 Posers and Puzzles

1. 06 Mar '08 01:101 edit
Carl Sagan is standing under a 1000 watt output streetlight on a pole 56 feet high. Carl is 6 feet high. Being an astronomer, he wonders how far away from the sun you would have to be for the sun to have the same illumination as Carl receives from that lamp.
Who can come up with the easiest way to figure it out?
2. 06 Mar '08 01:43
Originally posted by sonhouse
Carl Sagan is standing under a 1000 watt output streetlight on a pole 56 feet high. Carl is 6 feet high. Being an astronomer, he wonders how far away from the sun you would have to be for the sun to have the same illumination as Carl receives from that lamp.
Who can come up with the easiest way to figure it out?
by energy flux.
Energy flux from the sun in Earth's surface is 1300 W/m^2 . And sun is 1.5e8 Km from Earth.
Find the flux from the lamp (with solid angles -> a sphere has an angle of 4*Pi) at the given distance, and voila. (the flux coming from the lamp and the flux of a sphere with radius = 50 feet is the same - just find out the proportion between areas, and compare to the sun)
3. 06 Mar '08 02:01
by energy flux.
Energy flux from the sun in Earth's surface is 1300 W/m^2 . And sun is 1.5e8 Km from Earth.
Find the flux from the lamp (with solid angles -> a sphere has an angle of 4*Pi) at the given distance, and voila. (the flux coming from the lamp and the flux of a sphere with radius = 50 feet is the same - just find out the proportion between areas, and compare to the sun)
So give me an answer. BTW, here is a jennings constant: suppose you don't remember how to calculate the surface area of a sphere, take the surface area of a cube the same diameter as the sphere and multiply it by 0.52333333.
4. 06 Mar '08 02:351 edit
Originally posted by sonhouse
So give me an answer. BTW, here is a jennings constant: suppose you don't remember how to calculate the surface area of a sphere, take the surface area of a cube the same diameter as the sphere and multiply it by 0.52333333.
the flux is power / area
for the lamp 1000 / (4*Pi*r^2)
for the sun 3.9e26 / 4*Pi*r'^2)
(sun is like a 3.9e26 W lamp - if you multiply 1300W by 4*Pi*1.5e11 squared, you get the result - approximately)
Now... how far do you have to be from the sun the get the same flux per square meter of a lamp 50 feet away?
5. 06 Mar '08 02:38
Originally posted by sonhouse
So give me an answer. BTW, here is a jennings constant: suppose you don't remember how to calculate the surface area of a sphere, take the surface area of a cube the same diameter as the sphere and multiply it by 0.52333333.
it's easier to know the area of a sphere is 4*Pi*r^2 then memorize that strange number.
6. 06 Mar '08 03:42
Originally posted by sonhouse
...take the surface area of a cube the same diameter as the sphere ...
Cubes don't have diameters.
7. 06 Mar '08 10:06
Originally posted by Gastel
Cubes don't have diameters.
Side length then. The same kind of thing can be used in 2 dimensions, like you don't want to do PI R squared,etc for the area of a circle, do a square the same size and multiply by .78, gives the area of a circle. Anyway, nobody has given an answer yet!
The idea of the surface area of a sphere was the way I did it, that and knowing a couple of facts about light propagation makes it a pretty easy problem.
8. 06 Mar '08 12:24
Originally posted by sonhouse
Side length then. The same kind of thing can be used in 2 dimensions, like you don't want to do PI R squared,etc for the area of a circle, do a square the same size and multiply by .78, gives the area of a circle. Anyway, nobody has given an answer yet!
The idea of the surface area of a sphere was the way I did it, that and knowing a couple of facts about light propagation makes it a pretty easy problem.
The answer is just solve the ultra simple system I gave you. Doing by head the calculations, you should be around Mars to get the same energy as the guy under the lamp.
Unfortunately I didn't have the patience to investigate how many meters 50 feet are.
9. 06 Mar '08 13:20
50 feet = 15 metres is a pretty accurate approximation.
10. 06 Mar '08 13:26
Originally posted by wolfgang59
50 feet = 15 metres is a pretty accurate approximation.
either way, the answer is 6.3 e11 * R, where R is the distance to the lamp ( if I got the calculations right)
11. 06 Mar '08 22:05