# 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.

## 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 .

## 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.

## 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 .