Solar Constant: Earth's Receiving Intensity

[Peter G.]

Hi,

So, I understand the concept that the power radiated by the sun spreads out as a sphere with a radius equal to 1 AU. Therefore, in order to find the intensity, we have to divide the power radiated by the surface area of the sphere. I thought the result of this calculation would yield the intensity per meter squared, that is, in an area of 1 m by 1 m on the surface of the Earth.

However, I was reading some websites and I got really confused. They claim 1380 Wm^-2 is the solar constant provided we treat the Earth as a sphere but 340 Wm^-2 if we consider only a "disc" is receiving the energy at any given point.

So, is 1380 Wm^-2 received at every point on Earth or the Earth itself receives 1380 Wm^-2?

Thanks!

 

[Simon Bridge]

Hi,

So, I understand the concept that the power radiated by the sun spreads out as a sphere with a radius equal to 1 AU.

The energy spreads our spherically - but the radius depends on where you do the measurement.

Therefore, in order to find the intensity, we have to divide the power radiated by the surface area of the sphere.

That is to say, the intensity at the Earth's orbit.

I thought the result of this calculation would yield the intensity per meter squared, that is, in an area of 1 m by 1 m on the surface of the Earth.

Yes it will ... and the solar flux will be radial through each square meter.

However, I was reading some websites and I got really confused. They claim 1380 Wm^-2 is the solar constant provided we treat the Earth as a sphere but 340 Wm^-2 if we consider only a "disc" is receiving the energy at any given point.

These sites are taking into account the fact that not everywhere on the Earth is directly facing the Sun. (Not all the same distance from the Sun either.)