The orbital velocity calculator above gives the speed a satellite must travel to hold a circular orbit at a given height, along with the period that speed produces. It also runs the other way: give it the period you want and it returns the altitude that delivers it, which is how the geostationary altitude is derived rather than looked up.
Arb Digital builds free calculators that answer the question actually being asked. Orbital speed is not one number for a body; it is a curve against altitude, and the two ends of that curve behave in ways that surprise people the first time they see them.
What This Orbital Velocity Calculator Does
You pick a central body, or type in a mass and radius of your own, and either an altitude or a target period. It returns the circular orbital speed, the period, the number of complete orbits in a 24-hour day, the orbital radius measured from the centre of the body, and the escape speed at that same radius.
That last figure is there for calibration. Circular orbital speed is always escape speed divided by the square root of two, about 71 per cent of it, at every altitude around every body. The gap between the two is the margin between staying and leaving, and it is smaller than most people expect: gain 41 per cent on your orbital speed and you are on an escape trajectory.
Everything here assumes a circular orbit. Real orbits are ellipses, and an elliptical orbit has a speed that varies continuously around the path — fastest at closest approach, slowest at the far point. A circular orbit is the special case where those two coincide.
How to Use It
- Choose the central body first. The satellite's own mass never enters the calculation, so the only thing that matters about the two objects is which one you are orbiting.
- Decide whether you are working from an altitude or a period. Mission planning usually starts from an altitude; communications and navigation work usually starts from a period, because the orbit has to match a ground pattern.
- Enter altitude above the surface, not distance from the centre. The tool adds the body's radius for you and shows the result so you can check it.
- Look at the orbits-per-day figure. It is the fastest sanity check available. A low Earth orbit gives about 15 or 16, a geostationary orbit gives exactly one, and anything wildly outside that range points to an input error.
The Formula: How Orbital Speed Is Calculated
A circular orbit is the case where gravity supplies exactly the centripetal acceleration needed to bend the path into a circle. Setting the gravitational force GMm ÷ r2 equal to the centripetal requirement mv2 ÷ r lets the satellite's mass m cancel from both sides, leaving v = √(GM ÷ r).
The period follows from the circumference divided by the speed: T = 2πr ÷ v = 2π√(r3 ÷ GM). Turning that round to solve for radius from a chosen period gives r = (GM T2 ÷ 4π2)1/3, which is what the period mode uses.
Work the default. Earth's mass is 5.97217 × 1024 kg and its mean radius 6,371.0084 km, from NASA JPL's planetary physical parameters table. At 400 km altitude the orbital radius is 6,771.008 km. With G = 6.6743 × 10−11, GM comes to 3.9860 × 1014, so v = √(3.9860 × 1014 ÷ 6.771008 × 106) = 7,673 m/s, or 7.673 km/s. The period is 2π × 6.771008 × 106 ÷ 7,673 = 5,545 s, which is 92.4 minutes — about 15.6 orbits a day.
Now switch to period mode and ask for 1,436.07 minutes, one sidereal day. The radius comes out at 42,164 km from the centre, which is 35,793 km above the surface, and the speed is 3.075 km/s. That is the geostationary orbit, and NASA's catalog of Earth satellite orbits gives the same 42,164 km figure.
Higher Orbits Are Slower, and Take Longer Anyway
Speed falls as the inverse square root of radius, so doubling the orbital radius drops the speed by about 29 per cent. Period rises as the three-halves power of radius, so doubling the radius raises the period by 183 per cent. Both come from the same equation and they pull in opposite directions, which is why the relationship between altitude and period is not intuitive until you have seen the numbers.
The practical consequence runs through spaceflight. To catch a spacecraft ahead of you in the same orbit, you do not speed up — that raises your orbit and makes you slower and later. You slow down, drop to a lower and faster orbit, gain ground, and then raise back up. Every rendezvous works this way and it is thoroughly counter-intuitive.
It also explains the geostationary orbit's altitude. There is exactly one radius at which the period equals a sidereal day, and it is far out, 36,000 km above the surface, precisely because a period that long demands a slow orbit and a slow orbit demands a large radius.
Why the Satellite's Own Mass Does Not Appear
A communications satellite of several tonnes and a bolt shed from it during deployment travel at the same speed in the same orbit. The mass cancels because gravity scales with it and the inertia resisting the turn scales with it identically.
This is the orbital version of the fact that all objects fall at the same rate, and it is why debris is such a persistent problem: a fragment left at a given altitude stays on the same track as the object it came from, at the same speed, for as long as drag allows.
