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

  1. https://sunearthday.nasa.gov/2012/articles/ttt_75.php
  2. How did Halley calculate the distance to the Sun by measuring the transit of Venus?

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

.

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

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

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

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 and ? Or, if easier, show a clear and detailed method for the computation of the AU from the time measurements of the transit of Venus?

usernumber
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Brasil
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1 Answers1

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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.

Pulsar
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