Formula - Stars

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 (")
On some computers the one in the formula (1) looks like the small letter L (l) - it's not an "L", it's a one.

Typical Problems

1. A star is observed to have a parallax shift of 0.0246". How far away is the star?

Solution: d = 1/p = 1/.0246 = 40.7 pc

2. Alpha Centauri is the closest star to the Sun. It is only 1.3 pc away. What size is its parallax shift?

Solution: p = 1/d = 1/1.3 = 0.77"

3. A star has a parallax shift of 6.78 x 10-8 degrees. How far away is the star?

Solution: First of all, you need the angle in arc seconds, not degrees. 1 degree = 3600 arcseconds, so multiply the angle you were given by 3600
6.78 x 10 -8 x 3600 = 2.44 x 10-4 arc second.
Now put it into the formula: d = 1/p = 1/(2.44 x 10-4) = 4100 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 10x, 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. If a star is 830. pc away and has an apparent magnitude of +8.1, what is its absolute magnitude?

Solution: you are looking for M in this case. Put the numbers you're given into the formula
m - M = -5 + 5 log (d)
8.1 - M = -5 + 5 log (830)
8.1 - M = -5 + 5 x 2.92 (the log of 830 = 2.92)
8.1 - M = -5 + 14.6
8.1 - M = 9.6
-M = 9.6 - 8.1
-M = 1.5
M = - 1.5

2. If a star is 234,000 pc away and has an absolute magnitude of -4.89, what is it's apparent magnitude?

Solution: now you are looking for m. Again, put in the numbers and start calculating
m - M = -5 + 5 log (d)
m - (-4.89) = -5 + 5 log (234,000)
m + 4.89 = -5 + 5 x 5.37 (the log of 234,000 = 5.37)
m + 4.89 = -5 + 26.9
m + 4.89 = 21.9
m = 21.9 - 4.89
m = 17.0

3. A star has an apparent magnitude of +7.8 and an absolute magnitude of -4.2. How far away is it?

Solution: Uh-oh, now you're looking for the value of d. Put in the numbers -
m - M = -5 + 5 log (d)
7.8 - (-4.2) = -5 + 5 log (d)
7.8 + 4.2 = -5 + 5 log (d)
12.0 = -5 + 5 log (d)
12.0 + 5 = 5 log (d)
17.0 = 5 log (d)
17.0/5 = log (d)
3.4 = log (d)
Now you use the 10x key to undo the log function - to get the d all by itself.
103.4 = d
d = 2500 pc

Luminosity-Radius-Temperature - the formula that relates these three characteristics of a star. This formula is given in two ways, the general format (which we won't use) and the one where the values are given in terms of the Sun's values (we'll use this one).
Formula:L = R2 T4 where:

• L = luminosity given in terms of the Sun's luminosity
• T = temperature given in terms of the Sun's temperature
All of the values are in terms of the Sun's, so they would be presented to you as "A star's radius is 5 times greater than the Sun's" or "the star's luminosity is 1000 times less than the Sun's" or "the temperature is twice as great at the Sun's", and so on. No actual values of these things are used in this format, which just makes it simpler for everyone.

Typical Problems

1. Star Freddy has a radius that is 10 times smaller than the Sun's, while its temperature is 4 times greater. How does Freddy's luminosity compare to that of the Sun?

Solution: Put the values right into the formula -
L = R2 T4
L = (1/10)2 44 (since it is 10 times smaller, it has to go in as 1/10)
L = 1/100 x 256
L = 256/100 = 2.56
The luminosity is 2.56 times greater than the Sun's.

2. Star Jason has a luminosity that is 135,000 times greater than the Sun's and its radius is 33 times greater than the Sun's. What is its temperature like?

Solution: Put the values into the formula -
L = R2 T4
135,000 = 332 T4
135,000 = 1089 T4
135,000/1089 = T4
124.0 = T4
To take the fourth root of something, you can either take it to the power of 0.25, or take the square root of it twice.
T = (124.0) 0.25 = 3.33
Jason's temperature is 3.33 times greater than the Sun's.

3. Star Mike has a luminosity that is 1000 times fainter than the Sun, and a temperature that is 2 times smaller. What is its radius like?

Solution: Pop the values into the formula -
L = R2 T4
1/1000 = R2 (1/2)4
1/1000 = R2 (1/16)
16 x 1/1000 = 16/1000 = .016 = R2
Take the square root of both sides
R = 0.13
The radius is 0.13 times the Sun's, or you could say it is 13% of the Sun's.

Kepler's Third Law for Binary Stars - Just like the previous version of Kepler's third law, but here the value for "k" is defined as a function of the masses of the two stars.
Formula:M1 + M2 = a3 / P2 where:

• M1, M2 = masses of the two stars given 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
Usually this formula is used with the next formula in the list to get individual values for stars' masses. But as it stands you can't use only this formula to get the individual masses - you can get the sum of the two stars masses.

Typical Problems

1. Two stars, Fred and Ethel, are in orbit about one another with an average distance apart of 8.92 A. U. and they take 12.7 years to complete one orbit. What are their combined masses?

