Thursday, August 25, 2011
Solving Intermediate Astronomy Problems (3)
We now proceed to solve the two problems from the last set, one relatively easy, the other relatively more difficult!
1) Estimate the distance of Venus from the Sun at its most recent maximum elongation, if the angle of max. elong. was 46 degrees.
2) A recent observation of Mars 36.5 days after opposition showed an angle of elongation = 136 degrees. Find the distance of Mars from the Earth if Mars' orbital period = 687 days.
Solution for #1
The diagram for Problem 1 is shown, and the angle of maximum elongation.
We know, from the geometry:
SV/ SE = sin (SEV)
Then Venus' distance is:
SV = (sin(SEV)) SE
where: angle SEV = 46 deg and SE = a(E) = 1 AU
SV = sin (46) 1 AU = (0.719) 1 AU = 0.719 AU
Solution for #2
The diagram for the configuration is shown for Problem #2, and the elongation angle which is also angle SE1p1 = 136 deg.
We are given t = 36.5 d, and we need to find the "gain" angle:
Θ = 360 (1/P - 1/P') t
where P, P' are the sidereal periods, or P = 365.25 d (Earth) and P' = 687 d (Mars)
Θ = 360 (1/365.25 - 1/687) (36.5) = 16.8 deg
which is none other than angle E1Sp1.
Now, angle E1p1S is required in order to get the distance - as the information at the bottom of the depiction shows.
Ep1S = 180 - [SE1p1 - E1Sp1]
and SE1p1 = 136 deg
Ep1S = 180 - [136 deg - 16.8 deg]= 27.2 deg
Again, from plane trigonometry:
sin(p1E1S)/ Sp1 = sin(E1p1S)/ SE1
and, the ratio of planetary distances at time t:
Sp1/SE1 = sin (p1E1S)/ sin (E1p1S)
E1p1S = 27.2 deg
and p1E1S = SE1p1 = 136 deg
Sp1/SE1 = sin (136)/ sin (27.2)
Sp1 = (0.694/ 0.457) a(E) = 1.518 (a(E)) = 1.518 AU