ABC Tools

Escape and Orbital Velocity, Explained

After reading this you can compute how fast something must travel to circle a planet at a given altitude, how fast it must go to leave that planet entirely, and how long one orbit takes, all from the central body's mass and the distance to its centre.

What these speeds mean and one quick hook

Two speeds define motion around any massive body. The circular orbital velocity is the speed at which an object falls toward the body exactly as fast as the surface curves away beneath it, so it never gets closer or farther. The escape velocity is the speed that lets an object coast to infinite distance and arrive with zero speed to spare, never falling back.

Here is the hook. The International Space Station orbits about 400 km above Earth. It moves at roughly 7.67 km/s, which is about 27,600 km/h. To break free of Earth entirely from that same altitude it would need about 10.84 km/s. That larger number is exactly \sqrt{2} times the first. This article shows why, and how to reproduce both figures.

When this model applies and when it does not

The formulas here assume a single dominant central mass, a spherically symmetric body, and no drag or thrust. They describe a two-body problem: one large mass and one much smaller test object whose own mass you ignore.

They work well for a satellite around Earth, a moon around a planet, or a planet around the Sun. They break down in several concrete cases. Low orbits below about 200 km on Earth lose energy to atmospheric drag, so no stable circular orbit exists there for long. Near two comparable masses (a close binary star) the single-centre assumption fails. And escape velocity ignores the pull of every other body, so a spacecraft that reaches Earth escape speed is still bound to the Sun.

Escape velocity is a speed, not a direction, and it assumes no further thrust. A rocket does not need to reach 11.2 km/s at the launch pad. It can climb slowly under continuous thrust. The escape figure only tells you the speed a purely ballistic (unpowered) object would need at that radius.

The formulas and the intuition behind them

Both speeds come from Newton's law of gravitation and conservation of energy. Let G = 6.674 \times 10^{-11} in SI units, let M be the central mass in kilograms, and let r be the distance from the centre of the body in metres.

For a circular orbit, gravity supplies exactly the centripetal force. Setting \frac{GMm}{r^2} = \frac{mv^2}{r} and cancelling the small mass m gives the orbital speed:

v_{\text{orb}} = \sqrt{\frac{GM}{r}}

Here v_{\text{orb}} is the circular orbital speed in metres per second. Notice the object's own mass dropped out. A bolt and a bus at the same altitude orbit at the same speed.

For escape, set the total energy to zero. Kinetic energy \tfrac{1}{2}mv^2 must equal the depth of the gravitational well, \frac{GMm}{r}. Solving for speed:

v_{\text{esc}} = \sqrt{\frac{2GM}{r}}

The factor of 2 under the root is the entire difference. That is why v_{\text{esc}} = \sqrt{2} \, v_{\text{orb}} at the same radius. Reaching escape speed costs exactly twice the kinetic energy of a circular orbit at that height.

The orbital period follows from speed and circumference. One lap covers 2\pi r at speed v_{\text{orb}}:

T = \frac{2\pi r}{v_{\text{orb}}} = 2\pi \sqrt{\frac{r^3}{GM}}

The right-hand form is Kepler's third law. Period grows with r^{3/2}, so doubling the orbit radius multiplies the period by 2^{1.5} \approx 2.83.

A worked example: the 400 km low Earth orbit

Reproducing the demo numbers

Use the tool defaults: Earth as the central body, mass M = 5.972 \times 10^{24} kg, and altitude 400 km. Earth's mean radius is about 6371 km, so the orbit radius from the centre is r = 6.371 \times 10^{6} + 4.0 \times 10^{5} = 6.771 \times 10^{6} m.

  1. Compute GM = 6.674 \times 10^{-11} \times 5.972 \times 10^{24} = 3.986 \times 10^{14} m³/s².
  2. Divide by radius: \frac{GM}{r} = \frac{3.986 \times 10^{14}}{6.771 \times 10^{6}} = 5.887 \times 10^{7} m²/s².
  3. Orbital speed: v_{\text{orb}} = \sqrt{5.887 \times 10^{7}} = 7673 m/s, about 7.67 km/s.
  4. Escape speed: multiply by \sqrt{2} to get 7673 \times 1.4142 = 10850 m/s, about 10.85 km/s.
  5. Period: T = \frac{2\pi \times 6.771 \times 10^{6}}{7673} = 5544 s, which is 92.4 minutes.

The ISS really does circle Earth in roughly 92 minutes, so the model matches reality to within its rounding.

