Orbital Velocity Calculator

Calculate the orbital speed for a circular orbit using v_orb = √(GM/r).

🛸 Orbit 📐 v=√(GM/r) 🌍 Circular Orbit
Central body mass (M) kg
Orbital radius (r) m

Presets:

⚠️ Enter valid positive numbers.

Understanding Orbital Velocity

Orbital velocity is the speed required for a satellite to maintain a circular orbit at a given altitude. The formula v_orb = √(GM/r) comes from setting centripetal force equal to gravitational force: mv²/r = GMm/r² → v = √(GM/r). Like escape velocity, it is independent of the satellite's mass.

At lower orbits (smaller r), orbital velocity is higher — satellites must move faster to 'fall around' a tighter curve. The ISS orbits at 400 km altitude (r ≈ 6,771 km) at about 7.66 km/s. At geostationary orbit (35,786 km), the orbital velocity drops to 3.07 km/s — exactly matching Earth's rotation, so the satellite appears stationary.

Orbital period follows from v_orb: T = 2πr/v = 2π√(r³/GM). This is Kepler's Third Law — period squared is proportional to orbit radius cubed. From LEO (T ≈ 90 min) to GEO (T = 24 hours) to Moon (T = 27.3 days), the relationship holds precisely.

Orbital velocity relates to escape velocity by v_esc = √2 × v_orb. To leave orbit and escape Earth's gravity, a spacecraft must increase its velocity by a factor of √2 ≈ 1.414. This is the 'Δv budget' for trans-lunar injection and interplanetary missions.

Formula Reference Table

Solve ForFormulaNotes
Orbital velocityv = √(GM/r)G = 6.674×10⁻¹¹ N·m²/kg²
Orbital periodT = 2πr/v = 2π√(r³/GM)Kepler's Third Law
GEO altituder_GEO = (GM·T²/4π²)^(1/3)T = 86,400 s
ISS (LEO)v ≈ 7.66 km/sT ≈ 92 min
Moon orbitv ≈ 1.02 km/sT = 27.3 days
vs escapev_esc = √2 × v_orb1.414× orbital speed

3 Worked Examples

Example 1
ISS Low Earth Orbit

Find ISS orbital velocity. r = R_Earth + 400 km = 6.371×10⁶ + 4×10⁵ = 6.771×10⁶ m.

  • v = √(6.674×10⁻¹¹ × 5.972×10²⁴ / 6.771×10⁶)
  • v = √(5.887×10⁷) = 7,673 m/s = 7.67 km/s
  • Period T = 2π × 6.771×10⁶ / 7,673 = 5,541 s = 92.4 min
✓ v = 7.67 km/s; T = 92.4 minutes
Example 2
Geostationary Orbit Altitude

Find radius of GEO orbit (T = 24 hours).

  • r = (GM × T²/(4π²))^(1/3) = (3.986×10¹⁴ × (86400)² / 39.48)^(1/3)
  • r = (7.532×10²²)^(1/3) = 4.216×10⁷ m = 42,160 km
  • Altitude = 42,160 − 6,371 = 35,789 km
✓ GEO altitude = 35,786 km
Example 3
Moon's Orbital Speed

Moon orbit: r = 3.844×10⁸ m, M_Earth = 5.972×10²⁴ kg.

  • v = √(6.674×10⁻¹¹ × 5.972×10²⁴ / 3.844×10⁸)
  • v = √(1.036×10⁶) = 1,018 m/s = 1.02 km/s
  • Period = 2π × 3.844×10⁸ / 1,018 = 2.372×10⁶ s = 27.5 days
✓ v = 1.02 km/s; T = 27.5 days

Real-World Applications

🛰️
Satellite Design
Communication satellites must hit precise orbital velocities. Too slow → fall to lower orbit; too fast → climb higher. Station-keeping thrusters correct for atmospheric drag in LEO.
🌍
GPS Satellites
GPS satellites orbit at 20,200 km altitude (MEO). v_orb ≈ 3.87 km/s, T = 11 hours 58 min. Engineers calculate orbital velocities precisely for timing synchronization required for centimeter-accuracy GPS.
🚀
Mission Planning
Δv (velocity change) budget for missions is calculated from orbital velocity differences. Going from LEO to GEO requires Δv ≈ 3.9 km/s via Hohmann transfer — a key cost driver in satellite propellant mass.
📡
Geostationary Comms
GEO satellites at 35,786 km appear stationary because their orbital period exactly matches Earth's rotation. This requires v_orb = 3.07 km/s — directly calculable from v = √(GM/r).
🌙
Lunar Missions
To enter lunar orbit, Apollo spacecraft decelerated to v_orb ≈ 1.5–1.7 km/s at their chosen lunar orbit altitude. Too fast → flyby; too slow → crash. The rocket equation determined how much fuel was needed for orbital insertion.

