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.
- To calculate a circular orbit, specify zero eccentricity and 400 km altitude by typing:
e = 0 «enter» h = 400 «enter»
The model infers orbital speed v and period T.
- Then, to find out how far the satellite moves in 1 minute, type:
t = 1 «enter»
The model 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»).
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 and lowest altitude (periapsis) of 300 km by typing:
ha = 2000 «enter» hp = 300 «enter»
The model infers orbit eccentricity e and orbital period T.
Elliptical orbit: A harder example
- 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,
io = t «enter» as = t «enter»
The orbit has an apogee of 37,000 km, so it is a geosynchronous transfer orbit. The satellite passed perigee 3 minutes ago (because t=3 minutes was inferred) and has an orbital period of 11 hours (T.h «enter»), therefore it approaches apogee 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» io = t «enter»
Looking at the periapsis altitude hₚ and the time t.h relative to periapisis, the asteroid skimmed beneath 250 km altitude 19 hours ago.