## Wiim

MtanMM OM»fOI

QWfltll iCW MC1I ©» MftMlilt mm—nil is*, »«hoc niirttaiifsi OArt 0» '«ST foarT yntmeo

W* COUNT OtTIMI fO> tMf IMMM KH

met o* »«|t DMIMUMtt fMT W8»nct 0* T»« nrTV*»i MTWM D*T*

HK| au»T; o* Mncowi • eot OAT OJIMCI >■ uiursor m(fiast0*'a»aa>t »olkobia6t>«i.astvaijbo*ta*c

»M'c P> V'lldtlQuAl TOO e use : the are that time,

5-30) mean

action :er the thod).

where the starting value £„ = M. (Note that if M is identically 0, this method will fail; the solution in this case is trivial, that is, M=0 implies £=0.) In routine ORBGEN, the iteration proceeds until either the correction term is less than 1X 10"8 or until 25 iterations have occurred. Once the eccentric anomaly is found, the true anomaly, v, and the distance, r, may be found from (see Chapter 3):

Equations (5-33) through (5-35) thus give the position of the spacecraft in the orbit plane. We need to take into account the orientation of the orbit in space to find the position relative to an inertial system. Use of the spherical triangles shown in Figure 5-12 gives x = r [ cos(w + f)cos(B) - sin(w + p)sin(S2)cos(»') ] y = r [ cos(w + e)sin(S2) + sin(« + p)cos(S2)cos(i) ] z = r[sin(to + p)sin(i)]

Fig. 5-12. Denning the Orientation of an Orbit in Space. (See also Fig. 3-7.)

0 0