1. Mar 28, 2006

### astronomystudent

1.) If a star has a temperature only 7/10 that of our Sun but a radius 10 times our Sun's, how would its luminosity compare to our Sun?
The only equation I have for luminosity involves mass and apparent magnitude, so I'm not really sure what to do here.

2.) If a star has a mass 11 times our Sun's mass, what would be its lifetime on the main sequence? What if the mass were 0.09 that of our Sun?
Do I use the lumionisty equation here which includes mass which is:
L = M^3.5

3.) If the parallax for a star is 0.015 arc sec (as measured by the Hipparchus satellite) how far away is it in parsec? In Light Years? If it has an absolute magnitude of 3, how bright would it appear from Earth?
d = 1/P to find parsecs? and then convert to light years. I'm not sure about the absolute magnitude though.

4.) If the temperature of a star is 7800 K, where is the peak wavelength from Wien's Law? Assuming that it is a Main Sequence star, what spectral class and absolute magnitude would it have?
Use Wien's Law to saolve for the peak wavelength seeing as I have the temperautre. How then would I find the absolute magnitude?

5.) If a star has an Absolute Magnitude 50 times brighter than our Sun, what would it be? If it appears as a 5 magnitude star, how far away is it?
No idea here. I don't have an equation or notes on this i.e. absolute magnitude.

2. Mar 28, 2006

### SpaceTiger

Staff Emeritus
What are you working from? What's the text? Are there online notes? I have a feeling that you can take the problems a little further than you have here.

3. Mar 28, 2006

### tony873004

1. There is a convenient formula that gives you luminosity as a function of Temperature (T) and Raduis (R). Just look up luminosity in the index of your textbook. It should be there. I'm looking in my old text book. The index has an entry: luminosity-temperature-radius relationship. Turning to the listed page, there's the formula. You could also Google that term and probably get it too.

2. Luminosity tells you how fast the star is burning its fuel. What else might you need to know to determine how long it will take to burn all its fuel?

3. There's a formula that lets you convert between absolute and visual magnitude. Absolute magnitude is simply what the star's visual magnitude would be if it were 10 pc away. Your chapter on absolute magnitude should give you the convenient formula. It's got a log base 10 in it.

5. Do you know how to determine distince from luminosity and brightness? There's a formula that does it for you. If you know that it 's abs mag is 50x the Sun's, you can figure out how much more luminous it is than the Sun.

4. Mar 29, 2006

### astronomystudent

1.) L=4(pi)R^2T^4
sun's luminosity = 3.90 x 10^26 Watts
sun's temp = 5800 K
sun's radius = 7x 10^5 Km

STAR
temp = (5800 K)(7/10) = 406,000 K
radius = (7x10^5)(10) = 7,000,000 km

L = 4(pi)(7,000,000)^2 (406,000)^4
L = 1.67 x 10^37 Watts

Compared with the sun
1.67 x10^37 (3.90 x 10^26) = 6.513x10^63 Watts

I'm still working on the other four.

5. Mar 29, 2006

### tony873004

It's asking you to find its luminosity compared to the Sun's. So your answer will have the units of LSun. You should be able to solve this with ratios. Then you wouldn't need to know the Sun's temp or luminosity.

I'll give you an example. Pretend you had 2 cubical boxes. Box 1's height is 0.738 meters. Box 2's height is 3 times that of box 1's height. How would box 2's volume compare to box 1's volume?

$$\begin{array}{l} v = l^3 \\ \\ \frac{{v_1 }}{{v_2 }} = \frac{{l_1^3 }}{{l _2^3 }} \\ \\ \frac{{v_1 }}{{v_2 }} = \frac{{l_1^3 }}{{\left( {3l_1 } \right)^3 }} \\ \\ \frac{{v_1 }}{{v_2 }} = \frac{{l_1^3 }}{{27l_1^3 }} \\ \\ \frac{{v_1 }}{{v_2 }} = \frac{1}{{27}} \\ \\ v_2 = 27v_1 \\ \end{array}$$

Answer: Box 2's volume is 27 times greater than box 1's volume
Notice that the 0.738 meters was not used to find this answer. Likewise, you do not need the Sun's temp or radius or luminosity to solve your problem.

Luminositystar=10 LuminositySun or
in sentence form: "This star is 10 times as luminous as the Sun."
(this is not the right answer btw.)

** edit ** does anyone know how to make TEX bigger? I can barely read what I wrote.

Also, if you do choose to slove it like you're doing:
1. You're missing something from your formula. It's not 4pi r^2 T^4.
3. (5800 K)(7/10) = 406,000 K. This is wrong. check your math.

