Orbital Speed Calculator
Instantly convert orbital speed with our free converter. See conversion tables, formulas, and step-by-step explanations.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Orbital Speed 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.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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