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As in, from what the planet emits and re-radiates out into space (this is going to be important if we are to image any Earth-like planet)

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Size? Albedo? Illumination? Phase? Then it is a almost trivial calculation of the areal ratio between the planets radiating surface and the surface of a sphere centered on the planet that includes the Earth and division by energy to convert flux to photons. – dmckee Aug 29 '11 at 15:57
There is going to be a strong dependance on wavelength also. – Nic Aug 29 '11 at 18:08
Hm - just assume full phase, Earth-size, average Earth albedo, and 50% illumination over the entire visible to IR range – InquilineKea Aug 30 '11 at 2:23
perhaps, you could invert the question for better answers. Assuming an observer is 50 light years away, in the plane of the solar system (or perpendicular, depending on what you want) and observing earth, what is the light flux of earth he would see. – Jus12 Aug 30 '11 at 11:08
Okay, question inverted – InquilineKea Aug 30 '11 at 15:51
up vote 6 down vote accepted

A 0 mag star gives you 1000 photons/cm^2/second/Angstrom at the top of the atmosphere in the visible band (astronomers are nostalgic about units!)

So if you imagine an earth-like planet at an earth-like distance you can, from the ratio of the cross section area of the planet and the surface area of a sphere with a radius of the planets orbit, you could work out what proportion of the stars output hits the planet.

Then pick a reflectivity for the planet and you can work out how much like bounces off it.

Then you can use the normal magnitude distance relation to work out how much fainter it would be at a given distance

edit: some numbers
The earth is around 150M km from the sun and has a radius of 6400km, so it intercepts roughly 6400^2/150M^2 = 2E-9 of the sun's light. It has an albedo of 0.3 so in the best case about 5E-10 of the stars light is reflected into space.

Just to simplify slightly, assume the star is at a distance of 10pc (32 lyr) then a sun like star has an apparent and absolute magnitude of 4.8 in the V band.

This means it is 2.51^4.8 = 1/85 as bright as our 0mag example, or around 12 photons/s/cm^2/angstrom. The v band is 88nm (or 880angstroms wide) so we have around 10,000 photon/s/cm^2 or 100Million photons/s/m^2 from the star.

Now remember only 5E-10 of these are reflected from the planet so we would have to detect 0.05 photons/m^/s and somehow avoid the billions more coming directly from the star.

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Hi Martin, would be great if you can also fill in the missing details and answer the OP's question with more concrete value.. OP already mentioned "earth-like", so how about assuming a sun-earth system. Otherwise, IMO, the answer is not complete. – Jus12 Aug 30 '11 at 11:04

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