DISTANCE MODULUS

 

In the magnitude handout, we distinguished between two different magnitudes: the apparent magnitude, which indicates how bring an object appears to be, and absolute magnitude, which indicates a star’s true brightness, or luminosity.  The only reason those two numbers are different for various stars is because every star is not the same distance from us.  We can take advantage of this by using the difference between a star’s apparent magnitude and absolute magnitude to actually calculate the distance of the star.  This difference is called the distance modulus, m – M.

Recall that apparent magnitude is a measure of how bright a star appears from Earth, at its “true distance,” which we call D.  Absolute magnitude is the magnitude the star would have if it were at a standard distance of 10 parsecs away.  So this presents us with three general possibilities for the value of the distance modulus:

·         If the star is exactly 10 parsecs away (rare, but it does happen), the absolute magnitude will be the same as the apparent magnitude.  The apparent magnitude is actually a good indicator of true luminosity.  Thus, if m – M = 0, then the distance D = 10 pc.

·         If the star is closer than 10 parsecs, then the star will appear deceptively bright; its apparent magnitude will be too bright to tell us its true luminosity.  The star looks brighter than it actually is.  Remember that the magnitude system is “backwards,” in that lower numbers mean brighter stars.  Therefore, in the case where the star is closer than 10 parsecs, the apparent magnitude will be a lower number (brighter) than the absolute magnitude, and m – M will be a negative number.  So if m – M < 0, then the distance D < 10 pc.

·         If the star is farther than 10 parsecs, then the star will appear deceptively dim; its apparent magnitude will be too dim to tell us its true luminosity.  The star actually is brighter than it looks.  In the case where the star is farther than 10 parsecs, the apparent magnitude will be a higher number (dimmer) than the absolute magnitude, and m – M will be a positive number.  So if m – M > 0, then the distance D > 10 pc.

The above arguments are good for making a rough estimate of a star’s distance, but how can we find a more precise value?  We have options similar to those we explored in the magnitude handout: a quick method involving a single sophisticated equation, or a slower method involving simpler math

f we know the apparent magnitude of a star (just by looking at it) and have some way to deduce the absolute magnitude (by means of various techniques to be discussed later), we can compute a number known as the distance modulus, m-M. This distance modulus can be converted into an actual distance if we remember the following:

  • a difference of 5 magnitudes corresponds to a factor of 100 times in brightness.
  • apparent magnitude gives us an idea of the stars apparent brightness, how bright the star appears to be from Earth, at its true distance, D.
  • absolute magnitude gives us an indication of the star's true brightness, also called luminosity, by telling us how bright the star would appear if it were at a standard distance of 10 parsecs.
  • the dependence of apparent brightness on luminosity and distance B = L / D2

Remember that B is the apparent brightness of the object, L is the true brightness (the luminosity) of the object. Since the absolute magnitude is simply related to the brightness at a distance D = 10 parsecs (pc), the distance modulus can be related to the ratio between two values of B, one at D = the true distance from Earth (corresponds to m) and one at 10 pc (corresponds to M). Thus

m - M transforms into (B at 10 pc) / (B at true distance D)

Now (B at 10 pc) = L / (10 pc)2 and (B at D) = L / (D)2

And (B at 10 pc) / (B at true distance D) = (L /102) / (L / D2)

Since we are talking about the same star, the values for L cancel, and we can then solve for D:

(B at 10 pc) / (B at D) = D2 /102

D2 = 102 X (B at 10 pc / B at D)

so D = 10 parsecs X square root (brightness ratio)

Our most common method of using this technique is to somehow discover a value for M for a star, get its apparent magnitude m by observation, and use the distance modulus to get a distance. But we can also turn it around: if we know the distance and observe the apparent magnitude, then we can compute first a brightness ratio, then a distance modulus, then an absolute magnitude!

Some examples of how to use the distance modulus are given below:

Example 1: Absolute magnitude known, Distance unknown

We discover that a star with an apparent magnitude of +6 has an absolute magnitude of -1. How far away is this star?

Compute the distance modulus: m - M = 6 - (-1) = 7

Break this into 1's and 5's: 7= 5 + 1 + 1

Now do the transformation from a magnitude difference to a brightness ratio. For every 5 write 100, for every 1 write 2.5, for every + write a X.

Brightness ratio: B(at 10 pc) / B(at D) = 100 X 2.5 X 2.5 = 625

Now compute distance: D = (10 pc)(square root(625))

D = 10 pc X 25 = 250 parsecs

The star is 250 parsecs from us.

Example 2: Distance known, Absolute magnitude unknown

A star that we know to be at a distance of 6250 parsecs has an apparent magnitude of 16. What is the star's Absolute magnitude?

This is a little more challenging because we have to work backwards. Start with the distance formula, but solve for the brightness ratio:

D = 10 pc (square root(brightness ratio)) becomes (square root(brightness ratio)) = D / 10

Square both sides: brightness ratio = (D / 10)2 = (6250 / 10)2

brightness ratio = (625)2 = 390,625

 

Now the tricky part: we have to "break this down" into 100's and 2.5's. Here's how:

Divide 390,625 by 100: 390,625 / 100 = 3906 And again: 3906 / 100 = 39

Now we're below 100, so we start dividing by 2.5:

39 / 2.5 = 15.6... 15.6 / 2.5 = 6.24... 6.24 / 2.5 = 2.5... 2.5 / 2.5 = 1... we're done!

So we can break 390,625 into approximately 100 X 100 X 2.5 X 2.5 X 2.5 X 2.5

(It's not perfect, but we're not exactly working for NASA, either...)

 

Now do the transformation to get a distance modulus

Every 100 becomes 5, every 2.5 becomes 1, every X becomes +

m - M = 5 + 5 + 1+ 1+ 1+ 1 = 14 = (16) - M so M = 2

The star's absolute magnitude is 2. How does this star's luminosity compare to the luminosity of the Sun, at an Absolute magnitude of 5?

Compute the difference in Absolute magnitudes: MSun - Mstar = 5 - 2 = 3

Break this down and transform it: 3 = 1 + 1 + 1 becomes 2.5 X 2.5 X 2.5 = 15.85

So the luminosity ratio is 15.85 = Lstar / LSun and

Lstar = 15.85 LSun and the star is 15.85 times brighter than the Sun.

Example 3: Spectroscopic Parallax

Examining a star's spectrum, we find out that the star is a main sequence star with a temperature of 4000 K. According to the H-R Diagram, what is the Absolute magnitude of this star?

Look at the graph you made for Homework #2. The stars with a temperature of 4000 K tend to cluster around a value of Absolute magnitude +7.

How does the luminosity of such a star compare to the luminosity of the Sun?

Compute the difference in Absolute magnitudes: Mstar - MSun = 7 - 5 = 2 = 1 + 1

Transform it: 2 = 1 + 1 becomes 2.5 X 2.5 = 6.25

So the luminosity ratio LSun / Lstar = 6.25

LSun = 6.25 Lstar and the Sun is 6.25 times brighter than the star.

 

The star has an apparent magnitude of 10. How far away is the star?

Compute a distance modulus:

m - M = 10 - 7 = 3 = 1 + 1 + 1

Transform into a brightness ratio:

Brightness ratio = 2.5 X 2.5 X.2.5 = 15.625

This is close to 16. (Remember, we're not finding the cure for cancer here).

D = (10 pc)(square root(16))

D = (10 pc) (4) = 40 parsecs

The star is 40 parsecs away!

 
Updated 8/27/99
By James E. Heath
  
 
 

Copyright Ó 1999 Austin Community College