Satellite Orbital Period Calculator

How Satellite Orbital Period Is Calculated

The orbital period of a satellite depends only on its altitude and the mass of the body it orbits, not on the satellite’s own mass. This is Kepler’s Third Law applied to circular orbits.
A bolt and a space station at the same altitude keep exactly the same schedule, which is precisely why orbital debris is such a problem: it stays on the timetable long after the satellite that shed it is gone.

Orbital Period Formula: T = 2π × √(r³ / GM)

Where:

• T = orbital period in seconds
• r = orbital radius (planet radius + altitude) in meters
• G = gravitational constant = 6.674 × 10⁻¹¹ N·m²/kg²
• M = mass of the central body (Earth = 5.972 × 10²⁴ kg)

Worked Example: International Space Station:

• Altitude: ~408 km = 408,000 m
• Earth radius: 6,371,000 m
• r = 6,371,000 + 408,000 = 6,779,000 m
• GM = 6.674×10⁻¹¹ × 5.972×10²⁴ = 3.986×10¹⁴ m³/s²
• T = 2π × √((6,779,000)³ / 3.986×10¹⁴) = 2π × √(3.115×10²⁰ / 3.986×10¹⁴)
• T = 2π × √(781,624) = 2π × 884.1 = 5,555 seconds ≈ 92.6 minutes

The measured ISS period runs a little longer, around 92.9 minutes, because its orbit is slightly elliptical and it does not sit at a constant 408 km.

Common Orbital Periods:

• Low Earth Orbit (400–600 km): ~92 to 97 minutes
• GPS satellites (20,200 km): 12 hours
• Geostationary orbit (35,786 km): 23.934 hours, one sidereal day
• Moon: 27.3 days (at 384,400 km)

Geostationary Orbit Calculation: Set T to one sidereal day, 86,164 s, and solve for r: r = (GM × T² / 4π²)^(1/3) = 42,164 km from Earth’s center (35,786 km altitude).

One detail worth knowing before you compare numbers: that 42,164 km figure is quoted against Earth’s equatorial radius of 6,378 km, which is where a geostationary satellite actually flies. This calculator uses the mean radius of 6,371 km, so entering 35,786 km puts the satellite 7 km lower and returns 23.928 hours rather than 23.934. Seven kilometers out of 42,000 does not matter for anything you would use this page for, but it explains the discrepancy if you check the answer against a GEO reference table.

Using the 24-hour solar day of 86,400 s instead gives 42,241 km, which is 77 km too high. A geostationary satellite has to keep pace with the Earth’s rotation relative to the fixed stars, and that takes 23h 56m 04s, not 24 hours. The Earth then needs an extra four minutes of turning to bring the Sun back overhead, because it has also moved a degree along its orbit. That four minutes a day is the entire difference between a sidereal and a solar day, and it is why the stars rise four minutes earlier each night.