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 earthbased 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 nonzero 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 reply to
Interesting question. by broll)
Yes, this was more of a mathematical problem than strictly astronomical, in the presence of those simplifying assumptions. The orbits of neither Venus, nor, especially the Earth, are perfectly circular, and Venus's orbit is tilted over 3° to Earth's. But my found 118° for SVE does fall within the range of values you find in the actual data.

Posted by Charlie
on 20150410 07:35:28 