All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Bright light (Posted on 2015-04-09)
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.

 See The Solution Submitted by Kenny M Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: GSP Model disagrees with the previous posts | Comment 4 of 9 |
(In reply to GSP Model disagrees with the previous posts by Jer)

I think the difference between our solutions is your reflecting-toward-earth fraction is 1 - EVS/180, while mine is (1+cos(EVS))/2. The latter is based on a diagram projecting the terminator (day/night boundary) onto a plane perpendicular to that of the solar system.

 Posted by Charlie on 2015-04-09 16:15:01

 Search: Search body:
Forums (0)