Satellite Speed Calculator
Calculate the orbital speed of a satellite at any altitude above Earth, Moon, Mars, or Jupiter.
Satellite Speed Calculator
Calculator
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Formula: v = sqrt(GM/r)
Worked example โ Speed: 7,667 m/s (27,601 km/h) | Period: 92.5 min | 15.56 orbits/day
Formula
v = sqrt(GM/r)
Orbital speed v equals the square root of the gravitational constant G times the central body mass M divided by the orbital radius r (center of body to satellite). The orbital period T = 2*pi*r/v. Escape velocity = sqrt(2) times the orbital speed.
Worked Examples
Example 1: International Space Station Orbital Speed
Problem:Calculate the orbital speed and period of the ISS at 408 km altitude above Earth.
Solution:G = 6.674 x 10^-11, M(Earth) = 5.972 x 10^24 kg r = 6,371,000 + 408,000 = 6,779,000 m v = sqrt(GM/r) = sqrt(6.674e-11 x 5.972e24 / 6,779,000) v = sqrt(5.878e7) = 7,667 m/s = 27,601 km/h Period = 2*pi*r/v = 2 x 3.14159 x 6,779,000 / 7,667 T = 5,553 seconds = 92 minutes 33 seconds Orbits per day = 86,400 / 5,553 = 15.56
Result:Speed: 7,667 m/s (27,601 km/h) | Period: 92.5 min | 15.56 orbits/day
Example 2: Mars Reconnaissance Orbiter
Problem:Calculate the orbital speed of a satellite at 300 km altitude above Mars (M = 6.417 x 10^23 kg, R = 3,389.5 km).
Solution:r = 3,389,500 + 300,000 = 3,689,500 m v = sqrt(GM/r) = sqrt(6.674e-11 x 6.417e23 / 3,689,500) v = sqrt(1.160e7) = 3,406 m/s = 12,262 km/h Period = 2*pi*r/v = 2 x 3.14159 x 3,689,500 / 3,406 T = 6,806 seconds = 113 minutes 26 seconds
Result:Speed: 3,406 m/s (12,262 km/h) | Period: 113.4 min | 12.7 orbits/day
Frequently Asked Questions
How is the orbital speed of a satellite calculated?
The orbital speed of a satellite is calculated using the vis-viva equation simplified for circular orbits: v = sqrt(GM/r), where G is the gravitational constant (6.674 x 10^-11 N m^2/kg^2), M is the mass of the central body (5.972 x 10^24 kg for Earth), and r is the orbital radius measured from the center of the body (Earth's radius plus altitude). For the International Space Station at 408 km altitude, r = 6,371 + 408 = 6,779 km = 6,779,000 m. This gives v = sqrt(6.674e-11 x 5.972e24 / 6,779,000) = 7,661 m/s or about 27,580 km/h. The key insight is that orbital speed depends only on altitude and the central body's mass, not on the satellite's own mass.
Why do satellites in lower orbits travel faster than those in higher orbits?
Satellites in lower orbits travel faster because they are closer to the central body and experience stronger gravitational pull. Since orbital speed equals sqrt(GM/r), a smaller orbital radius r produces a larger speed. This seems counterintuitive at first because objects in higher orbits have more energy, but that energy goes into gravitational potential energy rather than kinetic energy. A satellite at 200 km altitude orbits at about 7,790 m/s, while one at 35,786 km (geostationary orbit) moves at only 3,075 m/s. This relationship was first described by Johannes Kepler in his Third Law of Planetary Motion, which states that the square of the orbital period is proportional to the cube of the orbital radius.
What is geostationary orbit and why is it important?
Geostationary orbit (GEO) is a specific circular orbit at approximately 35,786 km above Earth's equator where a satellite's orbital period exactly matches Earth's rotation period of 23 hours 56 minutes and 4 seconds. At this altitude, the satellite appears to hover motionlessly above a fixed point on the equator when viewed from the ground. This makes GEO invaluable for communications satellites, weather monitoring, and direct broadcast services because ground antennas can point at a fixed location in the sky without tracking. The concept was popularized by science fiction author Arthur C. Clarke in 1945. GEO satellites travel at approximately 3,075 m/s (11,070 km/h). The orbit must be equatorial and prograde for true geostationary positioning.