The one caveat is that the equation assumes the orbiting mass is small compared with the central one. Strictly, both bodies orbit their common centre of mass, and for a genuine two-body system of comparable masses that correction matters. For a satellite around a planet it is utterly negligible. The Kepler's third law calculator is the tool for the general case, since it takes the central mass explicitly and works from a semi-major axis rather than an altitude.
Orbiting at the Surface, and Why Nothing Does
Set the altitude to zero and this calculator will give you a perfectly valid answer: 7.910 km/s around Earth, with a period of 84.4 minutes. That is the theoretical minimum orbital period for any object orbiting Earth, and no satellite has ever been anywhere near it.
The reason is atmosphere, not gravity. At the surface, and for a long way above it, air resistance removes energy far faster than any spacecraft could carry fuel to replace. Even at 400 km the residual atmosphere is thick enough that the space station must be reboosted periodically or it would re-enter within a couple of years.
This calculator models gravity alone. It has no view on drag, on radiation pressure, on the fact that Earth is not a perfect sphere, or on the gravitational tugs of the Moon and Sun. All of those matter for a real mission, and all of them are outside the scope of a two-body circular calculation. The drag force calculator covers the atmospheric side of the problem separately.
How This Differs From the Site's Other Orbital Tools
This page holds the orbit circular and works between altitude, speed and period around a chosen body. The escape velocity calculator answers a different question: the speed at which an object stops being bound at all, which is the threshold this page's orbits sit below. The Kepler's third law calculator relates period, semi-major axis and central mass for orbits of any shape, and is the right tool when you have a semi-major axis rather than an altitude, or when you want to weigh the central body from an observed orbit.
For the force itself at a given separation, use the gravitational force calculator. For the circular-motion side of the derivation, the centripetal force calculator gives the acceleration a curved path demands, and the angular velocity calculator converts between period, angular rate and tangential speed. To express the result in other units, use the speed converter.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Entering distance from the centre as an altitude — the geostationary orbit is 42,164 km from the centre but only 35,793 km above the surface, and confusing the two changes the answer substantially.
- Including the satellite's mass — it cancels out. If a calculation seems to need it, something else has gone wrong.
- Using a solar day for a geostationary orbit — the period must match Earth's rotation relative to the stars, 1,436.07 minutes, not the 1,440 minutes of a solar day. The four-minute difference moves the required altitude by several kilometres.
- Assuming faster means a shorter trip to something ahead — speeding up raises the orbit, which slows you down and lengthens the period. Orbital rendezvous works the opposite way round.
- Treating a circular result as an elliptical orbit — on an ellipse the speed changes continuously, and neither the fastest nor the slowest point equals the circular figure for the same average distance.
Related Free Tools From Arb Digital
Compare the orbit against the unbound case with the escape velocity calculator, and handle elliptical orbits or unknown central masses with the Kepler's third law calculator. Get the force at a separation with the gravitational force calculator and the turning requirement with the centripetal force calculator. Convert between rotation rate and tangential speed with the angular velocity calculator, express results in other units with the speed converter, and see what the atmosphere does to a low orbit with the drag force calculator. Everything Arb Digital publishes is listed at the free online tools hub.
Frequently Asked Questions
No. The satellite's mass cancels out of the equation, because gravity scales with mass and so does the inertia resisting the turn. A multi-tonne spacecraft and a loose bolt travel at exactly the same speed in the same orbit.
Speed falls as the inverse square root of the orbital radius, while the circumference grows in direct proportion to it. The path lengthens faster than the speed drops, so the period rises as the three-halves power of the radius.
By setting the period equal to one sidereal day, 1,436.07 minutes, and solving the period equation for radius. That gives 42,164 km from Earth's centre, or about 35,793 km above the surface, at a speed near 3.07 km per second.
Circular orbital speed is always escape speed divided by the square root of two, roughly 71 per cent of it, at every altitude around every body. Adding about 41 per cent to a circular orbit speed puts an object on an escape trajectory.
The equation gives a valid answer, about 7.91 km per second around Earth with an 84-minute period, but the atmosphere makes it impossible in practice. Drag would remove energy far faster than any vehicle could replace it.
No. It assumes a circular orbit, where speed is constant. On an ellipse the speed varies continuously, fastest at the closest point and slowest at the furthest, and the shape is handled by Kepler's laws rather than by this single-radius calculation.
The mean radius is the usual choice and it is what the presets here use. For a markedly flattened body such as Jupiter or Saturn the equatorial and polar radii differ considerably, and a precise orbit calculation has to account for that flattening rather than treating the body as a point.
This tool is provided for educational and study use. It models a circular two-body orbit under gravity alone and does not account for atmospheric drag, orbital perturbations, the oblateness of the central body or the gravity of any third body.