As altitude rises from 200 km to 40,000 km above Earth, orbital speed falls from about 7.79 km/s to 3.07 km/s, while the period stretches from 88 minutes to nearly 24 hours. At roughly 35,786 km altitude the period equals one sidereal day, which is the geostationary orbit.

Reading and interpreting the results

Three numbers come out of the calculation. Read them together.

The orbital speed tells you how fast a stable circular satellite moves at that height. Lower orbits are faster: a satellite at 200 km moves at about 7.79 km/s, while one at 20,000 km moves at only about 3.89 km/s. Slow satellites are high satellites.

The escape speed is a threshold. Anything moving faster than v_{\text{esc}} at that radius, in any outward direction, will not return under gravity alone. Note that escape speed also decreases with radius, because you start higher in the well and have less depth left to climb.

The period is your clock. Compare it against a day to judge coverage. The table below shows how the three quantities vary with altitude above Earth.

Orbital quantities above Earth at four altitudes
Altitude (km)Radius (km)Orbit speed (km/s)Escape speed (km/s)Period
40067717.67310.8592.4 min
200083716.8989.755127.2 min
20000263713.8875.497710.6 min
35786421573.0754.34823.93 h
Orbital speed falls smoothly with altitude, from about 7.91 km/s at the surface to 3.01 km/s at 40,000 km. The marker sits at the 400 km ISS altitude.

Common mistakes

The errors below account for most wrong answers.

Using altitude instead of radius from the centre
The formulas need r measured from the body's centre, not the surface. For a 400 km Earth orbit you must add Earth's radius: 6371 + 400 = 6771 km. Forgetting this makes the speed far too large.
Mixing units
With G in SI units, mass must be in kilograms and radius in metres. The result is metres per second. Feeding kilometres directly gives a number off by a factor of about 31.6, which is \sqrt{1000}.
Confusing the two speeds
Escape speed is always \sqrt{2} \approx 1.414 times the circular speed at the same radius, never twice. If your escape figure is double the orbital figure, you doubled the speed instead of the energy.
Expecting the object's mass to matter
It does not. Both speeds depend only on the central mass and the radius. A 1 kg cube and a 400,000 kg station share the same 7.67 km/s orbit at 400 km.

A fast sanity check for Earth: escape speed from the surface is close to 11.2 km/s. If your surface escape result is far from that, recheck your radius and units before trusting the rest.

Related tools on this site

These calculators cover neighbouring parts of the same physics. The Newton's Law of Gravitation Calculator gives the raw force F = \frac{GMm}{r^2} that these speed formulas are built from. The Weight on Other Planets Calculator shows how the same GM/r^2 surface gravity changes your weight from body to body.

For distances rather than speeds, the Astronomical Distance Converter handles light-years, parsecs and AU. And if you want to observe the satellites and planets these orbits describe, the Telescope Calculator covers magnification and resolution.

Frequently asked questions

Why is escape velocity exactly root two times orbital velocity?

Circular orbit needs kinetic energy equal to half the well depth, while escape needs kinetic energy equal to the full well depth. Since kinetic energy scales with speed squared, doubling the energy multiplies the speed by \sqrt{2} \approx 1.414. At 400 km, 7.673 \times 1.414 = 10.85 km/s.

Does escape velocity depend on which way you launch?

No, not the speed itself. Escape velocity is derived from energy alone, which does not care about direction. A ballistic object reaching 11.2 km/s from Earth's surface escapes whether it goes straight up or at an angle, ignoring the atmosphere. Direction does affect the trajectory shape and how much you gain from Earth's rotation.

How high is a geostationary orbit?

Set the period equal to one sidereal day, 86,164 s, and solve r = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3}. That gives r \approx 42{,}157 km from the centre, so about 35,786 km above the surface, where the orbital speed is about 3.07 km/s.

Can I use this for orbits around the Sun or another star?

Yes, as long as one mass dominates. Set the central mass to the Sun's 1.989 \times 10^{30} kg and the radius to the orbit distance. At Earth's distance of 1.496 \times 10^{11} m the formula gives about 29.8 km/s, which matches Earth's actual orbital speed.

Why does a lower orbit mean a faster satellite?

Deeper in the gravity well, gravity is stronger, so a faster speed is needed to balance it. Since v_{\text{orb}} = \sqrt{GM/r}, speed falls as the square root of radius. Going from 400 km to 20,000 km altitude cuts speed from 7.67 to 3.89 km/s.

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