When the moon is full, it always looks brightest to us on Earth vs. any other phase since we see its full illuminated area. It’s the same for Mars and the other outer planets, as we see their “full” phase when they are also closest to the Earth, i.e. when they are aligned on the same side of the sun as the earth. Not so for Venus and Mercury. Because their orbital radii are smaller than Earth’s, they present less and less lit surface to an earth-based observer as they approach their closest distance to our planet. Therefore the position at which these two planets look the brightest from Earth is at some point where their orbital radius has some non-zero angle to that of Earth’s.
Develop the relationship on how bright Venus looks from the earth as a function of the angular difference in orbital positions, and solve for the angle where Venus looks the brightest from earth. Simplifying assumptions: the orbits are perfect circles in the same plane, and the orbital radius of Venus is 0.723 that of earth (a good avg. value).
Bonus for purely analytical answer.
In a hypothetical analysis, there are just two factors:
1. The angle between the Sun, Venus and Earth (SVE)
2. The distance between Earth and Venus (EV)
Two methods were attempted:
1. The proportion of the lit side that can be seen was estimated to be (SVE+2(90-SVE)/180
2. The proportion of the lit side that can be seen was estimated at the 'phase angle' offered in astronomical texts: 0.5*(1+cos SVE)
The Sun and Earth are both assumed to be still, so it is sufficient to start with the changing angle SVE, varying from 0, when Venus is opposite Earth, to 180, when Venus is between Earth and the Sun. Using Au, with SE=1, start with the visible proportion, P, and find sin(SVE). Sine SEV is then sin(SVE*(4/13^(2/3)), enabling ESV to be computed. Lastly, This in turn gives the range EV as sin(SVE)/sin(ESV). P is then divided by the square of the range.This method gives SVE= about 134 for maximum brightness.
Mostly the same, except that the starting point for the visible portion is 0.5*(1+cos SVE). This method gives SVE= about 118 for maximum brightness.
Actually, neither method performs particularly well comparative to actual observations. (520 weeks from May 5, 2010, to May 5, 2020).
I plotted apparent magnitude, Angle SVE, and range (chart available by e-mail if someone is prepared to host it). Brightness has a local peak when Venus is furthest, then falls a little before rising steadily to greatest brightness just before, and just after, closest approach. But the height of the peaks varies from cycle to cycle, and the pairs of peaks are not the same height, varying from magnitude about -4.72 to -4.9. The difference mainly comes from range effect, as the angle varies quite smoothly in each synodic period.
The SVE angle at maximum brightness from actual observation (calculated week to week) was 121, 118, 123, 116, 121, 123, 127,...
Daily observations were also looked at for some periods. Venus was of magnitude -4.87 from Nov 29 to Dec 5, 2010, when angle SVE varied from 117 to 124; of magitude -4.69 from July 6 to July 14 2012, when angle SVE varied from 115 to 125; of magnitude -4.89 from Dec 8 to Dec 12 2013, when angle SVE varied from 119 to 124; and of magnitude -4.86 from Feb 9 to Feb 14, 2014, when angle SVE varied from 120 to 126.
On average, then, the observed angle seems to fall somewhere between the two methods used, though a little closer to method 2 than method 1.
One interesting consequence of this is that a more accurate method of computation of 'when Venus is brightest' might be not to concentrate on the angle at all, but rather to analyse at what point of the synodic period Venus is brightest. Venus is actually very well-behaved in this respect, since a year is almost exactly 5/8 of the synodic period. Like the Irishman, if as an astronomer you wanted to know when to point your telescope at Venus, you wouldn't necessarily start with the angles in the first place.
Edited on April 10, 2015, 12:09 am
Posted by broll
on 2015-04-09 23:47:03