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Name: Brian
Status: student
Grade: 9-12
Location: MO
Country: N/A
Date: 6/26/2005

How powerful must a light on earth be to be seen in space? For example, how far away could you see a 1 kilowatt light versus a 350,000 kilowatt light?

Dear Brian,

This is a straightforward calculation except for the precise definition of what you mean by "seen". I will assume the detector is a normal human eye fully dilated to a diameter of 8 mm and sensitive to a light flux of 10 photons/second. The human eye at its most sensitive can sometimes detect a single photon, so a sensitivity to 10/s seems reasonable. I assume you know that light is made up of photons as postulated by Max Planck around the turn of the century and proven by Einstein's theory of the photoelectric effect published during his magic year of 1905. Each photon carries an energy E = hf where h is Planck's constant (6.6 E-34 Js) and f is the frequency of the light wave. Visible light is a spectrum centered near a wavelength of 500 nm (nm = nanometers; 1 nm = 1E-9 m) and a frequency of 6 E14/s. The energy carried by a single photon of that frequency is 4.0 E-19 J.

Note that I am writing large (and small) numbers using the old computer notation 100 = 1E2 and 0.001 = 1E-3. I will show powers by rr = r^2.

Assuming the light source sends the light out equally in all directions, at a distance R it will uniformly illuminate a sphere of radius R, which has an area of 4 pi R^2. An eye of radius r will have an area of pi r^2 so the fraction of photons hitting the eye is (r/R)^2/4. The number of photons of energy 4.0 E-19 J produced per second by a P kW = 1000P J/s light is 2.5 P E21 Ph/s. Finally, where N is the number of photons detected by the eye per second:

N Ph/s = 2.5 P E21 Ph/s (r/R)^2 /4 or

R^2 = 6.2 P

E20 ph/s r^2 m^2 / (N Ph/s)

Plugging in P = 1 kW, r = 0.004 m = 4 mm, we find R = 3.2 E7 m = 3.2E4 km = 20,000 mi.

Notice that the distance increases linearly with the radius of the detector and with the square root of the power and decreases like the inverse of the square root of the sensitivity of the detector. If you plug in a bunch of numbers, I think you will develop a good feeling for how this works. For example, using a 350,000 kW light (P = 350,000) increases R to 1.9 E10 m = 11,400,000 miles.

Best, Dick Plano, Professor of Physics emeritus, Rutgers University

This is mostly dependent upon the sensitivity of the detector and the length of time that the detector can be kept operational and pointed accurately pointed at the source. The smallest unit of "light"-- I use the quotes because light is just one small part of the electromagnetic spectrum -- is one quantum. Assuming that the detector is 100% efficient -- that is, it can "count" every photon that strikes it -- then it is (only!!) a matter of how many photons it takes to generate a signal in the detector. In the reverse configuration the Hubble telescope can "watch" a distant source for weeks to compile a picture of the source. Quite remarkable.

This remarkable telescope has produced more data about the Universe than all telescopes in all prior history. Let us hope it meets a better fate than to let it "die", which is the present intent of NASA and the federal government.

Vince Calder

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