- 19 Nov '06 00:12Hi all, here is a nice arithmetic problem: There is talk (Astronomy magazine, Dec. 2006 issue) of a probe to the sun. It will approach within 4 solar radii and has extensive heat shielding (understatment of the year!). The sensors are behind this heat shield cone which is 2.7 meters across and about 4 meters long. This cone will be pointed directly at the sun to shade the delicate equipment which will keep everything on the backside under about 45 degrees C. So here is the

question: Given there is about 1355 watts per square meter ariving on top of the earth's atmosphere, how many watts are impinging on the surface of the cone when it is at a distance of 4 solar radii at its closest approach? - 19 Nov '06 09:43Assuming it approaches within 4 solar radii of the surface of the sun

then it is on a sphere of radius 5 solar radii

or 3,475,000 km

Earth's distance from the sun ranges between 146-152,000,000 km

Assuming 149,000,000 km

and calculating the surface areas of the spheres (or just comparing 3.475 squared and 149 squared) you find the 1 AU sphere is a factor of 1838 times bigger than that of the 5 solar radii.

So about 2,500,000 watts per metre squared - 19 Nov '06 11:29 / 2 edits

I got a lot more than that.*Originally posted by aging blitzer***Assuming it approaches within 4 solar radii of the surface of the sun**

then it is on a sphere of radius 5 solar radii

or 3,475,000 km

Earth's distance from the sun ranges between 146-152,000,000 km

Assuming 149,000,000 km

and calculating the surface areas of the spheres (or just comparing 3.475 squared and 149 squared) you find the 1 AU sphere ...[text shortened]... f 1838 times bigger than that of the 5 solar radii.

So about 2,500,000 watts per metre squared

Using your method, work back to earth's orbit, How many watts/M^2 do you come up with?