Tackling Intermediate Astronomy Problems (3)

One of the more important applications in intermediate astronomy is obtaining the relative distance to an inferior or superior planet in relation to the Earth. As we've seen before, it's easy to apply Kepler's 3rd law to obtain the basic dimensions of an orbit, namely the semi-major axis of the orbit - or the mean distance from the Sun. But things become somewhat more difficult when we seek to find the distance say, of the Earth to the planet, or the planet at a specific time - say before it commences retrograde motion.

So, we consider two cases:

(A) Inferior planet (see the diagram A)

As seen in instalment (3), the maximum elongation occurs when the planet's geocentric radius vector (pE in diagram A) is perpendicular to the planet's heliocentric radius vector, pV. Then by a careful measurement of the angle SEp, say over a series of nights around maximum elongation, one can obtain a value for the angle of maximum elongation. At such time the angle SEp is right-angled hence:

Sp/ SE = sin(SEp)

The quantity Sp/SE is therefore the distance of the planet from the Sun in terms of Earth units (or AU, astronomical units). Let Sp = R and SE = a(E) the semi-major axis for Earth's orbit, then:

Sp = R = sin(SEp) [a(E)]

Example:

If the angle SEp = 60 deg, find the planet's distance from the Sun.

Then: sin(SEp)= sin (60) = [3]^½/2 = 0.866

So: R = 0.866[a(E)]= 0.866 AU


Case (B) Superior Planet

In this case, we apply diagram (B), showing the planet at two successive positions, p and p1 and the Earth at E and and E1. This is a more difficult case but can be worked if the planet's synodic period S is known.

Here, we let the planet p be in opposition at some given time with the Earth and Sun (e.g. showing the alignment S-E-p in diagram (B). As we know, with opposition, the elongation is a straight angle or 180 degrees. Then after t days have elapsed the Earth's radius vector SE has moved ahead of the planet's as shown in comparing SE1 to Sp1. As can be seen, this reduces the angle of elongation from 180 degrees at opposition to angle SE1p1. This is then measured.

Now, over t days, the angle ESp will have increased from 0 (at opposition) to a value Θ given by:

Θ = [n - n(p)] t

where n, n(p) are the mean daily motions of the Earth and the planet, respectively. Using relations for the periods seen in instalment (3) we may write:

Θ = 360 (1/P - 1/P') t

where P and P' are the sidereal periods for the Earth and the planet, respectively/ Then, it follows by the relations seen in instalment (3):

Θ = 360 (1/S)

So, since t and S are both known, Θ can be obtained - that is, angle E1Sp1 is calculated. Hence, angle E1p1S can be found from:

angle E1p1S = 180 - angle SE1p1 - angle E1Sp1

From plane trigonometry we then obtain:

sin(p1E1S)/ Sp1 = sin(E1p1S)/SE1

or:

Sp1/ SE1 = sin(p1E1S)/ sin(E1p1S)

again, giving the distance from the Sun in terms of Earth's distance unit.

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.