When the Venus transit method was first tried in 1769, precise observations of ΔT were foiled by an unexpected phenomenon. As the dark disk of Venus approached the edge ("limb") of the solar disk, a dark bridge seemed to form between it and the darkness beyond the disk, making it difficult to identify the moments of second and third contact. The cause of this "black drop effect" is still being debated, but it may be related to the steep decrease in the Sun's brightness near the Sun's limb. Such "limb darkening" (also evident in the Sun's image used in part 1 of this calculation) occurs because light reaching us from near the visible edge is by necessity emitted at a very shallow angle. It therefore has to travel through a great thickness of the cooler layers of the Sun (above the light-emitting photosphere but still below the chromosphere and corona), and part of it is absorbed again.
Putting all the above in (8), where D (like R–r) is expected to be given in minutes of arc,
D = (15.25)(0.76124)(0.02673) = 0.310306'
or just 1/3 the width of the disk of Venus in front of the Sun.
We will assume that the transit paths across the Sun are parallel to the ecliptic, and we will only use the projection of PP' perpendicular to the ecliptic.
The Earth's Orbit
We now need to introduce distances in the solar system as measured in astronomical units. The mean Sun-Earth distance is 1 AU, but it varies slightly because of the eccentricity of the Earth's orbit, e = 0.01673. (This value might in principle be deduced from the unequality of seasons, discussed in section #12a on Kepler's 2nd law, but the calculation is not a simple one.) The distance to any planet moving in a Kepler ellipse, in polar coordinates (r, θ), is given by
r = a(1-e2)/(1 + e cosθ)
(note this use of the symbol "r" differs from the earlier one). The (smallest, largest) distances occur when θ= (0°,180°) and cosθ = (1, –1). Given the identity (1–e2) = (1–e)(1+e) those distances become r(1–e), r(1+e), or for the Sun-Earth distance in AU, just 1–e and 1+e.
Earth is closest to the Sun (=at perihelion) around January 4, so its greatest distance (at aphelion) should be in early July. The transit of June 8, 2004, was close to the latter, so as a guess let's put the distance at 1.015 AU (considering it is 1.01673 at aphelion, where it has a maximum and varies slowly, and 1 AU at spring equinox, near March 21).
But what is the distance of Venus in AU? Here one can use Kepler's 3rd law, by which the square of the orbital period T is proportional to the cube of the mean distance a (the semi-major axis). The period of Venus is 0.616 years, hence its mean distance from the Sun is 0.723 AU.
At transit time Venus is between us and the Sun, and if both orbits were circular, this would put Earth at 0.277 AU from Venus. In the absence of more information, a circular orbit will be assumed for Venus (actually, a very good approximation). However, since Earth is near aphelion, at an assumed distance of 1.015 AU, we need add 0.015 AU to the Earth-Venus distance, and get 0.292 AU.
The Shift in the Sun's Apparent Position
|Fig. 5 The Sun's center viewed from P and from P'
Next a more subtle matter. Because P and P' are some distance apart and the Sun is not infinitely distant, the position of a point on the Sun, in relation to distant stars (i.e. as measured by coordinates on the celestial sphere) is slightly different at each position. Take the Sun's center O. If the Sun were transparent and we could see the stars behind it, viewing it from P and from P' would show slightly different backgrounds, and the two directions would form a small angle F (see drawing). The sky coordinates of any other point on the Sun (e.g. a small sunspot) would also have this difference when observed from P or P'.
We now denote by x the number of kilometers in one AU: that is the number we wish to derive! The long narrow triangle PP'O may be viewed as a sliver cut (like a pie section) from a circle with O at the center. The entire circle contains 360×60 = 21600 minutes of arc, and the ratio between F and that number is essentially the same as the ratio between the distance PP' (within a tiny error because PP' is straight) and the whole length of the circle, which is 2πx, with 2π=6.2832 to an accuracy of 4 figures past the decimal. Expressed in numbers
F / 21600 = PP'/(6.2832 (1.015 x))
The angle F is very small, because the Sun is very far. Nevertheless, it cannot be ignored, because the angle D with which we are working is also extremely small.
|Fig. 6 Correcting the viewing angle
for the Sun's finite distance
Let us now choose some instant during the transit, when P sees Venus at point Q (on line AB) and P' sees Venus at point Q' (on line A'B'). For this calculation, the sky-positions seen from P will serve as our "reference system," and "up" and "down" here will refer to the directions in the drawing. To get the "standard sky directions" of any point on the Sun seen from P' --including any point of A'B', and in particular Q'--its position needs to be "lifted upwards" by an angle F.
The Corrected Parallax Angle D'
|Fig. 7 Geometry
of the corrected
To obtain the "sky position" of A'B' and of Q' in the same celestial coordinates as AB and Q, we must (as noted before) "lift" it by an angle F towards the center of the Sun. It follows then that the angle PVP' (or QVQ', which equals it) is not D but
D' = D + F (11)
Earlier, when Figure (1b), was discussed, that angle was referred to as "D", which (as is now realized) is not completely correct. In fact the angle PVP' was never measured, it was only inferred from the observed positions of Venus in front of the Sun. Actually PVP' equals D', not the originally inferred angle D, because not just Venus but also the Sun behind it was observed from two different points.
Now in Figure 7 we have a long narrow triangle whose side is the Earth-Venus distance, estimated as Rv=0.292AU . Since the angle D' is very small, PP' may be regarded as small part of a circle, giving very nearly
PP'/ Rv = D'/360°
In Figure 5 the same argument can be applied, except the angle is now F and the side of the triangle is the Earth-Sun distance, estimated as Rs=1.015AU. Hence (very nearly)
PP'/ Rs = F/360°
Dividing right by right, left by left
Rs / Rv = D' / F = 1.015/0.292 = 3.476 (12a)
x = (PP' / F) [21600/(6.2832. 1.015)] =
3386.9 (PP' / F) (12b)
Combining equations (11) and (12a),
D = D' – F = F [(D'/F) – 1] = 2.476 F
The value of D was 0.310306 minutes of arc. From (12b)
x = 3386.9 PP' [2.476/D] = 3386.9 PP' [2.476/0.310306]
x = 27024.8 PP' (13)
(Thanks to Prof. Udo Backhaus of the Univ. of Essen for pointing out some shortcuts here.)