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 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).

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

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 my solution | Comment 1 of 9
let S be the sun, E be earth, and V be venus.

the brightness of venus (seen from earth) is directly related to the angle between earth and the sun, that is angle EVS.

Now let r1 be the orbital radius of venus and r2 be the orbital radius of earth.  Then the angle EVS is given by (when VSE is t)
2r1^2-2r1*r2*Cos(t)
-------------------------------------------
2*r1*sqrt(r1^2+r2^2-2r1*r2*Cost(t))

Taking the derivative, setting equal to zero, and solving for Cos(t) we get
Cost(t)=r2/r1

thus the angle is maximized when t=acos(r2/r1)

for the given example we have r2/r1=0.723
this gives

 Posted by Daniel on 2015-04-09 11:36:31

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