Kepler Third Law (General)
Orbital period from semi-major axis.
Formula
Variables
Example
Farther planets orbit slower
How it works
This law links an orbit's size to how long one full revolution takes around a much more massive central body. It comes from combining Newton's law of gravitation with the requirement that gravity provides the centripetal acceleration needed to keep an object in orbit; for elliptical orbits, the semi-major axis replaces the simple orbital radius. The result shows that larger orbits have much longer periods, and that the central mass controls the overall timescale of the motion.
When to use it
Use this when you know the semi-major axis of a planet, moon, asteroid, or binary system and want its orbital period, or when you know the period and want to infer the orbit size. It is especially useful for objects orbiting a dominant central mass, such as planets around a star or satellites around a planet.
Worked example, step by step
Find the orbital period of a satellite with semi-major axis a = 2.00 × 10^7 m around Earth, where M = 5.97 × 10^24 kg. Step 1: Write the formula: T^2 = (4π^2 / GM) a^3 Step 2: Substitute values: T^2 = [4π^2 / ((6.674 × 10^-11)(5.97 × 10^24))] (2.00 × 10^7)^3 Step 3: Compute the denominator: GM = 3.985 × 10^14 Step 4: Compute a^3: (2.00 × 10^7)^3 = 8.00 × 10^21 Step 5: Combine: T^2 = (39.48 / 3.985 × 10^14) × 8.00 × 10^21 ≈ 7.92 × 10^8 Step 6: Take the square root: T ≈ sqrt(7.92 × 10^8) ≈ 2.81 × 10^4 s Step 7: Convert to hours: 2.81 × 10^4 s ÷ 3600 ≈ 7.8 h Final answer: T ≈ 2.81 × 10^4 s, or about 7.8 hours.
Common mistakes to avoid
- Using the orbital radius instead of the semi-major axis for an elliptical orbit
- Forgetting that M is the central mass only, not the combined mass of the orbiting system
- Mixing units, such as using kilometers for a in one step and SI units for G and M
Did You Know?
Kepler's third law ties a planet's year to its distance, which is why distant Neptune takes 165 Earth years to orbit.
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