Orbital Period and Velocity Calculator
Compute the orbital period and circular velocity at any altitude around a solar system body using Kepler's third law, T = 2pi x sqrt(a^3 / mu).
🌍 What is the Orbital Period and Velocity Calculator?
Orbital period and velocity are the two fundamental quantities that describe a circular orbit. Orbital velocity is the speed a spacecraft must maintain to travel in a stable circular path around a body at a given altitude: too slow and it falls back, too fast and it climbs to a higher orbit. Orbital period is the time to complete one full revolution. Both quantities follow directly from Newton's law of gravitation and Kepler's third law, derived over 350 years ago but still the exact equations used by mission designers today.
The Orbital Mechanics mode computes the circular orbital velocity v_orb = sqrt(mu / r), the orbital period T = 2 x pi x sqrt(r cubed / mu), and the escape velocity v_esc = sqrt(2) x v_orb for any altitude above eight solar system bodies including Earth, Moon, Mars, Venus, Mercury, Jupiter, Saturn, and the Sun. Common use cases include: verifying satellite orbit design, computing ISS or GPS orbital parameters for physics courses, finding GEO altitude from its 24-hour period, and comparing orbital speeds across different planets.
The Kepler Third Law mode handles heliocentric orbits around the Sun. Kepler's third law states that T squared equals a cubed when T is in years and a is in astronomical units (AU). This applies to any body orbiting the Sun: planets, comets, asteroids, and spacecraft on interplanetary trajectories. Planet presets from Mercury to Neptune are included, plus Ceres for the asteroid belt. The mode also shows T squared and a cubed side by side to verify the law numerically and displays the mean orbital velocity for mission planning purposes.
These calculations are foundational: before you can design a transfer orbit (Hohmann transfer), compute a gravity assist trajectory, or plan an orbital rendezvous, you must know the orbital speeds and periods of the orbits involved. This calculator provides those starting values instantly for any body in the solar system.
📐 Formula
T = 2π √( a3 / μ )
T = orbital period (seconds)
a = semi-major axis = body radius + altitude (metres)
μ = gravitational parameter = G × M (m³/s²)
vorb = √(μ / a) (circular orbital velocity, m/s)
vesc = √2 × vorb (escape velocity at same altitude, always 1.4142 × vorb)
T2 = a3 (Kepler’s third law, T in years, a in AU)
T = orbital period in Earth years (yr)
a = semi-major axis in astronomical units (AU)
1 AU = 149,597,870.7 km = mean Earth-Sun distance
Example: Mars a = 1.524 AU → T = √(1.5243) = √(3.540) = 1.881 yr = 687 days
📖 How to Use This Calculator
Steps
1
Select the central body and altitude - Choose the Orbital Mechanics mode, select the planetary body from the preset list, and enter the orbital altitude in km using the number field or the slider.
2
Read the orbital parameters - The calculator shows circular orbital velocity, orbital period (in human-readable format and in seconds), escape velocity at that altitude, and the orbital radius from the body center.
3
Use Kepler mode for heliocentric orbits - Switch to the Kepler Third Law mode, select a planet preset or enter a custom semi-major axis in AU, and get the orbital period in years and days plus the mean orbital velocity in km/s.
4
Verify Kepler's law - The Kepler mode also displays T squared and a cubed side by side so you can confirm that they match, demonstrating Kepler's third law numerically for any body in the solar system.
