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.
E=earth, S=sun, V=venus
Setting SE = 1 au, SV = .723 au
Construct circles centered as S with radii SE and SV.
Measure angle EVS is degrees. The proportion of reflected light seen by earth (assuming perfect diffusion) is 1 - EVS/180.
Measure the distance VE.
Apparent brightness = proportion seen / distance-squared
Let V travel around its orbit. Noting when the above reaches a maximum.
[I interpolated to increase accuracy, because relatively large changes in the angle lead to small changes in brightness.]
The maximum brightness is 2.06421
(the unit would be 1= the brightness of a fully lit venus 1 unit away)
EVS = 133.4953 degrees
VES = 31.6341 degrees
ESV = 14.8706 degrees
VE = .35378 au
1 - EVS/180 = .25836
So even though you're only seeing a quarter of the lit face, the nearness more than makes up for it.
Posted by Jer
on 2015-04-09 14:19:29