# Do enough photons from Proxima b make it to Earth that its surface could be imaged directly with a huge, but not astronomically so, telescope?

I hope I didn't miss someone answering something similar. I was doing a small thought experiment. And I was wondering whether you kind people would check up on my math.

Say we want to build a telescope to see the surface of the extrasolar planet Proxima Centauri B. Well, it is $$4 \times 10^{13}\ \mathrm{km}$$ away and, if we assume it as big as Earth, approximately 12 742 km large in diameter. (*)

Let us not be greedy and just say we want from our telescope a 10 x 10 resolution, i.e. we need to be able to resolve 1274 km "points" on the planet's surface. The angular size of such a point is, therefore, $$3.14 \times 10^{-12}$$ radians.

Using $$D = \frac{\lambda}{R}$$ for the telescope's needed size, we get to rather comfortable 185 km, which is a lot for practical application, but theoretically reasonable.

Maybe we can see it in classical terms, but real light is quantum. Will we see it photon wise? That is, will enough photons get into our telescope to make a useful image in a time-frame we can afford to wait on? Let us forego reality for a second and say that PC b reflects as many photons per second per square meter as Earth or $$\approx 1.26 \times 10^{21}$$, considering our rather large "pixel" we should get somewhere around $$1.23 \times 10^{29}$$ photons per 1274 $$\mathrm{km^2}$$ area. Using area for a half-sphere at the distance of $$4 \times 10^{13}\ \mathrm{km}$$ we should still get about 12 photons per km^2/s.

Which again seem fairly reasonable.

Huh, I didn't expect it to be within the realm of realistic possibility. Is my conclusion correct, that we could still collect enough photons to get an image?

(*) My (The_Sympathizer) note: Proxima b is actually somewhat larger.

• I think I know what you mean and can make an edit to fix this up. Would you mind? – The_Sympathizer Apr 5 at 5:49
• @The_Sympathizer of course not, I would be very grateful. I am not 100% sure what is being asked of me – Dvorkam Apr 5 at 6:47
• Done. What do you think? – The_Sympathizer Apr 5 at 7:49
• @The_Sympathizer It is much more understandable, thank you :) I hope you didn't waste your time and the question will be "unclosed" – Dvorkam Apr 5 at 7:56
• you're welcome :) I also suspect the original problem was that this site apparently demands a very high level of English proficiency from those who submit their questions and write-ups, and I take it you are a little ❤️Foreign Peep❤️, right (i.e. foreign to the 'Anglosphere') (❤️#meep)? And that admittedly makes it very difficult for especially a (and growing) global audience whose English may not be excellent - which I find especially sad here given that your English was enough for me to make out the message. – The_Sympathizer Apr 5 at 7:57

I don't quite agree with your numbers. In order of magnitude terms, if the planet receives 1400 W m$$^{-2}$$ and reflects 30% (which is implicit in your question), then if the average photon is say 600 nm (questionable given spectral type of Proxima Cen), there are $$\sim 4\times 10^{21}$$ photons per second per square metre incident on the planet. The area this incident upon is $$\pi R^2$$, and with 30% albedo, means a luminosity of $$1.6\times 10^{35}$$ photons/s reflected into space.
The number heading towards us depends on the phase the planet is observed. A fully illuminated disc would be like a uniformly emitting sphere with twice the luminosity quoted above. The opposite phase would result in zero photons. I'll just assume half illumination and assume this is equivalent to the luminosity quoted above radiating into $$4\pi$$ solid angle.
In which case, dividing by $$4\pi d^2$$, we get 8 photons per square metre at the Earth.