}
}.
\]
For a circular orbit, \(a\) is the orbital radius. For an elliptical orbit, \(a\) is the semi-major axis.
3. Solving for semi-major axis
If the period and mass are known, solve for \(a\):
\[
T^2=\frac{4\pi^2a^3}{G(M+m)}.
\]
Rearranging gives
\[
a^3=\frac{G(M+m)T^2}{4\pi^2}.
\]
Therefore,
\[
\boxed{
a=\sqrt[3]{\frac{G(M+m)T^2}{4\pi^2}}
}.
\]
4. Solving for mass
If \(T\) and \(a\) are measured, the total gravitating mass is
\[
\boxed{
M+m=\frac{4\pi^2a^3}{GT^2}
}.
\]
If the orbiting mass \(m\) is negligible, this gives the central mass \(M\) directly.
In a binary system, it gives the total mass of the two-body system.
5. Comparing two orbits
For two orbits around the same central body,
\[
T_1^2=\frac{4\pi^2a_1^3}{GM},
\qquad
T_2^2=\frac{4\pi^2a_2^3}{GM}.
\]
Dividing the equations cancels the constants:
\[
\frac{T_2^2}{T_1^2}=\frac{a_2^3}{a_1^3}.
\]
Taking the square root gives
\[
\boxed{
\frac{T_2}{T_1}
=
\left(\frac{a_2}{a_1}\right)^{3/2}
}.
\]
This is the classic proportional form of Kepler’s third law.
6. Geostationary orbit example
A geostationary satellite has an orbital radius of about
\[
a=42164\ \mathrm{km}
=
4.2164\times10^7\ \mathrm{m}.
\]
Use Earth’s mass:
\[
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg}.
\]
The period is
\[
T=2\pi\sqrt{\frac{a^3}{GM_{\oplus}}}.
\]
Substitute:
\[
T
=
2\pi
\sqrt{
\frac{(4.2164\times10^7)^3}
{(6.67430\times10^{-11})(5.9722\times10^{24})}
}.
\]
\[
T\approx86164\ \mathrm{s}.
\]
Convert to hours:
\[
T\approx\frac{86164}{3600}\approx23.93\ \mathrm{h}.
\]
This is one sidereal day, which is why geostationary satellites stay above nearly the same longitude on Earth.
It is commonly rounded to 24 hours.
7. Circular-speed preview
For a circular orbit, the average orbital speed is
\[
\boxed{
v=\frac{2\pi a}{T}
}.
\]
Equivalently, for a circular Newtonian orbit,
\[
\boxed{
v=\sqrt{\frac{G(M+m)}{a}}
}.
\]
Kepler’s third law and circular orbital speed are two views of the same balance between gravity and orbital motion.
8. Binary systems
In a binary system, neither body is fixed at the center. Both bodies orbit their common center of mass.
The same law applies if \(a\) is the relative semi-major axis between the two bodies:
\[
\boxed{
T^2=\frac{4\pi^2a^3}{G(M_1+M_2)}
}.
\]
This is why observing binary-star orbits is one of the most important ways to measure stellar masses.
9. Summary table
| Goal |
Formula |
Use case |
| Period |
\(T=2\pi\sqrt{a^3/[G(M+m)]}\) |
Find how long one orbit takes |
| Semi-major axis |
\(a=\sqrt[3]{G(M+m)T^2/(4\pi^2)}\) |
Find orbital size from a known period |
| Total mass |
\(M+m=4\pi^2a^3/(GT^2)\) |
Find the mass from measured orbit data |
| Period ratio |
\(T_2/T_1=(a_2/a_1)^{3/2}\) |
Compare two orbits around the same mass |
| Circular speed |
\(v=2\pi a/T\) |
Preview orbital speed for circular orbits |
10. Important assumptions
- The orbit is treated using Newtonian gravity.
- The semi-major axis \(a\) is measured from the focus/central body for satellite-style problems.
- For binary systems, \(a\) is the relative semi-major axis of the two-body orbit.
- Air resistance, thrust, atmospheric drag, and orbital perturbations are ignored.
- The circular-speed preview is exact for circular orbits but only an average-style reference for elliptical orbits.
Key idea: when orbiting the same mass, a larger semi-major axis produces a much longer period because \(T\propto a^{3/2}\).
