Kepler's Third Law Harmonic Equation
Relates the orbital period T of a celestial body to the semi-major axis a of its elliptical orbit around a primary gravitational mass.
\[ T^2 = \frac{4\pi^2}{G(M_1 + M_2)} \cdot a^3 \approx \frac{4\pi^2}{\mu} \cdot a^3 \]
Plain Text Formula:
T^2 = (4 * pi^2 * a^3) / (G * (M1 + M2))
📐 Variable Definitions & Parameters
| Symbol | Variable Name | Standard Unit | Physical Definition |
|---|---|---|---|
| \[ T \] | Orbital Period | seconds (s) or years | Time required to complete one full revolution |
| \[ a \] | Semi-Major Axis | meters (m) or AU | Average distance from primary focus |
| \[ G \] | Gravitational Constant | 6.67430×10⁻¹¹ m³/(kg·s²) | Universal gravitational constant |
| \[ M_1 + M_2 \] | Combined Mass | kg or Solar Masses | Total mass of primary star/planet and secondary orbiting body |
💡 Worked Engineering Example
Problem Statement:
Using Kepler's Third Law in AU and Earth years (T² = a³), find the orbital period of Jupiter at a = 5.204 AU.
Step-by-Step Solution:
- Calculate a³: (5.204)³ = 140.93
- Take square root: sqrt(140.93) = 11.87 years
🎯 Result: Jupiter takes 11.87 Earth years (4,332.6 days) to orbit the Sun.