Suppose we have two stars of apparent magnitudes m1 and m2 and absolute magnitudes M1 and M2 that are located at distances d1 and d2 from us. Remember the two equations that relate their brightnesses, luminosities and magnitudes:
m2 - m1 = 2.5 log (b1/b2)
M2 - M1 = 2.5 log (L1/L2)where b stands for brightness and L stands for luminosity.
The most luminous stars have luminosities of 100,000 times that of the sun. Therefore, we say that their luminosity is 105Lsun. We can compute their absolute magnitude using:
Msun - Mbrightest = 2.5 log (105L sun/1 Lsun) = 2.5 log (105) = 2.5 x 5 = 12.5
Mbrightest = +4.77 - 12.5 = -7.73
The faintest stars have a luminosity of .01 times that of the sun. Therefore, we say that their luminosity is .01 Lsun. We can compute their absolute magnitude using:
Msun - Mfaintest = 2.5 log (.01 L sun/1 Lsun) = 2.5 log (0.01) = 2.5 x -2 = -5
Mbrightest = +4.77 + 5 = 9.77
How bright would the Sun be if it were at the distance of Alpha Centauri?
The Sun's apparent magnitude is -26.8. Its absolute magnitude is +4.77.
The distance to Alpha Centauri is 1.3 pc.
Therefore, the apparent magnitude that the Sun would have, if it were located at 1.3 pc is:
msun at 1.3 pc = Msun -5 + 5 log d = 4.77 - 5 + 5 log(1.3)
msun at 1.3 pc = 4.77 - 5 + 0.57 = +0.34
so that its apparent magnitude would be +0.34, instead of -26.8!
How far away could we detect a star like the Sun with the Hubble Space Telescope?
With HST, it is possible to see Cepheids in the galaxy M100 which have apparent magnitudes of +26. Let us adopt this as the limit of apparent brightness that HST can "see". Therefore, our problem reduces to: at what distance would the Sun have an apparent magnitude of +26?
Msun = mHST limit + 5 - 5 log d
5 log d = mHST limit + 5 - Msun = +26 + 5 - 4.77 = 26.23
log d = 26.23/5 = 5.25
d = 105.25 parsecs = 176,198 parsecs = 176 kiloparsecs (kpc).
Note that 176 kpc is about 3 time the distance to the Large Magellanic Cloud.