This model infers the characteristics of an orbit. It combines a model of conic sections, a model of motion in polar coordinates, and a Kepler solver. The orbit is represented within its plane where t=0 corresponds to θ=0, which corresponds to periapsis (closest approach to the central body). The default central body is Earth.
Press «enter» to see the model. To infer results, try one of the examples below, or your own inputs. To clear results, refresh your browser.
- Specifying eccentricity e=0 for a circular orbit and 400 km altitude:
e = 0 «enter» h = 400 «enter»
The model infers orbital speed v and period T.
- Choosing t=1,
t = 1 «enter»
infers that the satellite moves 4° in this time.
The change in velocity need to escape from this orbit around Earth is stored within the orbital invariants (I.circ «enter»).
- A satellite with a periapsis of 400 km is now measured to have an altitude of 500 km while at a speed of 10 km/s, what kind of orbit does that imply?
hp = 400 «enter» h = 500 «enter» v = 10 «enter»
- If the satellite altitude is increasing, how long ago was it at periapsis, and how long until it reaches apoapsis on this orbit? Setting is_outbound = true and assume_fractional_orbit = true,
iso = t «enter» af = t «enter»
The orbit has a periapsis altitude of 37,000 km, so it is a geostationary transfer orbit. The satellite passed periapsis 3 minutes ago (because t=3 minutes was inferred) and has an orbital period of 11 hours (T.h «enter»), therefore it approaches apoapsis in 5 hours.
- An asteroid enters Earth’s sphere of influence at a relative speed of 5 km/s. It misses Earth and is deflected by 90°; how close did it get?
I.vi = 5 «enter» I.d = 90 «enter»
- The asteroid has been detected as it passes the Moon at a distance of 238,000 miles, how long ago did it pass closest to Earth?
h.mi=238000 «enter» iso = t «enter»
The asteroid skimmed beneath 250 km altitude 19 hours ago.