Gravitation and circular orbits
v = √(G·M / r)
Enter the central body’s mass and the orbital radius (from the centre, not the surface: trap number one in these exercises) and the tool computes the orbital speed, the orbital period (Kepler’s third law), the escape velocity (√2 times the orbital speed, exactly) and the local gravity. Works for the ISS around Earth, the Moon, a geostationary satellite or an exoplanet around its star.
7.6725 km/s
- Calculation
- v = √(G·M/r) = 7.6725 km/s, G = 6.67430 × 10⁻¹¹ (CODATA 2018)
- Orbital period
- 1 h 32 min
Circular orbit at 7.6725 km/s, completed in 1 h 32 min. To escape from this altitude: 10.851 km/s, exactly √2 times the orbital speed, whatever the orbit.
Scientific dossier
What the tool computes, what it assumes, where it stops being valid, and where its data comes from.
Method & formulasv = √(G·M / r)
v = √(G·M / r)
T = 2π·√(r³ / G·M), Kepler: r³/T² = G·M/4π²
v_esc = √(2·G·M / r) = √2 · v
g(r) = G·M / r²
G = 6.67430 × 10⁻¹¹ m³/(kg·s²), codata 2018
Everything follows from one equality: gravity provides exactly the centripetal acceleration (G·M·m/r² = m·v²/r), and the satellite's mass cancels. A bolt and the Space Station orbit identically. Kepler's third law follows: r³/T² is the same constant for everything orbiting the same central body. The escape velocity comes from energy (½mv² = G·M·m/r) and is exactly √2 times the circular orbital speed, at any altitude. G is the least precisely known fundamental constant (relative uncertainty 2.2 × 10⁻⁵), your results inherit that precision.
- Orbital radius (r)
- · the distance to the centre of the central body, never the altitude. ISS: 6,371 km Earth radius + 400 km altitude = 6,771 km. Getting this wrong skews everything by ~6%^(3/2).
- Escape velocity
- · the minimum speed to coast away to infinity without further propulsion, from distance r. It does not depend on launch direction (it is an energy condition), 11.2 km/s from Earth’s surface.
- Kepler constant (r³/T²)
- · identical for all satellites of the same central body: the Moon and the ISS share the same one, Earth’s. Comparing two orbits is comparing their r³/T².
Validity domainCIRCULAR orbits around a spherically symmetric body, satellite mass negligible against M: elliptical orbits (perigee/apogee), perturbations (J2 flattening.
CIRCULAR orbits around a spherically symmetric body, satellite mass negligible against M: elliptical orbits (perigee/apogee), perturbations (J2 flattening. Atmospheric drag, third bodies) and relativistic effects are outside the model. The real geostationary period targets the sidereal day (86,164 s), not the 86,400 s solar day. The body masses given as hints are usual reference values, known far better than G itself, the uncertainty comes from G.
Common pitfall: counting from the surface“The ISS orbits at 400 km”, yes.
“The ISS orbits at 400 km”, yes. Of altitude. The r in the formulas starts at Earth’s centre: 6,771 km. Using 400 km would give an orbital speed of 31 km/s and a period of 81 seconds, absurd by a factor of 4. Another anchor worth keeping: the lower the orbit, the faster it is (v ∝ 1/√r), the ISS loops in 92 minutes, the Moon in 27 days; braking a satellite makes it descend and therefore speed UP, orbital mechanics’ favourite paradox.