Calculating a Planetary Position. We will compute the RA and DEC of the planet M

Question

Calculating a Planetary Position. We will compute the RA and DEC of the planet Mars on a particular day. The orbital elements of Mars at 00h UTC, September 24, 2008 (Julian Day 2454733.50000) are: a = 1.523592306645371 AU (1) e = .09344312528111040 (2) i = 1.849004686981594 (3) = 49.53243746552888 (4) = 286.6093852342062 (5) = 2454274.425603987 Days (6) n = 0.5240842808934711Day1 . (7) Those for the Earth are: a = 1.000321204431015 AU (8) e = .01629888341148748 (9) i = 0.0009628323502579264 (10) = 14.28804242135567 (11) = 87.85756134487137 (12) = 2454577.780568315 Days (13) n = 0.9851344662855015Day1 . (14)

a) Find Mars’s Mean Anomaly, M.

b) Use Kepler’s Equation to arrive at an approximate value of the eccentric anomaly, E. By “approximate,” one usually means a numerical result of about the same accuracy as the other values in the problem. Mathematica’s FindRoot function may be valuable for this. Note, however, that Kepler did not have the benefit of FindRoot.

c) Use the derived value of E to calculate the true anomaly, f.

d) Use the derived value of f to calculate the heliocentric radius, r.

Explanation / Answer

a)   Mars’s Mean Anomaly, M = (M0 + M1*(JJ2000))mod360°

                                              =    (19.3730 + 0.52402068*(24530972451545))mod360°

                                               =   832.6531° mod360°

                                               =   112.6531°

b)    tan(V/2) = sqrt((1 + e)/(1 - e)) * tan(E/2)

      tan(286.609/2) = sqrt((1 + .09344/(1 - .09344)) * tan(E/2)

=>   eccentric anomaly, E =   68.33 degree

c)      true anomaly, f = 244.921657 degrees

d)      heliocentric radius, r   = 1.524 AU

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