- #1

danunicamp

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- Homework Statement
- Hello, This is problem 13b, from chapter 1 of Zeilik, Gregory, Introductory Astronomy and Astrophysics.

Mars has a synodic period of 779.9 days and a sidereal period of 686.98 days.

On February 11, 1990, Mars had an elongation of 43° West.

The elongation of Mars 687 days later, on December 30, 1991, was 15° West.

What is the distance of Mars from the Sun in astronomical units?

- Relevant Equations
- Law sins, law cosines

I called the point E1 the point where Earth was at Feb,11 1990 and E2 at 30, Dec 1991, S for Sun and M for Mars and r for the Mars-Sun distance.

Since we got a whole sidereal period between both alongation, I assumed Mars was on the same point in space (wrt Sun).

I think I got the triangles wrong, since I already redid the calculations twice.

I defined the triangles E1, S, M and E2, S, M.

E1, S, M with a side r opposite to the angle of 43 deg, a side 1 opposite to the angle SME1.

E2, S, M with a side r opposite to the angle of 15 deg, a side 1 opposite to the angle SME2.

The sides SE1 and E2M crossed at a point P.

I calculated that in (730-687) 43 days, or 42.41 degrees, the point E2 would reach E1. And with that got the angle of 42.41 in E2SP.

From that i got the angle of 122.59 at SPE2 and the same angle at E1PM.

From that I got the angle of 57.41 at SPM.

The angle E1MP was 14.41 (180 - 43 - 122.69)

From the law of sins (1/sin 122.59 = SP/sin 15) I calculated the side SP to be 0,3 and PE2 to be 0.7 (1 - 0.3).

From the law of sins I calculated the side PM to be 1.98 (0.7/sin 14.41 = PM/sin 43).

With the sides 0.3, 1.98 and r and the angle 57.41 i used the law of cosines:

r^2 = 0.3^2 + 1.98^2 - 2 (0.3)(1.98)(cos57.41)

and got a r = 1.83, which is wrong, since the Sun-Mars distance is 1.52 AU.

I know it is hard to visualize without the picture. Is there any way to add it here?

Any help would be very welcome. Thank you in advance

Since we got a whole sidereal period between both alongation, I assumed Mars was on the same point in space (wrt Sun).

I think I got the triangles wrong, since I already redid the calculations twice.

I defined the triangles E1, S, M and E2, S, M.

E1, S, M with a side r opposite to the angle of 43 deg, a side 1 opposite to the angle SME1.

E2, S, M with a side r opposite to the angle of 15 deg, a side 1 opposite to the angle SME2.

The sides SE1 and E2M crossed at a point P.

I calculated that in (730-687) 43 days, or 42.41 degrees, the point E2 would reach E1. And with that got the angle of 42.41 in E2SP.

From that i got the angle of 122.59 at SPE2 and the same angle at E1PM.

From that I got the angle of 57.41 at SPM.

The angle E1MP was 14.41 (180 - 43 - 122.69)

From the law of sins (1/sin 122.59 = SP/sin 15) I calculated the side SP to be 0,3 and PE2 to be 0.7 (1 - 0.3).

From the law of sins I calculated the side PM to be 1.98 (0.7/sin 14.41 = PM/sin 43).

With the sides 0.3, 1.98 and r and the angle 57.41 i used the law of cosines:

r^2 = 0.3^2 + 1.98^2 - 2 (0.3)(1.98)(cos57.41)

and got a r = 1.83, which is wrong, since the Sun-Mars distance is 1.52 AU.

I know it is hard to visualize without the picture. Is there any way to add it here?

Any help would be very welcome. Thank you in advance