# Some HS math:

sonhouse
Posers and Puzzles 03 Sep '11 22:37
1. sonhouse
Fast and Curious
03 Sep '11 22:37
So I am reading this sci fi story, this device is close to the sun, so close that the sun subtends a full 30 degrees across the sky. 1: How far from the sun is this device,
and 2: how many watts per square meter would the device intercept? I don't say absorb because that would involve other problems, just what is the intensity of the light hitting the device in terms of one square meter?
2. forkedknight
Defend the Universe
08 Sep '11 16:011 edit
Originally posted by sonhouse
So I am reading this sci fi story, this device is close to the sun, so close that the sun subtends a full 30 degrees across the sky. 1: How far from the sun is this device,
and 2: how many watts per square meter would the device intercept? I don't say absorb because that would involve other problems, just what is the intensity of the light hitting the device in terms of one square meter?
Solving a 15,75,90 triangle with shortest side 696,000km (radius of sun) results in a distance of 2,598,000km to the center of the sun.

Subtracting the radius from that gives 1,902,000km to the nearest edge of the sun.

A bit too close for comfort ðŸ™‚

I don't feel like solving the insolation problem. It would be a matter of converting insolation at earth (1.366kW/m^2, according to wikipedia) from a distance of 1AU to 1.9E6 or 2.6E6 km. I'm not even sure which of those you would use, since the ratio of the radius of the sun vs the distance to the sun is so large in this problem.
3. joe shmo
Strange Egg
08 Sep '11 17:041 edit
D= distance to object
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
4. sonhouse
Fast and Curious
10 Sep '11 21:21
Originally posted by joe shmo
D= distance to object
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!