Last edited: Mar 29, 2006
6. Mar 29, 2006

### astronomystudent

1.) L=R^2 * T^4
L = (10)^2 (7/10)^4
L = 100 (.2401)
L = 24.01
The Luminosity is 24.01 times greater than the Sun's.

2.) L = M^3.5
L= (11)^3.5
L = 4414.427586

The star's luminosity is about 4.14 x10^3 times that of the Sun's.

L = M^3.5
L = (0.09)^3.5
L = 2.187 x 10^-4

The star's luminosity is about 2.187 x 10^-4 times that of the Sun's.

3.) d = 1/p
d = 1/0.015
d = 6.66 x10^1 parsecs

distance light years = 21.725 b/c the conversion is 1 parsec = 3.26163 light years

m - M = -5+5log(d)
m - (3) = -5+5log(6.66 x 10^1)
M = 7.11 x 10^0

4.) I'm still working on it

5.) I'm not really sure what I am doing here. I am confused as to whether I need to use a table from my book or actual math.

7. Mar 30, 2006

### tony873004

1. Your answer is correct. Your method is even a little more convient than ratios since it takes advantage of the fact that for the Sun, MSun=1 and LSun=1.

The ratio formula is

$$\frac{{L_1 }}{{L_2 }} = \frac{{R_1^2 }}{{R_2^2 }} \cdot \frac{{T_1^4 }}{{T_2^4 }}$$

If you use the Sun as either star 1 or star 2, it just multipies across as 1, while the other star multiplies across the formula exactly how you did it. But if you had to compare 2 stars and one was not the Sun, then you'd need both the numerator and denomenator of that formula.

2. M3.5 is only a very rough estimate of Luminosity. But if your teacher gave you that formula then go with it. You computed luminosities. You're only halfway there. What does the question ask for?

3. Check your math for you LY conversion.
Did it ask how bright it would appear from Earth? Or did they ask for its visual magnitude. There's a difference. Assuming they asked for its visual magnitude, does your answer seem reasonable? Check your formula.

8. Mar 30, 2006

### astronomystudent

2.) L = M^3.5
L= (11)^3.5
L = 4414.427596
L = 4.41 x 10^3

L = M^3.5
L = (0.09)^3.5
L = 2.187x10^-4

I think this is the equation for what it's lifetime would be on the main sequence.
T = M/L
T = 11/4.41 x 10^3
T = .0024918293

T = M/L
T = (0.09)/2.187 x 10^04
T = 411.5226337

3.) I checked the m-M=-5+5log(d) equation and I got the same answer. The conversion that I made from parsecs------->light-years was wrong now I got:
light years = 2.17x10^2
To find out how bright it was from earth could I use the brightness equation: b = L/4(pi)d^2

4.) Wien's Law = wavelength(t) = 2.898x10^7
wavlength(7800)=2.898x10^7/7800
wavelength = 3.72 x 10^3 angstroms

I'm not really sure what to do here after this.

5.) Still thinking, but I'm definitely stuck, could I also use the brightness equation here
b =L/4(pi)d^2

9. Mar 31, 2006

### tony873004

Check your math on the lifetime of the 0.09 solar mass star.

T=M/L is correct. Do you know what units to put after your answer?

Where did the -5 come from in your formula?
m = 5 log(d/10pc)+M, d is in parsecs

10. Apr 2, 2006

### astronomystudent

2.) The units would be (years), I think. And I checked my my on the lifetime of the star with the mass of 0.09. I still got the 4.115226337 which in scientific notation would be 4.12 x 10^2.

3.) 5log(d/10pc)+M=m
5log(6.66x10^1/10pc)+3 = m
m= 7.12 x 10^0

Does little "m" solve for "how bright would it appear from Earth?" I am confused because I thought maybe you could apply the F= L/4(pi)d^2 but you never responded to that.

d= 1/p
d= 1/0.015
d= 6.66 x 10^1

1 parsec = 3.26124 light years
6.66 x 10^1 parsec = 2.17 x 10^2 light years

4.) I did Wien's Law here and got the wavelength to equal = 3.72 x 10^3 and then I looked up spectral class online. Since it is 7800 K then it would be in spectral class "A" I think. I still don't know how to solve for "absolute" magnitude with simply wavelength, temperature. Help please?

5.) Can I use the equation: m-m = -5+5log(d) here and plug the 5 in as M to solve for distance?

