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Orbit Calculator

This model calculates Keplerian orbits. It 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

  1. 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 ⏎.
  2. 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.
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Elliptical orbit: Calculate eccentricity from highest and lowest altitudes

  1. 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 ⏎.
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Elliptical orbit: A harder example

  1. 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. 
  2. 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.
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Hyperbolic flyby

  1. 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.
  2. 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 ⏎.
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See also

3D Orbit Calculator

Orbit Transfer Calculators