Solar constant and the Earth surface

[pincopallino]

1. We all know that the solar constant is 1350Watt.m^-2, which is related to a unit surface perpendicular to the solar rays. The question is: which is the average solar intensity on the Earth surface?


2. first, the portion of the EArth surface is half the EArth sphere. second, we have to take into account that at different latitudes the solar rays have difference incidence angles.
to calculate the power reaching the surface I have integrated the solar constant over the half sphere, which is the same as multiply the solar constant to the maxiun disc, i.e. pi.R^2, where R s the Earth radius.
to have the average intensity on half the Earth surface I have (solar constant)*pi*R^2/(2.pi.R^2) = half solar constant =675watt.m^2

BUT THE TEXTBOOK REPORTs 610 watt.m-2


3. in your opinion , where is the error?
has anyone a clue?
why the factor is 45% instead of 50%?
thkx

 

[QuarkCharmer]

What did you put as the radius of the earth?

 

[pincopallino]

What did you put as the radius of the earth?

the radius is irrelevant, as you may notice that it simplifies n the last formula I have written

 

[Quinzio]

I am afraid we have to get very pedantic here.
You have assumed that the sun and the Earth are both ideal points, but they are not.
Or better, you assumed they have the same radius.
Although very far away, the sun is a very big ball out there.
I have done no exact calculation, but with just some operations on my pocket calculator, I think we can justify some % of difference.

I hope this makes sense.

 

[pincopallino]

I am afraid we have to get very pedantic here.
You have assumed that the sun and the Earth are both ideal points, but they are not.
Or better, you assumed they have the same radius.
Although very far away, the sun is a very big ball out there.
I have done no exact calculation, but with just some operations on my pocket calculator, I think we can justify some % of difference.

I hope this makes sense.

I have not made this assumption..
R is the radius of the Earth in the formula. no Sun radius in there...
I have just assumed that, as the Earth-Sun distance is 1.5exp11meters, the sun rays arrives in the Earth parallel (whish is true if we consider our shadows under the sun!...)

 

[Quinzio]

I have not made this assumption..
R is the radius of the Earth in the formula. no Sun radius in there...
I have just assumed that, as the Earth-Sun distance is 1.5exp11meters, the sun rays arrives in the Earth parallel (whish is true if we consider our shadows under the sun!...)

Well, I think my idea will not explain your % difference, but anyway.
The sun is roughly 100 times the Earth (radius), and the sun-earth distance is roughly 10.000 the Earth diameter.
The sun rays are not really parallel but they form a cone of 1/100 radians.
The Earth circle that sees a dawn or a sunset, they don't see the entire sun, but only a part, eg. a man that looks at the sun during the sunset sees only half of the sun disk, thus he doesn't get the full solar radiation.

But this is not enough to justify your difference, anyway. I don't know.

 

[willem2]

Your textbook is wrong. It's well known that the average over the entire Earth is 1/4 of the solar constant, because a unit sphere has a surface area 4 times bigger than a unit disk (see wikipedia article on solar constant). The surface area of the daytime section is obviously half that.

 

[thopsy]

Atmospheric attenuation!