11. Apr 2, 2006

### tony873004

Does that make sense to you? You're claiming that the main-sequence lifetime of a star with 11 solar masses would be 0.00249 years, or roughly 18 hours. And the star with 0.09 solar masses will have a main-sequence lifetime of 411 years?

Hint: You used the units:

LSun for luminosity
MSun for mass
TSun for temperature

12. Apr 2, 2006

### tony873004

What happened to big M inbween the 1st and 2nd lines. Also, there's no reason to put 10^0 after any answer. 10^0 = 1, so 7.12 x 10^0 is the same as 7.12 x 1 is the same as 7.12.
Little "m" is the apparent magnitude. When you look at the sky and see a magnitude 2 star, that's referring to little "m". Big M is absolute magnitude. It's the value for little "m" that the star would have if it were exactly 10pc from Earth.

13. Apr 2, 2006

### tony873004

5. If you know that the star's absolute magnitude is 50x that of the Sun, what might you need to know about the Sun to compute this star's absolute magnitude?

14. Apr 2, 2006

### tony873004

You need to put units on your answer or it is meaningless.

15. Apr 2, 2006

### astronomystudent

2.) Different equation so the units would equal yrs. I found it in my notes and it also says that M^4 = L so I guess I will use that thus:
L = (11)^4
L = 1.4641 x 10^4 W

L = (0.09)^4
L = 6.561 x 10^-5 W

Lifetime = [1/mass^4] (1 x 10^10 yrs)
L = [1/1.4641 x 10^4](1 x 10^10)
L = 2.24 x 10^5 yrs.

L = [1/6.561 x 10^-5](1 x 10^10 yrs)
L = 1.52 x 10^14 yrs.
3.) If I use this equation:
m - M = 5 log(d/10)
m - 3 = 5 log(6.66 x 10^1/10)
m = 3 + 5log (6.66x10^0)
m = 3 + (4.117371146 X 10^0)
m = 7.12 X 10^0 thus m = 7.12 X 10^0

4.) The units when using Wien's Law and solving for wavelength:
wavelength= 3.72 x 10^3 nm, still stuck on the absoulte mag and whether the spectral class is right.

5.) I think I may have this one.
Sun's abs. magnitude = 4.83
4.83(50) = 2.415 X 10^2
M = 2.415 X 10^2
m = 5
5 - (2.415 x 10^2) = -2.365 x 10^2
-2.365 x 10^2/5 = 4.73 x 10^1
10^-4.73 X 10^1 = 5.01 X 10^48 PARSECS
DISTANCE = 5.01 X 10^38 parsecs
What would it be is that meaning like white dwarf, that sort of thing?

16. Apr 3, 2006

### astronomystudent

Am I done with problem 3 or do I still need to look for an equation to figure out how bright it appears from Earth? I am confused on that point, sorry.

17. Apr 3, 2006

### astronomystudent

Answers: On the previous page you haven't seen those answers. But I think I did number 5 wrong maybe it should be:
5.) m = 5
M = 4.83
5 - 4.83 = 1.7 x10^-1
(1.7 x10^-1)/5 = 3.4 x 10^-2
10^3.4 x 10^-2 = 1.08 x 10^0
distance = 1.08 x 10^0 parsecs

i am still stuck on what would it be? is it asking like white dwarf or spectral class? i'm not really sure there. thanks

18. Apr 3, 2006

### tony873004

#5. The question is phrased strange:
"If a star has an Absolute Magnitude 50 times brighter than our Sun"
What is it asking? Does the star have an absolute magnitude 50 times that of the Sun's absolute magnitude, or is the star 50 times brighter than the Sun? One of these makes no sense. Can you figure out which one, and hence figure out what this question is trying to ask?
DISTANCE = 5.01 X 10^38 parsecs
should have stood out as unreasonable. That's
500000000000000000000000000000000000000 parsecs, much larger than the entire visible universe.

Does your 2nd answer 1.08x10^0 parsecs, seem reasonable? Again, there's no reason to include the 10^0 part of your answer since that equals 1.

This star is 50 times brighter than the Sun. The Sun has an apparent magnitude of about 5, which means its visual magnitude at 10 parsecs would be about 5. The star in question also has a visual magnitude of about 5. Since it is much brighter than the Sun, would it make sense that it be closer than 10pc or farther than 10pc to appear the same magnitude as the Sun?

19. Apr 3, 2006

### astronomystudent

These problems are due in an hour and I am still have no answer as to whether I am done with problem three or what I am to do now once I have completed Wien's Law on problem 4. I am working on problem 5.

20. Apr 4, 2006

### astronomystudent

I turned these questions in but I am still interested in figuring out how to solve them.