Earth's Surface Area that reflects light

[dsfranca]

Hi!
I was reading about the relationship between the surface area of Earth that reflects light and the one that radiates. As the whole Earth emits radiation, the area that would radiate is A=4\Pir², where r is the radius of earth. But the weird part is that the textbook says that only one quarter (\Pir) of the Earth's surface reflects light. Why isn't it half of the surface? Afterall half of the surface is illuminated by day!
Am I missing something or is the textbook wrong, as he uses this information many times!
Thanks

 

[Bob S]

The effective surface of the Earth that is being illuminated by sunlight at any given time is only pi r² (the cross section seen by the Sun), not 2 pi r² (half of 4 pi r²).

 

[dsfranca]

Thanks for the answer, but I still did not understand why the effective surface is only one quarter of the Earth's surface. Could you go into more detail please?
Thanks

 

[Andy Resnick]

I agree w/ Bob S, the book probably means that the *projected* area of a hemisphere exposed to a plane wave is pi*r^2 (i.e. a circle of radius 'r'). You have to take into account the orientation of the surface with respect to the incident wavefront.

Perhaps it makes more sense geometrically- when you are looking at the (full) moon you do not perceive it to be a hemispherical object, it appears to be a circle.

And, the area of a circle is 1/4 the area of a sphere.

 

[dsfranca]

Yes, I see what you mean, but wouldn't 2$2\pi r^{2}$ be a better approximation
in this case? Afterall, I am trying to determine the area of the earth's
surface that is reflecting light, and by common knowledge, one would
say that half of the Earth is always illuminated (half of the earth
is having a day while the other a night). Where is the flaw in this
argument and why is this not a better approximation?

 

[tiny-tim]

Hi dsfranca!

(have a pi: π and try using the X² tag just above the Reply box )

Suppose the Earth was a spherical mirror …

then it would reflect all the light falling on it …

but that would be the same amount of light as if the Earth was replaced by a flat disc of the same diameter …

so the "effective surface" is the area of that disc, which is πr², or A_Earth/4.

 

[dsfranca]

Hum! Everything makes sense now, how could I not have considered optics when talking about reflection! Thank you for the explanation tiny-tim!