Orbital Speed Calculator
Calculate orbital velocity, period, and circumference for any orbit around any celestial body.
Orbital Speed Calculator
Calculator
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Formula: v = sqrt(G * M / r), where r = body radius + altitude
Worked example — ISS orbital speed = 7.667 km/s, period = 92.6 minutes
Formula
v = sqrt(G * M / r), where r = body radius + altitude
Orbital speed v equals the square root of the gravitational constant G (6.674e-11) times the central body mass M (in kg) divided by the orbital radius r (in meters). The orbital radius is the sum of the central body radius and the orbit altitude. The orbital period T = 2*pi*r / v gives the time for one complete orbit.
Worked Examples
Example 1: ISS Orbital Speed
Problem:Calculate the orbital speed of the ISS at 408 km above Earth (mass 5.972e24 kg, radius 6,371 km).
Solution:r = 6,371,000 + 408,000 = 6,779,000 m v = sqrt(G * M / r) v = sqrt(6.674e-11 * 5.972e24 / 6,779,000) v = sqrt(5.878e7) = 7,667 m/s
Result:ISS orbital speed = 7.667 km/s, period = 92.6 minutes
Example 2: Earth Orbital Speed around Sun
Problem:Find Earth's orbital speed around the Sun (mass 1.989e30 kg, distance 1.496e11 m).
Solution:r = 6.957e8 + 1.496e11 = 1.503e11 m v = sqrt(G * M / r) v = sqrt(6.674e-11 * 1.989e30 / 1.503e11) v = sqrt(8.828e8) = 29,712 m/s
Result:Earth orbits the Sun at about 29.7 km/s or 107,000 km/h
Frequently Asked Questions
What determines orbital speed?
Orbital speed is determined by the mass of the central body being orbited and the distance from its center. A higher orbit means a lower speed because the gravitational pull weakens with distance. The formula v = sqrt(GM/r) shows that speed depends only on these two factors, not on the mass of the orbiting object. This is why the ISS at 408 km altitude orbits at about 7.66 km/s while geostationary satellites at 35,786 km orbit at only 3.07 km/s.
What is a geostationary orbit and why is it useful?
A geostationary orbit is a circular orbit at approximately 35,786 km above Earth's equator where the orbital period exactly matches Earth's rotation period of 23 hours 56 minutes 4 seconds. A satellite in this orbit appears stationary relative to a point on the ground. This makes it ideal for communications satellites, weather monitoring, and television broadcasting because ground antennas can point at a fixed position in the sky without needing to track the satellite's movement.
How does orbital speed relate to orbital period?
Orbital speed and period are inversely related through the orbital circumference. The period T equals 2*pi*r / v, where r is the orbital radius and v is the orbital speed. As altitude increases, orbital radius grows (increasing the path length) while speed decreases (due to weaker gravity), so the period increases rapidly. This relationship follows Kepler's Third Law: the square of the period is proportional to the cube of the semi-major axis.
Why do astronauts on the ISS float if they are still within Earth's gravity?
Astronauts on the ISS experience weightlessness not because gravity is absent but because they are in continuous free fall around Earth. At the ISS altitude of about 408 km, gravity is still about 89% as strong as on the surface. The station and everyone inside are falling toward Earth at the same rate, but their horizontal speed of 7.66 km/s means they continuously fall around the curved Earth rather than into it. This state of perpetual free fall creates the sensation of zero gravity.
What happens if an orbiting object speeds up or slows down?
If an orbiting object speeds up (gains kinetic energy), it moves to a higher orbit with a larger radius, counterintuitively ending up at a slower average speed in the new orbit. If it slows down, it drops to a lower orbit with a faster speed. This paradox occurs because orbital mechanics involves trading between kinetic and potential energy. To reach a higher orbit, spacecraft use a Hohmann transfer: they fire engines to speed up, coast to the higher altitude, then fire again to circularize the new orbit.
How does the vis-viva equation extend this to elliptical orbits?
The circular formula is a special case of the vis-viva equation, which states that v squared equals GM times the quantity two over r minus one over a, where r is the current distance from the centre of the body and a is the semi major axis. In a circular orbit a equals r, the bracket reduces to one over r, and the expression collapses back to the square root of GM over r. On an ellipse the speed varies continuously, peaking at periapsis where r is smallest and falling to a minimum at apoapsis.
What is the standard gravitational parameter and why is it used?
The standard gravitational parameter, written as mu, is simply the product of the gravitational constant G and the central body mass M. It is used because mu can be determined from spacecraft tracking far more precisely than either G or M can be measured on its own, so quoting the product avoids importing the relatively large uncertainty in G. For Earth mu is about 3.986 times 10 to the 14th cubic metres per second squared. Substituting mu for GM leaves the orbital speed formula otherwise unchanged.
How is escape velocity related to circular orbital speed?
Escape velocity at a distance r is the square root of two GM over r, while circular orbital speed at the same distance is the square root of GM over r. The escape value is therefore larger by a factor of the square root of two, about 1.414 times, at every radius. A satellite in a circular orbit needs roughly 41 percent more speed to reach a parabolic escape trajectory. Because both quantities scale with one over the square root of r, that ratio holds at low altitude and far out alike.
Why is a low Earth orbit faster than a geostationary orbit?
Speed falls off as one over the square root of the orbital radius, so a closer orbit must move faster to supply the centripetal acceleration that the stronger local gravity demands. At about 400 kilometres altitude the orbital radius is roughly 6,778 kilometres and the speed is about 7.7 kilometres per second. At the geostationary radius of 42,164 kilometres the speed is only about 3.07 kilometres per second. The much larger circumference combined with the lower speed is what stretches the period from about 90 minutes to 24 hours.
How does Kepler's third law follow from the orbital speed formula?
Period is circumference divided by speed, so T equals two pi r divided by the square root of GM over r, which simplifies to two pi times r to the power three halves divided by the square root of GM. Squaring both sides gives T squared equal to four pi squared r cubed over GM. That is Kepler's third law: the square of the period is proportional to the cube of the orbital radius, with the constant of proportionality set only by the mass of the central body.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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