Illuminated fraction of the Moon

[JeffOCA]

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 $f_i=\frac{1}{2}(1+\cos E_s)$" where Es is the phase angle.

Thanks for helping...

Jeff

 

[Redbelly98]

The phase angle E_s is $\frac{t}{T} \cdot 360^o$, 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, E_s is 0 at full moon, 90^o at first quarter, 180^o 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 360^o with 2π in the formula for E_s.

Hope that helps.

EDIT: my expression for E_s is an approximation, assuming a circular orbit for the moon.

 

[JeffOCA]

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

 

[JeffOCA]

Anyone ?

 

[Redbelly98]

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.

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 $r + r \cos {E_s} = r(1 + \cos{E_s})$. So the illuminated fraction is their ratio,

$$\frac{r(1 + \cos{E_s})}{2r} = \frac{(1 + \cos{E_s})}{2}$$

 

[JeffOCA]

It's very clear ! Thanks a lot Redbelly98 !

Best regards