Orbital Period Calculator

T = 2π√(a³/GM): Kepler’s Third Law connects orbital size to orbital period

Worked example — Orbital Period Calculator

Orbital period	1.54 h

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Altitude does not set the period 🖖

The field is labelled semi-major axis for a reason: eccentricity never appears in T = 2π√(a³/GM). Enter 42164 km and the tool returns 23.94 h, the geostationary circle. An ellipse that skims 6771 km at perigee and swings out to 77557 km at apogee has the same semi-major axis, so it returns the same 23.94 h — while crossing the low point at 10.41 km/s and crawling through the high point at 0.91 km/s. The ISS, circular at that same 6771 km, is back in 1.54 h.

The orbiting object's mass drops out 🖖

The formula T = 2π√(a³/GM) contains M, the central body's mass, but not the satellite's. A feather and a fully loaded space station at the same altitude circle Earth in exactly the same time. Orbital period depends only on how big the orbit is and what you orbit, never on how heavy you are.

A tunnel through Earth keeps orbital time 🖖

Drop a ball into a frictionless tunnel bored straight through Earth and it reaches the far side and returns in about 84.5 minutes. That is exactly the period of a satellite skimming Earth's surface. Both are governed by the same √(R³/GM), so the 'gravity train' and the lowest possible orbit share one clock.

Problem solved in full

Time for one lap of the ISS flying 400 km above the ground 5 steps
The ISS flies 400 km above the ground. Find how long one lap takes, and then find the shortest orbit Earth allows at all. This is the ISS state — semi-major axis 6771 km about Earth — with G = 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻², M = 5.972 × 10²⁴ kg and Earth's radius R = 6371 km.
1. A circular orbit is a balance: gravity supplies exactly the centripetal force the circle needs, no more and no less. The satellite's own mass cancels out of that equation, which is why a loose bolt and the station it fell off stay side by side.
  
  $\frac{GM m}{a ^{2}} = \frac{m v ^{2}}{a} ⟹ v = \frac{GM}{a}$
2. Period is distance over speed once the speed is known. Nothing about the satellite survives into the result — only the size of the orbit and the mass of the planet, and those two arrive multiplied together as GM. That product is what orbits measure; G and M are known separately to nowhere near the same precision.
  
  $T = \frac{2 π a}{v} = 2 π \frac{a ^{3}}{GM}, GM = (6.674 \times 1 0^{- 11}) (5.972 \times 1 0^{24}) = 3.986 \times 1 0^{14} m^{3} s^{- 2}$
3. 6771 km is Earth's 6371 km radius plus 400 km of altitude, because a is measured from the centre and never from the ground. Fifteen and a half of these laps fit in a day — 86 400/5545 = 15.58 — and the leftover 0.58 is why the station crosses a different strip of ground on every pass instead of retracing the same one.
  
  $T = 2 π \frac{( 6.771 \times 1 0 ^{6} ) ^{3}}{3.986 \times 1 0 ^{14}} = 5545 s = 1 h 32 m 25 s = 1.54 h$
4. The formula's only grip on distance is a^3/2, so a ratio of two periods needs neither G nor M. That is Kepler's third law, and it is why the scaling table above can be built from bare multipliers: four times the semi-major axis is exactly eight times the period, for any satellite of any planet.
  
  $T \propto a^{3/2} : 2^{3/2} = 2.83, 4^{3/2} = 8 ⟹ 8 \times 1.54 = 12.32 h$
5. Now run the orbit downwards. Below a = R there is no orbit left to have, so that radius fixes a floor under the period — and the useful move is to write the mass as volume times density before evaluating it. R³ then appears both on top and inside M, and it cancels.
  
  $M = \frac{4}{3} π R^{3} ρ ⟹ T_{m i n} = 2 π \frac{R ^{3}}{GM} = \frac{3 π}{Gρ} : ρ_{\oplus} = \frac{5.972 \times 1 0 ^{24}}{\frac{4}{3} π ( 6.371 \times 1 0 ^{6} ) ^{3}} = 5513 kg m^{- 3} \Rightarrow 5061 s = 84.4 min$
Answer

The tool prints 1.54 h for the ISS and 12.32 h at four times the semi-major axis, a factor of exactly 8 because 4^3/2 = 8, with ×2.83 for a doubling because 2^3/2 = 2.828. Step 5 is the part it does not compute: nothing can circle Earth in less than 84.4 minutes, and that floor is set by Earth's mean density of 5513 kg/m³ and by nothing else. Saturn is nine times wider than Earth, yet a spacecraft skimming its cloud tops takes 4.0 hours rather than a day, because Saturn's mean density is 687 kg/m³ — an eighth of Earth's. The fastest possible orbit is therefore a measurement of the ground beneath it: time a low satellite and you have weighed the planet per unit volume, without ever needing to know how big it is.

$T_{m i n} = \frac{3 π}{Gρ} ρ = 687 kg m^{- 3} \Rightarrow T_{m i n} = 3.98 h$

Learning path

Orbits from two numbers

Leads to Hohmann transfer the time to go round once, from the semi-major axis alone.

References (1)

Insight block 3 — the gravity train sharing a clock with the lowest orbit: P. W. Cooper, "Through the Earth in Forty Minutes." American Journal of Physics 34(1), 68–70, 1966.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

Example problems

ISS - ISS at ~400 km altitude: T ≈ 92 min — circles Earth 16 times a day
GPS satellite - GPS satellites: T ≈ 12 h — each satellite orbits twice per day
Geostationary - Geostationary orbit: T ≈ 24 h — appears fixed above one point on Earth
Moon - Moon around Earth: T ≈ 27.3 days — the origin of the word "month"
Mercury - Mercury around Sun: T ≈ 88 days — fastest planet
Earth (around Sun) - Earth around Sun: T ≈ 365.25 days — defines the year
Mars (around Sun) - Mars around Sun: T ≈ 687 days — nearly two Earth years