What is escape velocity and how does it relate to orbital speed?
Escape velocity is the minimum speed an object must reach to break free from a celestial body's gravitational pull without further propulsion. It is calculated as v_esc = sqrt(2GM/r), which is exactly sqrt(2) (approximately 1.414) times the circular orbital speed at the same altitude. For Earth's surface, escape velocity is about 11,186 m/s (40,270 km/h), while orbital speed at the surface would theoretically be 7,910 m/s. At the ISS altitude of 408 km, escape velocity drops to about 10,834 m/s. This means a satellite already in orbit needs to increase its speed by only about 41.4% to escape the planet entirely. This mathematical relationship holds true for any altitude around any gravitationally bound body.
How do different orbital altitudes serve different purposes?
Satellite orbits are categorized by altitude, each serving distinct purposes. Low Earth Orbit (LEO, 160-2000 km) is used by the ISS, Earth observation satellites, and mega-constellations like Starlink. LEO offers low latency (useful for communications) and high-resolution imaging but requires many satellites for global coverage. Medium Earth Orbit (MEO, 2000-35786 km) hosts navigation constellations like GPS at 20,200 km, balancing coverage area with signal strength. Geostationary Orbit (GEO, 35,786 km) serves telecommunications and weather monitoring with fixed-point coverage. Highly Elliptical Orbits (HEO) like Molniya orbits provide extended coverage over high-latitude regions. Sun-synchronous orbits cross the equator at the same local solar time, ideal for consistent lighting conditions in Earth imaging.
When should I use the vis-viva equation instead of v = sqrt(GM/r)?
Use v = sqrt(GM/r) only when the orbit is genuinely circular, so that radius and semi-major axis are the same number. For any elliptical orbit the speed changes continuously and you need the full vis-viva relation, v squared = GM times (2/r minus 1/a), where r is the instantaneous radius and a is the semi-major axis. That form gives the fast perigee speed and the slow apogee speed of a transfer or Molniya orbit, and it reduces to the circular case exactly when r equals a.
What does the specific orbital energy tell you about an orbit?
Specific orbital energy is the total mechanical energy per kilogram of spacecraft, equal to minus GM divided by twice the semi-major axis. Because the satellite mass cancels, it characterises the orbit itself rather than the vehicle. A negative value means the orbit is gravitationally bound, zero corresponds to a parabolic escape trajectory at exactly escape velocity, and a positive value means a hyperbolic departure with leftover speed at infinity. Raising an orbit means making this number less negative, which is what every prograde burn accomplishes.
Why does atmospheric drag make a decaying satellite speed up rather than slow down?
Drag removes mechanical energy, which lowers the semi-major axis. But circular speed goes as one over the square root of the radius, so a smaller orbit is a faster orbit. The lost energy comes out of the much larger gravitational potential term while the kinetic term actually grows, a result often called the drag paradox. Below roughly 500 kilometres the effect is fast enough to matter operationally, which is why the ISS needs periodic reboost burns to hold its altitude near 400 kilometres.
How much velocity change does it take to move a satellite from low Earth orbit to geostationary orbit?
The standard route is a Hohmann transfer: one prograde burn at low orbit raises apogee to geostationary radius, then a second burn at apogee circularises there. From a typical 200 kilometre parking orbit the first burn is roughly 2.4 kilometres per second and the second roughly 1.5, for a total near 3.9 before any plane change. Launching from a low-latitude site matters because the inclination change to reach the equatorial plane is added at apogee, where the orbital speed is lowest and the manoeuvre is cheapest.
What makes a sun-synchronous orbit possible, and why is it retrograde?
Earth is not a perfect sphere; its equatorial bulge, described by the J2 term of about 1.0826 times ten to the minus three, exerts a torque that precesses the orbital plane. Choosing altitude and inclination together can tune that precession to 0.9856 degrees per day, matching Earth's mean motion around the Sun so the orbital plane keeps a fixed angle to sunlight. Achieving eastward precession requires an inclination greater than 90 degrees, so these orbits are slightly retrograde, typically near 98 degrees for altitudes around 700 to 800 kilometres.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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