Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've read it several places that the apparent retrograde motion of planets (during which they seem, as viewed from Earth, to move in the opposite sense of their normal "direct" orbital motion against background stars at infinite distance) occurs between the two quadrature points (at which the planet-Earth-Sun angle is 90°). I have always assumed that "between" is an approximation, since, at quadrature, though the Earth's motion is contributing nothing to the planet's apparent motion (since it is moving directly along the line of sight to the planet), the planet's true motion is providing apparent "direct motion", so that retrograde motion, though bounded by the quadrature points, does not begin or end there.

enter image description here

I've never been able to prove this to my satisfaction, and often come across descriptions that make me wonder whether I've had it wrong all along.

Some of these sources would seem to be quite authoritative, such as a translator's footnote to Copernicus, in which it is stated that the angular extent of a superior planet's retrograde motion observed from Earth is defined by the tangents to the Earth's orbit that pass through the planet:

enter image description here

While these tangents certainly bound the angular extent of retrograde motion — in fact, they define the retrograde (and direct) motion of an unmoving planet (since they correspond to the maximum parallax for the planet seen from Earth) — isn't the actual extent of retrograde motion smaller, for the reasons stated above?

How can it demonstrated geometrically (assuming circular orbits with common centers and uniform angular velocities, and given periods and radii for those orbits for Earth and the Planet) what the rate of change of the apparent angular position of an orbiting planet is as a function of the planet-Earth-Sun angle?

share|improve this question
1  
As a practical astronomical matter, it should be possible to do this given only the orbital periods ($p_E$ and $p_P$) orbital radii ($r_E$ and $r_P$) and the planet-Earth-Sun angle, or "elongation" ($\varepsilon$). –  raxacoricofallapatorius Sep 22 '12 at 20:41
add comment

1 Answer 1

The tangents do not bound the retrograde motion, but the range is close. You are right they define the range for an unmoving outer planet. The boundaries of retrograde motion are found when the velocity of each planet projected on the perpendicular to the line joining them is the same. For a first approximation, you can consider the velocity of the outer planet to be perpendicular to the joining line (valid if the outer planet is much farther from the sun than Earth.) In this approximation, if $\theta$ is the angle at the Earth from the sun vector to the Earth vector is $v_P=v_E \cos \theta$ This supports the tangent for an unmoving outer planet-if we set $v_P=0$ we want $\cos \theta =0,$ so $\theta = \frac \pi 2.$ The boundary of retrograde motion for a moving outer planet will be beyond the tangent when Earth is leading, and behind it when Earth is lagging, but the range will be about the same.

Added: the following assumes circular orbits. Draw the triangle from the planet to the earth and sun. Let $\theta$ be the angle at the earth and $\phi$ the angle at the planet. The projection of the planet velocity perpendicular to the planet-earth line is $v_P \cos \phi$ The projection of the earth velocity perpendicular to the planet-earth line is $v_E \cos \theta$ At the boundaries of the retrograde motion these are equal, so $v_P \cos \phi=v_E \cos \theta$ $\phi$ will very close to the half-width of the Earth's orbit as seen from the planet, so you can solve for $\theta=\arccos \frac {v_P \cos \phi}{v_E}$. You can iterate the solution (improve $\phi$ from the $\theta$ you find) if you need more accuracy.

share|improve this answer
    
Yes, it's clearly close, and the tangent lines are bounding, but what are the actual beginning and end values of $\varepsilon$? (Also, I'm not sure I understand the last sentences.) –  raxacoricofallapatorius Sep 26 '12 at 22:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.