Escape Velocity and Orbital Velocity Calculator
Escape Velocity and Orbital Velocity Calculator
Pick a body and an altitude and you get the escape speed, the circular orbital speed, the orbital period and the field strength at that height — with the √2 that is always between the first two. Built on GM rather than on a mass, because GM is the quantity that is actually known.
Escape speed, orbital speed and period
Earth at the IAU nominal values, 400 km up — roughly the Space Station
√(2GM/r) and √(GM/r), and why the ratio is always √2
- GM
- the standard gravitational parameter, in m³/s². The quantity that is actually measured: watch anything orbit a body and GM comes straight out of the period and the radius, to nine or ten significant figures. The mass M requires dividing by G, which is known to only six, so M is always the less certain number. Every option on this page is therefore a GM and a radius, never a mass.
- r
- the distance from the centre of the body, not the height above its surface. This is the most common error on the topic: at 400 km altitude the Earth’s r is 6778 km and not 400 km. The page adds the altitude to the radius for you, and prints the r it used.
- the escape speed
- from setting the total energy to zero: ½mv² = GMm/r, so v = √(2GM/r). The mass of the escaping object cancels out, which is why there is only one escape speed per radius and a feather and a satellite share it. It is a speed and not a velocity — the direction does not matter at all, as long as the ground is not in the way — which is why the CBSE syllabus calls it escape speed.
- the orbital speed
- from equating the gravitational force with the centripetal force needed: GMm/r² = mv²/r gives v = √(GM/r). Also mass-independent. Put it together with the circumference and you have T = 2π√(r³/GM), which is Kepler’s third law with the constant written out.
- √2
- the ratio of the two, 1.414 213 56, and the same for every body in the universe. It has no physics in it: both speeds have the same GM/r under the root and one has an extra factor of 2. The useful consequence is that a satellite in a circular orbit is at 70.7 per cent of the escape speed for its altitude, so raising its speed by 41.4 per cent sends it away for good.
Worked example
Earth at the IAU nominal values, 400 km up — roughly the Space Station
THE RADIUS FROM THE CENTRE, WHICH IS WHERE THE MARKS ARE LOST. The IAU nominal equatorial radius is 6378.1 km, so at 400 km altitude r = 6378.1 + 400 = 6778.1 km = 6.778 1 × 106 m. Using 400 km as r gives an escape speed of 44.6 km/s, which is four times too big and is the standard wrong answer on this question.
THE ESCAPE SPEED. v = √(2GM/r) = √(2 × 3.986 004 × 1014 / 6.778 1 × 106) = √(1.176 14 × 108) = 10 845.0 m/s, which is 10.845 km/s. Note that it is lower than the 11.180 km/s from the surface: you are further out, so there is less to climb out of.
THE ORBITAL SPEED, FROM THE SAME GM AND r. v = √(GM/r) = √(5.880 7 × 107) = 7668.6 m/s, or 7.669 km/s — the familiar “about 7.7 kilometres a second” for low Earth orbit. Divide the two speeds and you get 1.414 213 562, which is √2 to ten figures and would be √2 for any body at any radius.
THE PERIOD. T = 2πr/v = 2π × 6.778 1 × 106 / 7668.6 = 5553.6 s = 92.56 minutes, so about 15.5 orbits a sidereal day. Equivalently T = 2π√(r³/GM), which is the Kepler form and gives the same number.
AND THE FIELD STRENGTH UP THERE, WHICH SURPRISES PEOPLE. g = GM/r² = 3.986 004 × 1014 / (6.778 1 × 106)² = 8.676 m/s², which is 88.5 per cent of the surface value. Gravity at the Space Station is almost as strong as on the ground. Astronauts float because they are in free fall along with the station, not because gravity has run out — which is why “microgravity” is the better word than “zero gravity”.
