An orbital period is the time a satellite takes to go once around its orbit — about 92 minutes for the ISS and almost exactly one day for a geostationary satellite.
The orbital period is one lap time. Start a stopwatch when a satellite crosses the equator going north. Stop it when the satellite next crosses the equator going north. That is its period.
For the International Space Station, NASA gives 90 to 93 minutes, depending on the station's exact height. That is about 16 orbits every day.
For a circular orbit, the period depends only on the distance from Earth's centre. The mass and shape of the satellite do not matter. A tiny student satellite and a large station at the same height take the same time.
ISRO gives Cartosat-3's mean altitude as 509 km. Its distance from Earth's centre is 6,378 + 509 = 6,887 km.
In one day (1,440 minutes) that is about 15.2 orbits.
A geostationary satellite has to keep pace with Earth's spin, measured against the distant stars. NASA's Earth Fact Sheet gives that sidereal rotation period as 23.9345 hours — about 23 hours 56 minutes. Our 24-hour solar day is a little longer because Earth also moves along its path round the Sun each day.
Note when the station's ground track crosses the equator on the map, then check again about 92 minutes later — it will be crossing again, farther west. Open the ISS tracker
For a circular orbit, the period is the circumference divided by the speed:
T = 2πr / √(GM/r) = 2π √(r³ / GM)
The same formula holds for ellipses if r is replaced by the semi-major axis a — this is Kepler's third law in Newton's form.
Finding the geostationary radius. Set T equal to the sidereal day: 23.9345 h = 86,164 s. Rearranging, r³ = GM T² / (4π²).
NASA rounds this to "about 36,000 kilometres".
Checking the ISS. r = 6,778 km: r³ = 3.1139 × 10¹¹, divided by GM = 781,200, square root = 883.9 s, times 2π = 5,553 s = 92.6 minutes. That sits inside NASA's 90–93 minute range.
Inverting a period. Tracking data often gives mean motion n in revolutions per day instead of a height. Then T = 1,440 / n minutes and a = (GM T² / 4π²)^(1/3). An ISS mean motion of 15.5 rev/day gives T = 92.9 minutes and a ≈ 6,795 km — about 417 km above the equator.
No. The period depends only on the size of the orbit and Earth's mass, not on the satellite's own mass.
It must match Earth's rotation relative to the stars (the sidereal day, 23.9345 h). The 24-hour solar day is longer because Earth also moves around the Sun.
√(6,887³ ÷ 398,600) ≈ 905 s; times 2π ≈ 5,690 s, or about 94.8 minutes.