I will solve a problem similar to one done earlier: how far can a sun-like star be so it is visible to us?
5. The eye needs 10 photons to send a signal to the brain. From Pf Johnson's earlier post on OoM we calcuate the eye's area as .25cm^2 (correspond to an entrance diameter of .5 cm for a fully dilated square-approximation eye). Thus we need to define our sun-like star's distance such that it emits
10 photons/ .25cm^2 =40 photons/cm^2
By this we mean to take star's distance to us, R, and create a sphere of radius R with the sun at the center such that, given the star emits isotropically, we have 40 photons on each cm^2 of the sphere.
Then the total number of photons emitted by the sun's double equals 40 photons/cm^2 *R^2. We want the twin to be just far that we can still see it from a dark site (no light pollution). The energy of a single photon is E_n=h*c//Lambda, were h is Plank's constant and /Lambda~ 550 nm, in the middle of the visible spectrum.
The energy of a single photon h*c/(/lambda))=3.67*10^-12 ergs.
Total energy emitted by sun's double is n*E_n where n is number of photons emitted by sun's double. We can find luminosity from this since we have a time interval. Our eye has a read rate of ~24 fps.
To be visible, luminosity= n*E_n (in ergs)*24 (frames/second).
We know the sun's luminosity is 4*10^33 ergs/s.
4*10^33=n*3.67*10^-12*24, so n=4.5*10^43 photons
But n= 40 photons/cm^2 * 4pi*R^2, and we calculate R=28*10^19cm, or approximately 317ly. The actual distance the sun can be away and still visible is ~100 ly, so this is correct to a factor of 3.
I compared this to Pro Johnson's blog post OoM where he calculated the number of photons to be 10^41 not 10^43. However, Professor Johnson assume read rate of 100 fps not 24, and also reached luminosity 10^33 while I used 4*10^33. That introduces one order of magnitude difference. The 2nd comes because there was a calculation error in OoM calculating the total energy. It should come out to 10^30 not 10^31 erg. But these are all order of magnitude estimations and mistakes, which is why our answer is correct to an OoM.