Solution: Again, you can't get the individual values of the masses, but you can get the sum of the masses:
M1 + M2 = a3 / P2
M1 + M2 = 8.923 / 12.72
M1 + M2 = 710 / 161
M1 + M2 = 4.4
The stars' combined mass is 4.4 times the mass of the Sun.

2. Two stars with a combined mass of 12.3 solar masses orbit one another with a period of 3.33 years. What is their average separation?

Solution: A rather unlikely question, but what the heck, let's try it - so you're basically looking for "a".
M1 + M2 = a3 / P2
12.3 = a3 / 3.332
12.3 = a3 / 11.1
12.3 x 11.1 = 137 = a3
Take the cubed root of both sides (or take both sides to the 0.33333 power)
137 0.333333 = 5.15 = a
The stars are on average 5.15 AU apart.

Center of Mass formula - used for binary star or anything orbiting around anything else. Again, this is like the teeter-tooter in the playground.
Formula:M1 a1 = M2 a2 where:

• M1, M2 = masses of the two stars given in terms of the Sun's mass
• a1, a2 = the average distance each star is from the center of mass, given in A. U.s
It should be noted that a1 and a2 add up to the value of "a" used in the Kepler's third law formula. This formula is usually used with the previous formula in the list to get individual values for stars' masses.
The formula can also be written in the following way
M1/M2 = a2/a1
This indicates that the ratio of the masses is the inverse of the ratio of the average distances.

Typical Problems

1. Star Bobby is 2 times the mass of star Teddy. How do their distances to the center of mass compare?

Solution: You actually don't need to do the math here, but you can if you want.
Using the second version of the formula (with B and T for the two stars) -
MB/MT = aT/aB
2 = aT/aB
This indicates that star Bobby is two times closer (1/2 the distance) to the center of mass compared to Teddy, or you could say Teddy is two times further from the center of mass compared to Bobby's distance. You're not getting any specific values, just their ratios.

2. Star Fred and Ethel (we looked at these two stars before) are the following average distances from the center of mass
Fred = 2.54 AU
Ethel = 6.38 AU
How do their masses compare?

Solution: Put them into the second version of the formula
MF/ME = aE/aF
MF/ME = 6.38/2.54 = 2.51
This indicates the Fred is 2.51 times more massive than Ethel, since it is 2.51 times closer to the center of mass. It is also interesting to note that when you add their individual distances together you get 8.92 AU, which is exactly the distance used in the Kepler's third law example for these two stars.

3. What are the individual masses of Fred and Ethel?

Solution: Well you have two different results - from Kepler's law you have that
MF + ME = 4.4 times the Sun's mass
And from the center of mass relation you have
MF/ME = 2.51
Take the second relation and isolate one of the variables, I'll isolate MF
MF = 2.51 ME
Now substitute this into the first formula (Kepler's Law result)
2.51 ME + ME = 4.4 times the Sun's mass
ME (2.51 + 1) = 4.4
ME (3.51) = 4.4
ME = 4.4/3.51 = 1.3
This tells us that Ethel's mass is 1.3 solar masses. You know that Fred is 2.51 times more massive, so that gives you 2.51 x 1.3 = 3.1 solar masses for Fred.
You can double check your results by using the two formula's you started with - Kepler's law says the sum of the masses should be 4.4 solar masses, and they are (3.1 + 1.3 = 4.4 solar masses), while the center of mass formula says the ratio of the two should be 2.51 and they are (3.1/1.3 = 2.51). Of course your results could vary slightly depending upon how much you round things.

Mass-Luminosity Relation - used for Main Sequence stars to estimate their luminosity.
Formula:L = M3.5 where:

• L = luminosity of the star given in terms of the Sun's luminosity
• M = mass of the star given in terms of the Sun's mass
This formula is really an approximation since it doesn't work very well for very large or very small values of mass, and is really only applicable for Main Sequence stars, not the other types.

Typical Problems

1. If a star has a mass that is 8.8 times the Sun's mass, what would its luminosity be on the Main Sequence?

Solution: just plug in the number and use a "power key" on your calculator to get the result
L = M3.5
L = (8.8)3.5 = 2021
The star's luminosity is about 2000 times that of the Sun's.

2. If a main sequence star has a mass that is 0.087 times the Sun's mass, what is its luminosity?

Solution:
L = M3.5
L = (0.087)3.5 = 1.9 x 10-4
The star's luminosity is 1.9 x 10-4 that of the Sun's, which is much, much smaller than the Sun's.

3. If a star has a Main Sequence luminosity that is 15,000 times that of the Sun, what is the star's approximate mass?

Solution: This is a little trickier, since you have to take the 3.5-root of both sides
L = M3.5
15,000 = M3.5
To take the 3.5-root, take both sides to the power of 1/3.5 = 0.286.
15,000 0.286 = M
M = 15.6
The star is 15.6 times more massive than the Sun.
You can check this by also just sticking this value back into the formula
L = M3.5 = 15.63.5 = 14,995
This isn't exactly the same luminosity we started with, but that's due to rounding.