Mathematical Applications

Formula - Milky Way

The first two formulas described below were already mentioned in the course, but they tend to pop up again in this part of the course, since there is a lot of talk about distance and magnitudes.

Parallax - the apparent shifting of stars caused by the motion of the Earth around the Sun.
Formula: d=1/p or p=1/d where:

d = distance measured in parsecs (pc)
p = parallax shift measured in arc seconds (")

Typical Problems

1. A star is observed to have a parallax shift of .029 milli-arcseconds. How far away is the star?

Solution: First you'll have to convert the parallax value from milli-arcseconds to arc-seconds, by dividing the number by 1000. So p = .029/1000 = 2.9 x 10^-5
d = 1/p = 1/(2.9 x 10^-5) = 34,000 pc

Magnitude - Distance Formula - used to give the relationship between the apparent magnitude, the absolute magnitude and the distance of objects.
Formula: m - M = -5 + 5 Log (d) where:

m = apparent magnitude
M = absolute magnitude
d = distance measured in parsecs (pc)

The formula use the Log function. This is the base-10 log function. Many calculators also have what's called a Natural Log, usually abbreviated Ln - don't use this. When using this formula, you have to remember the order of multiplying things and adding/subtracting things. Multiplication always is before addition or subtraction.
To get values for d, you'll need to use the "un-log" function, which is 10^x, or "10 to the power of a number".
Also be careful about the signs in this formula (+/-) and make sure you remember to treat them properly.

Typical Problems

1. A Cepheid has an apparent magnitude of +15.2. For a star with its pulsation period it should have an absolute magnitude of -4.2. How far away is it?

Solution: you are looking for d in this case. Put the numbers you're given into the formula
m - M = -5 + 5 log (d)
15.2 - (-4.2) = -5 + 5 log (d)
15.2 + 4.2 = -5 + 5 log (d)
19.4 = -5 + 5 log (d)
19.4 + 5 = 5 log (d)
24.4/5 = log (d)
4.88 = log (d)
To "un-log" (d), take 10 to the power of 4.88
d = 10 ^4.88 = 75,900 pc

2. A type II supernova just went off in the Andromeda galaxy. These supernova usually reach an absolute magnitude of around -17 when they are their brightest. The Andromeda galaxy is 730,000 pc away. How bright will the supernova be in the night sky (or what is its apparent magnitude)?

Solution: now you are looking for m. Again, put in the numbers and start calculating
m - M = -5 + 5 log (d)
m - (-17) = -5 + 5 log (730,000)
m + 17 = -5 + 5 x 5.86 (the log of 730,000 = 5.86)
m + 17 = -5 + 29.3
m + 17 = 24.3
m = 24.3 - 17
m = 7.3

Kepler's Third Law for Galaxy Masses - Just like the previous version of Kepler's third law, but here the value for "k" is defined as the mass within a certain distance from the center of the galaxy ("a" in the formula). This formula is really only useful for estimating the mass of a galaxy that has organized motion, like our Milky Way.
Formula:M_galaxy = a³ / P² where:

M_galaxy = mass of the galaxy contained within the limit given by "a", values are in terms of the Sun's mass
a = the average distance they are apart, given in A. U.s
P = the period of their orbits given in years

The only complication with this formula is that the units for "a" and "P" are rather inconvenient for measuring distances within a galaxy. They are just too small! But that's the way it works.

Typical Problems

1. If the outermost part of a galaxy is 16 kpc from the center and it takes 350 million years for that part of the galaxy to orbit around once, how much mass is contained within that galaxy?

Solution: Before you can do anything, you need to convert the distance (16 kpc) into AU. 1 pc = 206265 AU, so multiply this with 16,000 (since a kpc = 1000 pc) to get the distance in AU.
a = 16,000 x 206265 = 3.3 x 10⁹ AU.
Also make sure you use 350,000,000 for the value of "P", not just 350
Let's put these into the formula:
M_galaxy = a³ / P²
M_galaxy = (3.3 x 10⁹)³ / (3.5 x 10⁸)²
M_galaxy = (3.6 x 10²⁸) / (1.2 x 10¹⁷)
M_galaxy = 3 x 10¹¹ solar masses