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!

To recap:

Problems:

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

then:

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)

Then:

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.

We know:

Ep1S = 180 - [SE1p1 - E1Sp1]


and SE1p1 = 136 deg

So:

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)

But:

E1p1S = 27.2 deg

and p1E1S = SE1p1 = 136 deg

Thus:

Sp1/SE1 = sin (136)/ sin (27.2)


Sp1 = (0.694/ 0.457) a(E) = 1.518 (a(E)) = 1.518 AU