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Bright light (Posted on 2015-04-09) Difficulty: 4 of 5
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)

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what I get | Comment 2 of 9 |
The lit fraction of the apparent disk of Venus depends on the angle at Venus in the Sun-Venus-Earth triangle. The disk is fully lit when the angle is zero, half lit when the angle is 90° and not lit at all when the angle is 180°. The actual function is (1+cos(theta))/2, where theta is the angle at Venus.

The visible area of the sky covered by Venus's bright portion also depends on the area of the full circle of Venus's disk, then this can be multiplied by the fraction from the preceding paragraph.

The full disk area covered can be considered inversely proportional to the square of the distance between Venus and the earth.

If theta is the angle at Venus, the distance of Venus to Earth will be found from

1 = x^2 + .723^2 - 2*.723*x*cos(theta)

x^2 - 2*.723*x*cos(theta) + .723^2 - 1 = 0

x = (2 * 0.723 * Cos(theta) + sqrt(4 * (0.723) ^ 2 * (Cos(theta)) ^ 2 - 4 * (0.723) ^ 2 + 4)) / 2

The brightness would be proportional to the inverse of the square of this.

 pi = 4 * Atn(1)
 
 
 For thdegrees = 117.903 To 117.906 Step 0.00001
   theta = thdegrees * pi / 180
   x = (2 * 0.723 * Cos(theta) + Sqr(4 * (0.723) ^ 2 * (Cos(theta)) ^ 2 - 4 * (0.723) ^ 2 + 4)) / 2
   bright = ((1 + Cos(theta)) / 2) / (x * x)
   Text1.Text = Text1.Text & mform(thdegrees, "###0.00000") & mform(x, "##0.00000") & mform(bright, "###0.000000000000") & crlf
   DoEvents
 Next
 
 produces a table, the central part of which is:
 
   angle at  distance     relative
    Venus      E-V       brightness
 117.90490  0.43089   1.432637732905
 117.90491  0.43089   1.432637732906
 117.90492  0.43089   1.432637732907
 117.90493  0.43089   1.432637732908
 117.90494  0.43089   1.432637732909
 117.90495  0.43089   1.432637732909
 117.90496  0.43089   1.432637732910
 117.90497  0.43089   1.432637732910
 117.90498  0.43089   1.432637732910   *
 117.90499  0.43089   1.432637732910   *
 117.90500  0.43089   1.432637732910
 117.90501  0.43089   1.432637732910
 117.90502  0.43089   1.432637732909
 117.90503  0.43089   1.432637732909
 117.90504  0.43089   1.432637732908
 117.90505  0.43089   1.432637732907
 117.90506  0.43089   1.432637732906
 117.90507  0.43089   1.432637732905
  
  So the angle at Venus looks like 117.904985°.
  
  We want the angle at the sun, say alpha. 
  
  sin(theta)/1 = sin(alpha)/x
  
  sin(alpha)= sin(117.904985)*0.43089
  
  alpha = 22.382516°
  
  Complicating the brightness could be variations in the brightness of various portions of the visible portions of Venus's surface (actually the tops of the clouds).
  

  Posted by Charlie on 2015-04-09 13:03:18
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