Orbital Period and Velocity Calculator

Kepler's third law states that the orbital period squared is proportional to the semi-major axis cubed: T squared = (4 x pi squared / mu) x a cubed, giving T = 2 x pi x sqrt(a cubed / mu), where mu = G x M is the gravitational parameter. For the Sun, mu_Sun = 1.327e20 m cubed/s squared. With a in AU and T in years, this simplifies to T squared = a cubed. For Earth at a = 1 AU: T = 1 year. For Mars at a = 1.524 AU: T = sqrt(1.524 cubed) = sqrt(3.540) = 1.881 years.

The International Space Station orbits at about 400 km altitude, which gives an orbital radius of 6,778.1 km. Using v = sqrt(mu_Earth / r) = sqrt(3.986e14 / 6,778,100) = 7,668 m/s = 7.668 km/s. The orbital period is T = 2 x pi x sqrt(6,778,100 cubed / 3.986e14) = 5,551 s = 92.5 minutes. The ISS completes about 15.5 orbits per day. At this speed, it experiences a sunrise and sunset roughly every 45 minutes.

Geostationary orbit (GEO) is the altitude at which the orbital period equals one sidereal day (86,164 s). Setting T = 86,164 s in T = 2 x pi x sqrt(r cubed / mu): r = (T squared x mu / (4 x pi squared)) to the one-third power = 42,164 km from Earth's center = 35,786 km altitude. Orbital speed at GEO = sqrt(3.986e14 / 42,164,000) = 3,075 m/s. Satellites at GEO appear stationary from Earth's surface and are used for communications, weather, and direct broadcast television.

Orbital period scales as r to the 3/2 power: T proportional to r to the power 1.5. Doubling the orbital radius multiplies the period by 2 to the 1.5 = 2.83. For Earth: LEO at 400 km has T = 92.5 min; GEO at 35,786 km has T = 1,436 min (24 hr). The period increases much faster than altitude because both the circumference and the required lower speed both increase together. This is why satellites at higher orbits are slower, cover less ground per orbit, and take longer to complete each orbit.

Earth orbits the Sun at a mean semi-major axis of 1.000 AU = 149.6 million km. Mean orbital speed = 2 x pi x 1.496e11 / (365.25 x 86400) = 29,785 m/s = 29.785 km/s. This speed varies between 30.29 km/s at perihelion (closest approach to Sun in January) and 29.29 km/s at aphelion (July). The period is exactly 1.000 tropical year by definition of the astronomical unit.

Mars has a semi-major axis of 1.5237 AU. By Kepler's third law: T squared = 1.5237 cubed = 3.540, so T = 1.881 years = 686.97 Earth days = 668.60 Martian sols. The mean orbital speed is about 24.08 km/s. A mission to Mars launched on a Hohmann transfer takes about 8.5 months (258 days) and must wait about 26 months for the next launch window (the synodic period, when Earth and Mars realign for departure).

The Moon's mean orbital semi-major axis around Earth is 384,400 km. Using v_orb = sqrt(mu_Earth / r) = sqrt(3.986e14 / 384,400,000) = 1,018 m/s = 1.018 km/s. The Moon's orbital period is T = 2 x pi x sqrt((3.844e8) cubed / 3.986e14) = 2.36e6 s = 27.32 days (sidereal month). This is also the calculation used in transfer orbit planning: a spacecraft must arrive at lunar distance with a speed matching the Moon's orbital velocity (plus or minus the hyperbolic capture burn).

Orbital speed at a given altitude scales as sqrt(mu), so heavier bodies require higher orbital speeds. At 400 km altitude: Earth v_orb = 7.668 km/s, Mars v_orb = 3.448 km/s, Moon v_orb = 1.612 km/s. Jupiter at 400 km altitude (above cloud tops): v_orb = 43.0 km/s. These speeds determine propellant requirements for orbit insertion burns: arriving at Jupiter and braking into a 400 km orbit requires decelerating from entry speed to 43 km/s, which demands enormous propellant mass or aerobraking.

No, in the two-body approximation where the satellite mass is negligible compared to the central body. The orbital period T = 2 x pi x sqrt(r cubed / mu) depends only on the orbital radius and the central body's gravitational parameter. This is why all satellites at the same altitude have exactly the same orbital period regardless of whether they weigh 1 gram or 10 tonnes. This property (period independence from satellite mass) is one of the most elegant consequences of Newton's law of gravitation.

At any altitude r, v_esc = sqrt(2) x v_orb, always. This is derived from their formulas: v_orb = sqrt(mu/r), v_esc = sqrt(2 x mu/r) = sqrt(2) x v_orb. For Earth at LEO: v_orb = 7.784 km/s, v_esc = 11.012 km/s, ratio = 1.4142. This means a spacecraft already in circular orbit needs only a 41.4% increase in speed to escape the gravitational field, which corresponds to a delta-V of about 3.23 km/s from LEO for Earth, or 1.43 km/s from low lunar orbit for the Moon.

Yes, but you must use the correct gravitational parameter. For Earth, mu = 3.986e14 m cubed per s squared. The formula T = 2 x pi x sqrt(a cubed / mu) gives the period in seconds for a in metres. The simplified T-squared = a-cubed version (with T in years and a in AU) only applies to heliocentric orbits around the Sun. For Earth-orbiting satellites, use T = 2 x pi x sqrt(r cubed / mu_Earth) where r is in metres. For Moon-orbiting objects, use mu_Moon = 4.905e12 m cubed per s squared.

Rearrange Kepler's third law: a = (mu x T squared / (4 x pi squared)) to the one-third power. For T in seconds and mu in m cubed/s squared, this gives orbital radius a in metres. Subtract the body radius to get altitude. For a 2-hour Earth orbit (T = 7200 s): a = (3.986e14 x 7200 squared / (4 x pi squared)) to the 1/3 = (2.093e21) to the 1/3 = 12,790 km, altitude = 12,790 minus 6,378 = 6,412 km. This is in the middle of the Van Allen radiation belt, so few satellites are placed there.

Jupiter has a semi-major axis of 5.2029 AU. By Kepler's third law: T squared = 5.2029 cubed = 140.83, so T = sqrt(140.83) = 11.87 years. This matches the observed Jupiter orbital period of 11.862 years closely. Jupiter's mean orbital speed is v = 2 x pi x 5.2029 x 1.496e11 / (11.862 x 365.25 x 86400) = 13.07 km/s. The long period means Jupiter gravity assists occur about every 13 months (synodic period of about 13.1 months relative to Earth).