*Originally posted by joe shmo*

**D= distance to object
**

R_sun = radius of sun

P_sun = Power of sun

q" = heat flux at D

A = Surface area of sphere at distance D

D= R_sun/tan(15 deg) = R_sun/(2-sqrt(3))

q" = P_sun/A = P_sun/(4*pi*D^2) = P_sun/(4*pi*(R_sun/(2-sqrt(3)))^2)

...= (7-4*sqrt(3))/(4*pi)*P_sun/(R_sun)^2

Here are the figures given by this link:

1370 W/M^2 reaches Earth (on top of the atmosphere) and the sun puts out 63E6 W/M^2 at the surface.

The square root of the ratio of those two number, 45985.4^-2= 214.44 and using the center of the sun as reference, 695500 Km times 214.4= 149.1E6 km, very close to the known distance from Earth to sun.

The surface area of the sun is about 6.08E18 M^2. The surface area of a sphere 2.6E9 M radius is 8.511E^19 M^2. The ratio is ~14:1 so 63E6 W/M^2 / 14 gives the energy intercepted by the device as 4.5 E6 W/M^2.

4.5 Megawatts per square meter. So a panel getting 100 % conversion efficiency could generate 1 Gigawatt with a panel 15 by 15 meters!

Panels about 1.7 Km by 1.7 Km running 100% efficiency of conversion would supply all the energy used by everyone on Earth (~14 Terawatts).

At 25% all it would take is a set of PV's 3.4 Km X 3.4 Km.

We wouldn't even be able to see it very well!

If that was a Fresnel lens able to take the heat, can you imagine that as a weapon?

So the next question: What would be the temperature of the device if it intercepted one square meter at that distance and was a 100% efficient black box?

14 TW energy beam hitting the Earth!