💡 Example Calculations
Example 1 - ISS Orbital Speed and Period (400 km LEO)
International Space Station at 400 km altitude above Earth
1
Orbital radius: r = (6,378.1 + 400) km = 6,778,100 m
2
Orbital velocity: v = sqrt(3.986e14 / 6,778,100) = sqrt(5.880e7) = 7,668 m/s = 7.668 km/s
3
Orbital period: T = 2π × sqrt(6,778,100³ / 3.986e14) = 2π × 883.7 s = 5,552 s = 92.5 min
4
Orbits per day: 86,400 / 5,552 = 15.6 orbits per day
ISS orbital speed: 7.668 km/s, orbital period: 92.5 min (~15.6 orbits/day)
Try this example →Example 2 - Geostationary Orbit (GEO)
Communications satellite in geostationary orbit at 35,786 km altitude
1
Orbital radius: r = (6,378.1 + 35,786) km = 42,164,100 m
2
Orbital velocity: v = sqrt(3.986e14 / 42,164,100) = sqrt(9.453e6) = 3,075 m/s = 3.075 km/s
3
Orbital period: T = 2π × sqrt((42,164,100)³ / 3.986e14) = 86,164 s = 23.934 hr (one sidereal day)
GEO orbital speed: 3.075 km/s, period: 23.934 hr (appears stationary above equator)
Try this example →Example 3 - Kepler's Third Law for Mars
Mars orbital period from Kepler's third law with semi-major axis a = 1.524 AU
1
Semi-major axis: a = 1.5237 AU (Mars mean orbital radius)
2
Kepler's third law: T² = a³ = 1.5237³ = 3.540
3
Period: T = sqrt(3.540) = 1.881 yr = 686.9 days
4
Mean orbital speed: v = 2π × 1.5237 × 1.496e11 / (686.9 × 86,400) = 24,080 m/s = 24.08 km/s
Mars orbital period: 1.881 years (687 days), mean speed: 24.08 km/s
Try this example →❓ Frequently Asked Questions
What is the formula for orbital period using Kepler's third law?
Kepler's third law states T squared = (4 x pi squared / mu) x a cubed, giving T = 2 x pi x sqrt(a cubed / mu), where mu is the gravitational parameter. For the Sun, mu = 1.327e20 m cubed per s squared. With a in AU and T in years, it simplifies to T squared = a cubed. For Earth at 1 AU: T = 1 year. For Mars at 1.524 AU: T = sqrt(1.524 cubed) = 1.881 years. This law applies to any elliptical orbit, not just circular ones.
What is the orbital velocity of the ISS?
The ISS orbits at about 400 km altitude, giving v_orb = sqrt(3.986e14 / 6,778,100) = 7,668 m/s = 7.668 km/s. The period is T = 2 x pi x sqrt(6,778,100 cubed / 3.986e14) = 5,552 s = 92.5 minutes. At this speed the ISS completes 15.6 orbits per day, experiencing about 16 sunrises and 16 sunsets every 24 hours. Astronauts on the ISS are in free fall (weightlessness) because the station and everything inside are falling together around Earth's curvature.
What altitude is geostationary orbit and what is the orbital speed?
Geostationary orbit (GEO) is at 35,786 km altitude where the orbital period equals one sidereal day (86,164 s). Orbital speed = sqrt(3.986e14 / 42,164,000) = 3,075 m/s = 3.075 km/s. Satellites at GEO appear stationary above a fixed point on the equator because they complete one orbit in the same time Earth completes one rotation. This is used for communications, weather satellites, and direct broadcast TV. Reaching GEO from LEO requires a Hohmann transfer with about 3.9 km/s total delta-V.
How does orbital period change with altitude?
Orbital period scales as r to the 1.5 power: T = 2 x pi x sqrt(r cubed / mu). Doubling the orbital radius multiplies the period by 2 to the 1.5 = 2.83. For Earth: LEO (400 km) T = 92.5 min; GPS orbit (20,200 km) T = 717 min = 11.97 hr; GEO (35,786 km) T = 1,436 min = 23.93 hr; Moon distance (384,400 km) T = 27.3 days. The period increases faster than altitude because both the circumference and the required lower speed both increase together as altitude rises.
What is the orbital period of Mars around the Sun?
Mars has a semi-major axis of 1.524 AU. Kepler's third law gives T = sqrt(1.524 cubed) = sqrt(3.540) = 1.881 years = 686.97 Earth days = 668.6 Martian sols. The mean orbital speed is about 24.08 km/s. The next launch window to Mars from Earth occurs every 26 months (the synodic period), which is when Earth and Mars realign for the minimum-energy Hohmann transfer. The Hohmann transfer itself takes about 259 days (about 8.5 months).
Does orbital period depend on satellite mass?
