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I was wondering how the distance in Mpc could be calculated between Earth and a galaxy using the luminosity provided? Do not be shy to use all math necessary :)

Clarification:

Luminosity is inversely proportional to the distance squared. I am wondering how that is used to calculate the distance to distant galaxies. They take the intrinsic luminosity of various stars within the galaxy to find the distance. But I wonder how.

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    $\begingroup$ I'm not sure what you're asking. Could you add more details? $\endgroup$
    – user21
    Jun 27 '19 at 16:20
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    $\begingroup$ @barrycarter Luminosity is inversely proportional to the distance squared. I was wondering how people would use that equation to calculate the distance to distant galaxies. They take the intrinsic luminosity of various stars within the galaxy to find the distance. But I wonder how.... $\endgroup$ Jun 28 '19 at 4:16
  • $\begingroup$ “Luminosity” is intrinsic brightness, and does not depend on distance. Observed (or apparent) brightness is inversely proportional to distance squared. $\endgroup$ Jul 2 at 0:43
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I can show you an ESA Series exercise I did a few months ago for an astronomy class.

Given the light curve of 12 cepheid variable stars in the galaxy M100 (which are very nice standard candles to measure large distances):

enter image description here

We can find the distance between us and the M100, a spiral galaxy in the Virgo cluster, using the apparent magnitude m and the absolute magnitude M.

First, we have to find the stars' absolute magnitude, but we can easily do so since we have their period of variation:

Period-luminosity relation in Cepheid variables: \begin{equation} M= -2.78\log_{10}P - 1.35 \end{equation}

Now we just need the apparent magnitude, which can simply average out by taking the highest and lowest value in each light curve. Once we have the two magnitudes, we can define the distance modulus as: \begin{equation} m-M=5\log_{10}\frac{d}{10pc} \end{equation}

Solve for the distance d and we have: \begin{equation} d = 10^{\frac{m-M}{5}+1} \end{equation}

Do this for all the 12 variable stars (the more you have the better), and average out the distance between us and M100!

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  • $\begingroup$ Is this how distance is computed using supernovae? And how about TRGB stars? $\endgroup$ Jul 1 at 21:03
  • $\begingroup$ @DaddyKropotkin this is essentially how most long distances are measured - that's why we like standard candles so much: we already have informations on their absolute magnitude, so by looking at their apparent magnitude we have intrinsic information on distance. Supernovae are much brighter so we can use them to calculate distance for even further away galaxies; further than that and we can only rely on estimations of redshift in light. $\endgroup$ Jul 1 at 21:43
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Distance to galaxies are measured using "standard candles": these objects are well-known astrophysical sources which have a known luminosity. Examples include Cepheid variables, RR Lyrae stars and Type 1a supernova.

The standard candle used differs based on the distance to the source we are observing.

Distances to relatively close galaxies are determined using Cepheid variables. These stars have a regular periodicity in their brightness, which is directly correlated to their luminosity. By observing them over a range of time, one can obtain their period, which in turn gives their luminosity (or absolute magnitude). This is in turn can give the distance using the distance modulus equation.

For galaxies much farther, it is very difficult to observe individual stars. So, the same technique is applied on Type 1a supernovae, which is caused by runaway thermonuclear fusion of a white dwarf due to its mass exceeding the Chandrashekar limit by accretion from a donor star. As the luminosity of all Type 1a supernovae is same, the distance modulus equation can be used to get the distance to the galaxies.

For galaxies even farther, Baryonic Acoustic Oscillations (BAOs) are used (also known as standard rulers), but the technique is well beyond the scope of this answer.

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