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.
The sun is magnitude 4.8 at 10 pc. For it to be barely visible, it would be 6th mag, or 1.2 mags fainter. This corresponds to being 10^(0.4*1.2) = 2.5^1.2 = 3 times fainter. To be three times fainter, the star would need to be sqrt(3) further away, which corresponds to sqrt(3) * 10 pc = 1.7 * 10 pc = 17 pc away =~ 51 light years.
ReplyDeleteThat's a good exercise to try to find the sun's distance using magnitudes rather than just luminosities and read rates. I will try to recreate that independently. But using this method, did I make some ridiculously incorrect estimate? My answer is then a factor of 6 off not 3 but its still order of mag away. Or is the difference just due to intrinsic differences in the assumptions used in the different methods?
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