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Circular Orbital Velocity

This gives the speed a satellite needs to stay in a circular orbit at a given radius. It equals the square root of the Earth's gravitational parameter divided by the orbital radius. For example, the International Space Station at 400 km altitude orbits at about 7.66 km/s. Higher orbits are slower: the Moon orbits at about 1 km/s. If the satellite goes slower than this, it falls toward Earth; if faster, it rises. It is the most basic orbital mechanics formula.

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Circular Orbital Velocity Calculator

Solve for circular orbital velocity, altitude or a body's gravitational parameter, with step-by-step math, an orbit diagram, velocity and period curves, and a comparison across planets.

v = √( μ / r ),   r = R + h
Central body
Common Earth orbits
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Orbit metrics
ORBITAL VELOCITY (v)
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Step-by-step solution
Orbit diagram (schematic, not to scale)
Orbit curves (altitude on a log scale)
Current central bodyCurrent orbit
Same altitude, different bodies

Circular orbital velocity

A satellite in a circular orbit is in continuous free fall: gravity supplies exactly the centripetal force needed to bend its path into a circle. Setting gravitational acceleration μ/r² equal to centripetal acceleration v²/r gives v = √(μ/r). The speed depends only on the central body and the orbital radius, not on the satellite's mass, and lower orbits are faster than higher ones.

v = √( μ / r )
r = μ / v² (so h = r − R)
μ = v² · r
T = 2π · √( r³ / μ )
vescape = √2 · v
SymbolMeaningUnits used here
vCircular orbital velocitykm/s
μGravitational parameter G·M of the central bodykm³/s²
RRadius of the central bodykm
hAltitude above the surfacekm
rOrbital radius from the body's centre (R + h)km

Worked example

Earth: μ = 398,600 km³/s², R = 6,371 km, h = 400 km
r = 6,371 + 400 = 6,771 km
v = √(398,600 / 6,771) = 7.673 km/s
T = 2π √(6,771³ / 398,600) ≈ 5,545 s ≈ 92.4 min

Reference orbits around Earth

OrbitAltitudeVelocity
ISS (LEO)≈ 400 km≈ 7.67 km/s
GPS (MEO)≈ 20,200 km≈ 3.87 km/s
Geostationary≈ 35,786 km≈ 3.07 km/s
Moon's orbit≈ 378,000 km≈ 1.02 km/s

FAQ

Why doesn't the satellite's mass matter?

The satellite's mass appears on both sides of Newton's second law (gravity ∝ m, inertia ∝ m) and cancels, so every object at the same radius has the same circular velocity.

Why are lower orbits faster?

Gravity is stronger closer in, so a higher speed is needed to keep falling around the body rather than into it. The period also gets shorter as the orbit shrinks.

What is the difference between orbital and escape velocity?

Escape velocity at a given radius is √2 times the circular velocity there. Speeds between the two give elliptical orbits; at or above escape velocity the path is open and the satellite leaves the body.

Does this include atmospheric drag?

No. It assumes a point-mass central body and a perfectly circular orbit above the atmosphere. Very low altitudes also need extra velocity to counter drag over time.

Variables

Symbol Quantity Unit
v Orbital velocity m/s
μ Standard gravitational parameter m3/s2
r Orbital radius m

What it means

The circular orbital velocity is the speed needed to maintain a circular orbit at a given radius, assuming a two‑body problem. It is derived from equating gravitational force to centripetal force. This velocity is also the minimum speed to sustain orbit (any lower and the object would re‑enter). For Earth’s surface, it is about 7.9 km/s (first cosmic velocity). This formula is used in satellite design, launch vehicle trajectory planning, and mission analysis. It also appears in the calculation of orbital periods and in the design of space rendezvous. Understanding orbital velocity is fundamental for spaceflight mechanics.

Worked example

Circular Orbital Velocity – Two Examples

Real‑World
Scenario: Find orbital velocity at LEO altitude (r = 6.678×10⁶ m, μ = 3.986×10¹⁴).
ParameterValue
μ3.986×10¹⁴ m³/s²
r6.678×10⁶ m
1v = √(μ/r) = √(3.986e14/6.678e6) = √(59,688,000) = 7726 m/s
Result 7,726 m/s ✓ LEO
Scenario: Find orbital velocity at GEO (r = 4.216×10⁷ m).
ParameterValue
r4.216×10⁷ m
1v = √(3.986e14/4.216e7) = √(9,454,000) = 3075 m/s
Result 3,075 m/s ✓ GEO
Key insight: Orbital velocity decreases with altitude – higher orbits mean slower speeds.

Common mistakes

  • Circular orbital velocity: v = √(μ/r).
  • μ: Gravitational parameter of central body.
  • r: Distance from centre (radius of orbit).
  • For Earth: at r = 6371 km + altitude.
  • Units: m/s.

Applications

Circular orbital velocity, v = √(μ/r), is the speed required for a spacecraft to maintain a stable circular orbit at distance r from the central body. This velocity is fundamental for all space missions, as it determines the required launch energy and the orbital period. Engineers use it to design satellite orbits, to compute rendezvous parameters, and to plan Earth‑to‑orbit missions. The velocity depends on the gravitational parameter μ and the orbital radius. By understanding circular velocity, aerospace engineers can design launch trajectories, plan station‑keeping manoeuvres, and evaluate the performance of propulsion systems for orbital insertion.

  • Design of Earth and planetary orbits for satellites and spacecraft
  • Launch vehicle performance requirements (injection velocity)
  • Rendezvous and proximity operations planning
  • Orbital period and ground track calculations
  • Spacecraft orbit maintenance and station‑keeping

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  • Published July 10, 2026
  • Last updated October 22, 2026