No. In the two-body approximation where satellite mass is negligible compared to the central body, T = 2 x pi x sqrt(r cubed / mu) depends only on orbital radius and the body's gravitational parameter, not on satellite mass. This is why a 1 kg CubeSat and a 400-tonne space station at the same altitude have exactly the same orbital period. This result follows directly from the equivalence principle: gravitational and inertial mass cancel in the equations of motion, making all orbiting objects fall at the same rate.
What is the relationship between orbital velocity and escape velocity?
At any altitude r, v_esc = sqrt(2) x v_orb = 1.4142 x v_orb. This follows from v_orb = sqrt(mu/r) and v_esc = sqrt(2 x mu/r). A spacecraft already in circular orbit needs only a 41.4% increase in speed to reach escape velocity. For Earth at 400 km LEO: v_orb = 7.668 km/s, v_esc = 10.844 km/s, delta-V needed = 3.176 km/s. For Moon at 100 km: v_orb = 1.630 km/s, v_esc = 2.306 km/s, delta-V = 0.676 km/s. This explains why returning from the Moon requires so much less propellant than leaving Earth.
What is the orbital speed of Earth around the Sun?
Earth's mean orbital speed around the Sun is 29.785 km/s. It varies from 30.29 km/s at perihelion (closest to the Sun in January) to 29.29 km/s at aphelion (farthest in July) due to Earth's slightly elliptical orbit (eccentricity 0.017). The variation follows Kepler's second law (equal areas in equal times): Earth moves faster when closer to the Sun. This speed is also the starting speed for any interplanetary spacecraft launched from Earth, before adding the departure hyperbolic excess velocity from Earth's sphere of influence.
How do I find orbital altitude for a given period?
Rearrange Kepler's third law: a = (mu x T squared / (4 x pi squared)) to the one-third power. For T = 24 hours = 86,400 s around Earth: a = (3.986e14 x 86,400 squared / (4 x pi squared)) to the 1/3 = 42,241 km. Altitude = 42,241 - 6,378 = 35,863 km (close to GEO, the small discrepancy is because a sidereal day is 86,164 s, not 86,400 s). For T = 12 hours (GPS): a = (3.986e14 x 43,200 squared / 39.48) to the 1/3 = 26,559 km, altitude = 20,181 km.
What is GPS orbital altitude and period?
GPS satellites orbit in Medium Earth Orbit (MEO) at about 20,200 km altitude. Orbital speed = sqrt(3.986e14 / 26,578,100) = 3,874 m/s = 3.874 km/s. Period = 2 x pi x sqrt((26,578,100) cubed / 3.986e14) = 43,082 s = 11.97 hours, roughly half a sidereal day. This exact half-day period means the GPS satellites return to the same position above the ground every 23 hours 56 minutes, creating a very predictable geometry for receivers. 24 active GPS satellites in six orbital planes ensure global coverage at all times.
How does orbital period compare between planets?
Planet periods by Kepler's third law: Mercury 0.241 yr (88 days), Venus 0.615 yr, Earth 1.000 yr, Mars 1.881 yr, Jupiter 11.862 yr, Saturn 29.457 yr, Uranus 84.01 yr, Neptune 164.8 yr. These follow precisely from T squared = a cubed. Outer planets have both longer orbital radii (larger circumference) and lower orbital speeds, compounding the period increase. Neptune takes 165 Earth years to complete one orbit, meaning it has completed fewer than two full orbits since its discovery in 1846.
Can I calculate orbital period for a moon orbiting a planet?
Yes, use the Orbital Mechanics mode with the planet as the central body, and enter the moon's orbital altitude. For example, to find the period of the ISS if it were a moon orbiting Mars: select Mars, enter an altitude equivalent to the moon's orbital radius minus Mars' radius. For Phobos, Mars' inner moon: orbital radius = 9,376 km, altitude above Mars = 9,376 - 3,390 = 5,986 km. Entering 5,986 km in Mars mode gives T = 2 x pi x sqrt((9,376,000) cubed / 4.283e13) = 27,553 s = 7.65 hours, matching Phobos' observed period closely.