Transit of Venus and the computation of the Astronomical Unit

I have searched for the computation of the AU. The two best websites I found about it were

In the first one, the author declares that the angular speed of Venus is

$\inline \small \omega \approx 0,055^\circ/\mathrm{h}$.

However, I can't get this result neither by using the fact that the planet has a sideral period of 225 days which leads to

$\inline \small \omega = \frac{360^\circ}{225 \cdot 24\ \mathrm{h}} = 0,067^\circ/\mathrm{h}$

nor by using the fact that the synodic period is 584 days, which leads to

$\inline \small \omega = \frac{360^\circ}{584 \cdot 24\ \mathrm{h}} = 0,026^\circ/\mathrm{h}$

I really believe that the synodic period is the one that must be used for this computation (if I'm wrong, please tell me), since it results from the relative motion of Venus observed from Earth.

The second reference isn't clear about how to reach the relation

$\inline \small \Delta \beta = (R - r) \frac{\sin^2\theta}{\cos\theta} \frac{\Delta T}{T}$

I would like to highlight that these references complete each other, since the second one explains how to determine the solar parallax, not explained in the first one, while the first one makes clear how to compute the angular sizes of the chords seen as the paths of Venus to each observer.

So, my question is: can anyone derive the computation of $\inline \small \omega$ and $\inline \small \Delta \beta$? Or, if easier, show a clear and detailed method for the computation of the AU from the time measurements of the transit of Venus?

• The angular speed of planets as perceived by the Earth is time-dependent, and oscillates based on the relative orientation of Earth and Venus. Most likely the author is referring to the angular speed calculated specifically during that transit interval, which would most likely match up to neither of those numbers, as the numbers you calculated assume a constant angular speed. Oct 25 '18 at 13:29
• I agree that angular speed is variable, @probably_somene. But the variation is not critical. Using Kepler's first two laws and the known values for Venus (Ra = 109 Gm, Rp = 107 Gm, orbital period To = 225 days) we find that it cruises an elliptical orbit at an areal rate of Ah = pi.Aa.Ap/To = v.r/2, leading to w = 720.Ra.Rp/To.r^2, r being the Sun-Venus distance at any moment. The constant 720.Ra.Rp/To = 781 Gm²/h, so that the maximum angular speed is wp = 0,068°/h, while the minimum is wa = 0,066°/h, meaning that tha variation on angular speed is not the actual reason of the discrepancy. Oct 25 '18 at 14:19
• Your calculation just now was the angular velocity as seen from the Sun's reference frame. What the author is likely calculating is the angular velocity of Venus as seen from the Earth's reference frame. Since the Earth is moving with respect to the Sun, these two quantities are different. Oct 25 '18 at 14:47

Yes, you have to use the synodic angular speed of Venus (which is simply the difference between the sidereal angular speeds of Venus and Earth), but this will give you the angular speed of Venus with respect to the Sun. What you need is the angular speed of Venus across the sky, as seen from Earth. Let's call the distance of Venus to the Sun $$d$$, in astronomical units. At transit, the distance between Venus and Earth is then $$1-d$$, and Venus is moving perpendicular to the Sun-Earth axis. The orbital velocity of Venus is $$v \sim d\omega = (1-d)\omega',$$ where $$\omega'$$ is the angular speed of Venus across the sky, as seen from Earth. Using $$\omega = 0.026^\circ/\text{h}$$, and from your first article, $$d = 0.69\$$AU, we get $$\omega' = \frac{d}{1-d}\omega\approx 0.058^\circ/\text{h}.$$ There are additional effects (e.g. the rotation of the Earth which adds a parallax effect to the observations), but this the gist.