Circular orbits
For a spherical central body with gravitational parameter μ, a circular orbit of radius r has
v = √(μ / r), T = 2π √(r³ / μ), vesc = √(2μ / r).
Earth’s μ is 3.986004418×10¹⁴ m³/s² (IERS / Vallado). At 400 km above the mean radius 6,371 km, v ≈ 7.669 km/s and the period is about 92.6 minutes. Geosynchronous radius is the circular radius whose period is one sidereal day (86,164.0905 s), about 42,164 km from Earth’s center, at about 3.075 km/s.
Vis-viva on an ellipse
The speed at radius r on an ellipse of semi-major axis a is v = √( μ (2/r − 1/a) ). Periapsis is the faster end. A 400 × 800 km Earth ellipse (mean radius) is a simple check of that identity on this page.
Sources
Vallado, Fundamentals of Astrodynamics and Applications; Curtis, Orbital Mechanics for Engineering Students; Bate, Mueller and White, Fundamentals of Astrodynamics. For the two-impulse transfer between circular orbits, see the Hohmann transfer calculator.
Frequently asked questions
Should I use the mean or the equatorial Earth radius?
For a quick LEO estimate either is fine (the difference is about 7 km). WGS-84 equatorial 6,378.137 km is the usual choice when you are quoting a geodetic altitude; the mean radius 6,371 km is common in first-cut orbital-mechanics problems. This page lets you pick.
What is a sidereal day doing in the GEO figure?
A geostationary orbit must match Earth’s rotation relative to the stars, not the 24-hour solar day. The sidereal day is 86,164.0905 s, which fixes the GEO radius for a given μ.
Is escape speed the speed to leave the solar system?
No. The escape speed on this page is escape from the selected central body at that radius, as a two-body problem. Leaving the solar system from Earth orbit also has to account for the Sun.