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Orbital Speed Calculator

Instantly convert orbital speed with our free converter. See conversion tables, formulas, and step-by-step explanations.

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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\nv = sqrt(G * M / r)\nv = sqrt(6.674e-11 * 5.972e24 / 6,779,000)\nv = 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\nv = sqrt(G * M / r)\nv = sqrt(6.674e-11 * 1.989e30 / 1.503e11)\nv = 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.

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.

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You may use the results for reference and educational purposes. For professional reports, academic papers, or critical decisions, we recommend verifying outputs against peer-reviewed sources or consulting a qualified expert in the relevant field.

References