# Orbit Calculator

Calculate circular, elliptical, or hyperbolic Keplerian orbits, including orbital periods, apoapsis, and periapsis about the Earth, Moon, Sun, and others — This Keplerian orbit model is made by joining a model of conic sections, a model of motion in polar coordinates, and a Kepler solver. By convention, time t=0 and angle (true anomaly) θ=0 correspond to periapsis, the orbit’s closest approach to its central body. The default central body is Earth.

Press **⏎** to see the model. Try your own inputs or the examples below. Refresh your browser to clear results.

## Circular orbit

- Calculate a circular orbit by asserting zero eccentricity e = 0
**⏎**and 400 km altitude h = 400**⏎**. The model infers orbital speed*v*and period*T*. Earth escape velocity can be found in the orbital invariants I.circ**⏎**. - Then, to find out how far the satellite moves in 1 minute type t = 1
**⏎**. The model infers that the satellite moves 4° in this time.

Aside from Earth, you can calculate orbits around the Sun, Moon, and other planets. To see a list of pre-configured central bodies type I.C **⏎** (tip: type wrap **⏎** to wrap subsequent output if the full list is truncated by your screen size). To swap the central body to the Moon, for example, type I.C = 10 **⏎**. Given the inputs of the example above, the model infers a lower orbital speed *v* yet longer period *T* than for Earth.

## Elliptical orbit: Calculate eccentricity from highest and lowest altitudes

- To calculate the eccentricity of an orbit, specify a highest altitude (apoapsis) of 2,000 km ha = 2000
**⏎**and lowest altitude (periapsis) of 300 km hp = 300**⏎**. The model infers orbit eccentricity*e*and orbital period*T*. Specific orbital energy, C₃, velocity at periapsis and apoapsis, as well as other inferred temporal invariants of the orbit can be seen by typing I**⏎**.

## Elliptical orbit: A harder example

- A satellite with a periapsis of 400 km hp = 400
**⏎**is now measured to have an altitude of 500 km h = 500**⏎**while at a speed of 10 km/s v = 10**⏎**, what kind of orbit does that imply? Answer: The orbit has an apogee ha**⏎**of 37,000 km so it is a geosynchronous transfer orbit.

- If the satellite altitude is increasing isOutbound = true
**⏎**, how long ago was it at periapsis, and how long until it reaches apoapsis on this orbit? Answer: The satellite passed perigee 3 minutes ago (because t=3 minutes was inferred) and has an orbital period T.h**⏎**of 11 hours, therefore it approaches apogee in 5 hours.

## Hyperbolic flyby

- An asteroid enters Earth’s sphere of influence at a relative speed of 5 km/s I.vi = 5
**⏎**. It misses Earth and is deflected by 90° I.d = 90**⏎**; how close did it get? Answer: The periapsis altitude hₚ shows the asteroid skimmed beneath 250 km altitude. - The asteroid has been detected as it passes the Moon at a distance of 238,000 miles h.mi = 238000
**⏎**from Earth, how long ago did it pass closest to Earth isOutbound = true**⏎**? Answer: The asteroid skimmed the atmosphere 19 hours ago based on the time since periapsis t.h**⏎**.