Escape speed, orbital speed and period at altitudes that get asked about
| Altitude above the Earth | r from the centre | Escape speed | Orbital speed | Period | g at that height |
|---|---|---|---|---|---|
| 0 km — the surface | 6378.1 km | 11.180 km/s | 7.905 km/s | 84.49 min | 9.7984 m/s² |
| 400 km — about the ISS | 6778.1 km | 10.845 km/s | 7.669 km/s | 92.56 min | 8.6760 m/s² |
| 800 km — a polar imaging orbit | 7178.1 km | 10.538 km/s | 7.452 km/s | 100.86 min | 7.7363 m/s² |
| 2000 km | 8378.1 km | 9.754 km/s | 6.897 km/s | 127.20 min | 5.6786 m/s² |
| 20 200 km — a GPS orbit | 26 578.1 km | 5.479 km/s | 3.874 km/s | 11.967 h | 0.5643 m/s² |
| 35 786 km — geostationary | 42 164.1 km | 4.348 km/s | 3.075 km/s | 23.934 h | 0.2242 m/s² |
| 378 022 km — about the Moon’s distance | 384 400 km | 1.440 km/s | 1.018 km/s | 27.45 d | 0.0027 m/s² |
Why the escape speed gets smaller as you go up, and why it is a speed
Escape speed is √(2GM/r), and the r is measured from the centre of the body. That single point accounts for most of the wrong answers on this topic: at 400 km altitude the Earth’s r is 6778 km, not 400. Once that is right, the formula follows from one line of energy bookkeeping. To escape, a body must arrive at infinity with nothing left over, so its kinetic energy at the start must exactly match the depth of the gravitational well it is in: ½mv² = GMm/r. The m cancels, which is why there is one escape speed per radius rather than one per object, and a feather and a spacecraft need exactly the same number.
The counter-intuitive consequence is that the escape speed falls as you climb — 11.18 km/s from the surface, 10.85 from the Space Station, 4.35 from geostationary orbit. That is not a paradox: escape speed is the cost of climbing out from where you already are, and 36 000 km up you have paid most of it.
It is a speed and not a velocity, and the CBSE syllabus is right to call it escape speed. The direction makes no difference at all, as long as you do not hit the ground: fire a projectile at 11.2 km/s straight up, at forty-five degrees, or horizontally, and in all three cases it never returns. The reason is that the derivation used energy, and energy has no direction in it. What the number is not is a launch requirement. No rocket is fired at 11.2 km/s from the pad. A rocket burns continuously, so it only needs to be moving fast enough at each height to keep climbing, and the escape speed it is working against gets lower the whole way. The ballistic version is impossible for a quite separate reason: anything moving at 11.2 km/s in the lower atmosphere is destroyed by air resistance.
The orbital speed is the other half of the page, and the √2 between them is worth more than either. A circular orbit needs gravity to supply exactly the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). Put that beside the escape speed and the ratio is √2 = 1.414 2, identically, for every body at every radius in the universe, because the two expressions differ only by a factor of two under the square root. So anything in a circular orbit is already at 70.7 per cent of its local escape speed, and a 41.4 per cent increase in speed is enough to leave for good. Combine the orbital speed with the circumference and you get T = 2π√(r³/GM), which is Kepler’s third law with its constant spelled out — and the route by which GM itself is measured, since a period and a radius give it directly.
Why this page uses GM and never a mass. GM is measured from the motion of satellites and is known to nine or ten significant figures; the IAU’s nominal terrestrial value is 3.986 004 × 1014 m³/s² and the WGS 84 defining parameter is 3.986 004 418 × 1014. The mass of the Earth is not measured — it is GM divided by G, and G is known to only about six figures, so the mass inherits an uncertainty two or three orders of magnitude worse than the product it came from. The IAU resolution behind the numbers here makes the point itself: it recommends no value for G, and asks instead that any publication needing a mass state which estimate of G it used. The gravitation and field strength page works the other way round, from a mass and a G, and prints what that choice costs.
Frequently asked questions
Why is the escape velocity 11.2 km/s when no rocket goes that fast at launch?
Because escape speed is the speed an unpowered projectile would need at that radius to never come back, and a rocket is not unpowered. A rocket thrusts continuously, so at every height it only needs to be moving fast enough to keep climbing, and the escape speed it is working against falls as it rises — 11.18 km/s at the ground, 10.85 at 400 km, 4.35 at geostationary. A launch to low Earth orbit therefore needs about 7.7 km/s of orbital speed plus a few more to fight gravity and air resistance, and not 11.2. The ballistic version is impossible anyway: anything travelling at 11.2 km/s through the lower atmosphere would be destroyed by drag.
Is escape velocity a speed or a velocity?
Physically it is a speed: the direction does not matter. The derivation is an energy balance and energy has no direction, so a projectile fired at the escape speed straight up, at forty-five degrees or horizontally all escape equally — provided nothing is in the way. The CBSE syllabus calls it “escape speed” for precisely this reason and the name is the better one. The phrase “escape velocity” is universal in practice and is what people search for, so both are used here, but if a question asks you whether direction matters, the answer is no.
Why is the escape speed exactly √2 times the orbital speed?
