The Moon Seen by Earthshine

https://isaacphysics.org/questions/moon_earthshine?stage=a_level
For part a, why is the answer not just 0.3*1370*(2*pi*r_m^2/4*pi*d^2)*0.12
Where r_m = 1737km and d= 384400km.

Reply 1

Joinedup

Did you convert units to keep everything consistant?

Reply 2

mosaurlodon

Yeah I believe so - I get the answer as 5.0x10^-4 but it still appears to be incorrect.

Reply 3

mqb2766

Original post by mosaurlodon

Yeah I believe so - I get the answer as 5.0x10^-4 but it still appears to be incorrect.

You may be overcomplicating it a bit. The inverse square rule is basically
intensityA / intensityB = distanceB^2 / distanceA^2
https://en.wikipedia.org/wiki/Inverse-square_law

Why would the moons radius be involved? If the question simply asked for the flux density reflected off the earth (at the earths surface), what would that be?

(edited 1 month ago)

Reply 4

mosaurlodon

Thank you I've managed to get the answer now!
For part b, however I employed the same working in part a, but got this:

I just did 0.3x0.12x1370*(r_e^2)/(d^2) - which was the answer in part a - * (r_m^2)/(d^2)
and I dont really know what it means by 'some light is reflected before it reaches the earth' - haven't I accounted for this?

(edited 1 month ago)

Reply 5

mqb2766

Original post by mosaurlodon

Thank you I've managed to get the answer now!
For part b, however I employed the same working in part a, but got this:

Right idea for the inverse square, but in part A you use the fact that 0.3 is reflected away from the earth, so in part B, how much will get to the surface?

Reply 6

mosaurlodon

Ohhh I understand the specific wording they used now.
Thank you so much :smile:

Reply 7

Hihobyu

Original post by mqb2766

Right idea for the inverse square, but in part A you use the fact that 0.3 is reflected away from the earth, so in part B, how much will get to the surface?