Friday, 22 May 2009

orbit - Determining distance from semi-major axis and eccentricity

Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period T from the 3rd Keplero's law:



fracT2a3=frac4pi2G(mE+mb)approxfrac4pi2GmE,



where a is the semi-major. With T, for each time istant you also know the mean anomaly M, given by (suppose t=0 at perigee):



M(t)=frac2piTt.



Solving numerically the Keplero's equation for the eccentric anomaly E (where e is the eccentricity)



M=EesinE



and then use the following equation to derive the true anonaly nu, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:



cosnu=fraccosEe1ecosE
and
sinnu=fracsqrt1e2sinE1ecosE.



The distance from the Earth is just given by the orbit equation



r=fraca(1e2)1+ecosnu.



If I'm not wrong with calculation, it should be:



T=9364s=2.6h.



M(90min)=207.60°



E(90min)=203.11°



nu(90min)=198.95°



r=11366km



To get the covered distance you should calculate the line integral of the orbit equation.

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