Because the two formulas differ only by a factor of two under the square root: vesc = √(2GM/r) and vorb = √(GM/r). The ratio is √2 = 1.414 213 6 and it contains no physical constants at all, so it is the same for the Earth, for Jupiter, for the Sun and for any body that has ever existed, at every radius. The useful form of the fact is that anything in a circular orbit is already at 70.7 per cent of the local escape speed, so increasing its speed by 41.4 per cent is enough to leave permanently — and that this page prints the ratio as a row, which must read 1.414 213 562 whatever you type.
Does the mass of the rocket or satellite affect the answer?
No. The mass of the escaping or orbiting object cancels out of both derivations, because gravity’s pull is proportional to that mass and so is the inertia resisting it. One escape speed per radius, one orbital speed per radius, whatever you put there — a feather, a satellite, a space station. What the mass does affect is the energy and therefore the fuel: the speed is the same, and the number of joules needed to reach it is proportional to the mass, which is why launch costs scale with payload and not with the destination’s escape speed alone.
Why do astronauts float if gravity at the Space Station is still 8.7 m/s²?
Because they are in free fall, and so is everything around them. At 400 km the field strength is 8.676 m/s², which is 88.5 per cent of the surface value — gravity has barely weakened. The station and everybody in it are accelerating towards the Earth at that rate continuously; they are also moving sideways at 7.67 km/s, fast enough that the ground curves away as fast as they fall towards it. Nothing is pressing them against a floor, so there is no sensation of weight. “Microgravity” is the honest word; “zero gravity” describes the feeling and not the field. It is the same reason you feel light in a falling lift.
Related calculators
References
- IAU 2015 Resolution B3 on recommended nominal conversion constants for selected solar and planetary properties, XXIXth General Assembly of the International Astronomical Union (read 11 November 2026). Four numbers on this site come from its table: the nominal terrestrial equatorial radius 6.3781 × 106 m, the nominal terrestrial polar radius 6.3568 × 106 m, the nominal terrestrial mass parameter (GM) 3.986 004 × 1014 m3 s−2, and the jovian and solar pairs. Two things about it are worth a student’s attention. The values are conversion factors, not measurements — they are fixed numbers close to the best estimates, so that everyone converting to SI gets the same answer. And the resolution gives no value for G: it notes that G is among the least precisely determined constants and asks that a publication needing a mass in kilograms state which estimate of G it used. That is exactly why this site tabulates GM and not M.
- World Geodetic System 1984 — the defining parameters of WGS 84, read 11 November 2026 from the UNOOSA International Committee on GNSS reference copy, which cites NIMA Technical Report 8350.2 and identifies itself as WGS 84 (G1674) at reference epoch 2005.0. The four defining parameters are the semi-major axis a = 6 378 137.0 m, the reciprocal flattening 1/f = 298.257 223 563, the geocentric gravitational constant GM = 3.986 004 418 × 1014 m3 s−2 with the atmosphere included, and the angular velocity ω = 7.292 115 × 10−5 rad s−1. Length-checked rather than trusted: the superscripts in that copy did not survive text extraction, so the exponents above were reconstructed from the parameter definitions and the document is cited for the mantissas it printed. It gives no value for the Earth’s mass and no normal-gravity coefficients, so neither appears on these pages.
- PHYSICS (042), Class XI-XII, Central Board of Secondary Education senior secondary curriculum document for the session 2025-26 (cbseacademic.nic.in, read 11 November 2026). Cited for what the syllabus asks for, never as any kind of endorsement. Unit VI, Chapter 7 Gravitation, asks for “Gravitational potential energy and gravitational potential, escape speed, orbital velocity of a satellite, energy of an orbiting satellite”. The syllabus writes escape speed and orbital velocity in the same line, which is a distinction worth keeping: the escape figure genuinely is direction-independent and the orbital one is not.
- Cambridge International AS & A Level Physics 9702, syllabus for use in 2025, 2026 and 2027, Version 1 (cambridgeinternational.org, read 11 November 2026). Quoted by outcome number so that the difference from the Indian syllabus is checkable rather than asserted. Topic 13 covers the orbital half and not the escape half. 13.2.3 asks the candidate to “analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it causes”, which is the derivation of √(GM/r) used here, and 13.2.4 covers the geostationary orbit, specifying “an orbital period of 24 hours, orbiting from west to east, directly above the Equator”. The word “escape” does not appear in the syllabus at all — checked by searching the document, not assumed. Note also the 24 hours: the exact geostationary radius follows from the sidereal day of 86 164.09 s, and the 50 km difference between the two is below the level the outcome is written at.