Illuminated fraction of the Moon

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JeffOCA
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Dear all,

It's late in the night and I have some trouble in deriving the expression of the ratio of the illuminated disk to the whole disk. Is it a formula "by definition" ?

See http://docs.google.com/viewer?a=v&q...vuuO5&sig=AHIEtbQ4C_gHumCJFdfEylxojg0t1MD6Vw" at page 16, figures 6 and 7.
It's written "Thus the ratio (...) can be expressed as [itex]f_i=\frac{1}{2}(1+\cos E_s)[/itex]" where Es is the phase angle.

Thanks for helping...

Jeff
 
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The phase angle Es is [itex]\frac{t}{T} \cdot 360^o[/itex], where t is the elapsed time since the last full moon and T is the period (elapsed time between full moons, about 29.5 days). For example, Es is 0 at full moon, 90o at first quarter, 180o at new moon, etc.

If you want to work in radians rather than degrees (for example, using Excel or google for the calculation), then replace 360o with 2π in the formula for Es.

Hope that helps.

EDIT: my expression for Es is an approximation, assuming a circular orbit for the moon.
 
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I understand your approximation for Es. What I don't understand is the expression of fi given in my first post. Why it is 0.5 * (1 + cos Es) ?

Thanks
Jeff
 
Sorry for the delay. This is best explained with a figure, and it was not until just now that I had time to make a decent one.

MoonFraction.gif

r is the radius of the moon, so of course the entire disk has a diameter of 2r. And, as the figure shows, the illuminated portion viewed from Earth is [itex]r + r \cos {E_s} = r(1 + \cos{E_s})[/itex]. So the illuminated fraction is their ratio,

[tex]\frac{r(1 + \cos{E_s})}{2r} = \frac{(1 + \cos{E_s})}{2}[/tex]
 
It's very clear ! Thanks a lot Redbelly98 !

Best regards