Common Mistakes to Avoid

⚠️
Confusing orbital and escape velocity

v_esc = √2 × v_orb. Orbital velocity keeps you in orbit; escape velocity breaks free of gravity. They differ by a factor of 1.414.

⚠️
Using altitude instead of orbital radius

r in the formula is distance from the planet's CENTER, not altitude. Always add planet radius: r = R_planet + altitude.

⚠️
Wrong G constant units

G = 6.674×10⁻¹¹ N·m²/kg². Using G = 6.674×10⁻¹¹ with M in solar masses or r in AU gives wrong answers — must use consistent SI units.

⚠️
Assuming all orbits are circular

v = √(GM/r) gives circular orbit velocity. Elliptical orbits have varying speed (faster at perigee, slower at apogee) — v = √(GM(2/r − 1/a)) where a = semi-major axis.

⚠️
Forgetting that orbital velocity decreases with altitude

Counter-intuitively, higher orbit = slower speed. This is because at higher altitude, less centripetal acceleration is needed (weaker gravity), so lower orbital speed is required.

Frequently Asked Questions

How is orbital velocity derived? ▾
For circular orbit: gravitational force = centripetal force → GMm/r² = mv²/r → v = √(GM/r). The satellite's mass cancels, so orbital velocity is independent of satellite mass — a 1 kg cubesat and the ISS orbit at the same speed at the same altitude.
What is Kepler's Third Law? ▾
T² ∝ r³: orbital period squared is proportional to semi-major axis cubed. Precisely: T² = 4π²r³/(GM). For solar system planets orbiting the Sun: T²/a³ = 4π²/(GM_sun) is constant. This allowed Newton to verify his inverse-square gravity law against Kepler's observations.
What is geostationary orbit? ▾
GEO is the altitude where orbital period = Earth's rotational period (24 hours = 86,400 s). Solving T = 2π√(r³/GM): r_GEO = 42,164 km from Earth's center = 35,786 km altitude. At this altitude and 0° inclination, the satellite appears fixed in the sky — ideal for TV, weather, and communications satellites.
Why do satellites decay from orbit? ▾
Even at LEO altitudes, trace atmosphere exerts tiny drag, slowly removing kinetic energy. The satellite spirals inward, paradoxically speeding up as it descends (lower orbit = higher speed). Eventually atmospheric heating destroys it. The ISS loses ~2 km of altitude per month and requires regular reboosts.
What is delta-v (Δv)? ▾
Δv is the change in velocity a spacecraft must execute to change orbits. It determines propellant needed via the rocket equation: Δm/m₀ = 1 − e^(−Δv/Isp·g). LEO→GEO requires Δv ≈ 3.9 km/s; Earth escape from LEO ≈ 3.2 km/s; Mars surface launch ≈ 3.6 km/s.
What is the Lagrange point? ▾
Lagrange points are positions in a two-body orbital system where a small object maintains a stable position relative to both large bodies, combining the gravity and orbital motion of both. There are 5 Lagrange points (L1–L5) for each two-body system. The James Webb Space Telescope orbits Earth-Sun L2 at 1.5 million km from Earth.
Can a satellite orbit at any altitude? ▾
Practically speaking, below ~160 km atmospheric drag rapidly decays any orbit. Above this, theoretically any altitude works, but GEO is special for communications. Some satellites use sun-synchronous orbits (polar, ~600 km) to always pass over the same location at the same local time, useful for Earth observation.
What determines how long a satellite lasts in orbit? ▾
Drag is the main decay mechanism at LEO. Drag depends on atmospheric density (varies with solar activity), satellite area-to-mass ratio, and Cd. Satellites at 800 km altitude can last decades; at 400 km (ISS altitude) only 1–2 years without reboost. GEO satellites last 15+ years with station-keeping propellant.

Related Calculators

🚀 Escape Velocity 🌍 Gravitational Force 🔄 Angular Velocity ⏱️ Kepler's Third Law 📡 Centripetal Force

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