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Density of a star

  1. Jan 18, 2008 #1
    [SOLVED] Density of a star

    1. The problem statement, all variables and given/known data
    For a star of mass M and radius R, the density increases from the centre to the surface as a function of radial distance r, according to

    [tex]\rho = \rho_{c}[1-(\frac{r}{R})^2][/tex]

    where [tex]\rho_{c}[/tex] is the central density constant.

    a) Find M(r).
    b) Derive the relation between M and R and show that the average density of the star is [tex]0.4\rho_{c}[/tex].


    2. Relevant equations

    See above and below.

    3. The attempt at a solution

    Right. Firstly, I believe there is a mistake in the question paper - I don't think that the (r/R) is meant to be squared, as I've seen the formula used many times before and it's never had it.
    Part a I've done (without the squared bit), and got [tex]M_{r}=\frac{4\pi}{3}\rho_{c}r^3(1-\frac{3r}{4R})[/tex], which I know to be right.
    Part b I've done most of, ending up with [tex]\rho_{c}=\frac{3M}{\pi(R^3)}[/tex], which I also know to be right. The bit I'm having trouble with is getting to the [tex]0.4\rho_{c}[/tex]. Can anyone offer any assistance?
     
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  3. Jan 18, 2008 #2

    malawi_glenn

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    How do you define average density? Find a formula for that first of all.
     
  4. Jan 18, 2008 #3
    Ah, I assumed that the [tex]\rho[/tex] equation given in the problem was for the average density of the star...(this is actually a question to recap last year's work, our lecturer thought it may be a good idea - seems she was right!)
    Also, do you know if the (r/R) in the first equation is actually meant to be squared? In all our work last year, it never was, so that's thrown me as well!
     
  5. Jan 18, 2008 #4

    malawi_glenn

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    that relation is just for calculation, it doesn't follow that 100%, so I dont know.

    try both =)
     
  6. Jan 18, 2008 #5

    malawi_glenn

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    it must be [tex]\rho = \rho_{c}[1-(\frac{r}{R})][/tex]

    otherwise:

    [tex]M_{r}=\frac{4\pi}{3}\rho_{c}r^3(1-\frac{3r}{4R})[/tex]

    cant be right.
     
  7. Jan 18, 2008 #6
    Yeah, I figured that must be right.
    Been trying with the regular density equations - was looking good. Using [tex]\rho=\frac{M}{V}[/tex] and [tex]V=\frac{4\pi}{3}r^3[/tex], and the equation worked out at the start of part b (after rearranging to get M), I end up with [tex]\rho=\frac{\rho_{c}R^3}{4r^3}[/tex], which certainly looks nicer. However, [tex]\frac{R^3}{4r^3}[/tex] isn't equal to 0.4 - even if we say that 1/r > 1/R so they can basically cancel, I still end up with 0.25. Can't really see where to go!
     
  8. Jan 18, 2008 #7
    The worrying thing is, if I use the M calculated by the equation given that I think is wrong, I actually end up with [tex]\rho=0.4\rho_{c}[/tex]...
     
  9. Jan 18, 2008 #8

    malawi_glenn

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    ok so if [tex]\rho = \rho_{c}[1-(\frac{r}{R})^2][/tex]

    then[tex]\rho=0.4\rho_{c}[/tex]?

    But then the other two things that you said "i know these are right" are wrong? =)
     
  10. Jan 18, 2008 #9
    She's given me exactly the same question before (but without the squared bit and the 0.4 bit), so I assumed she'd make a mistake. She makes them constantly in the lecture handouts.
    I'd also assumed that the equation given in the question is a generalised equation for star density, because she never told us whether it was or wasn't...!
     
  11. Jan 18, 2008 #10

    malawi_glenn

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    Go and talk to her.

    It doesent matter here if [tex]\rho = \rho_{c}[1-(\frac{r}{R})^2][/tex] is reasoanble in real stellar models or not, the important things is that the answers are consistent with the info given in the problem and that you use right formulas etc.
     
  12. Jan 19, 2008 #11

    dynamicsolo

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    I suspect the [tex]0.4\rho_{c}[/tex] is from a different model entirely. For the linear density model, I agree with your mass function and the average density of [tex]0.25\rho_{c}[/tex]. For the quadratic density model, [tex]\rho = \rho_{c}[1-(\frac{r}{R})^2][/tex] ,

    I get a mass function

    [tex]M_{r}=\frac{4\pi}{3}\rho_{c}r^3(1-\frac{3r}{2R}+\frac{3r^2}{5R^2})[/tex] ,

    yielding an average density of

    [tex]0.1\rho_{c}[/tex] .

    [In fact, if you "normalize" the mass functions by removing the [tex]\frac{4\pi}{3}\rho_{c}[/tex] factor, set R = 1, and plot the functions (r^3) times the polynomials, you get the cumulative mass, relative to a sphere of radius R with constant density [tex]\rho_{c}[/tex], as a function of the fractional radius out from the center. For the linear model, the value is 0.25 at r/R = 1; for the quadratic model, the value is 0.1 at r/R = 1.]

    I seem to recall that there is a density model that gives the "0.4 central density" result, but it isn't one of this form... (I'd have to look it up.)
     
    Last edited: Jan 19, 2008
  13. Jan 20, 2008 #12
    Ah, it's the quadratic density model...I didn't know that. For that model, I got the mass function:

    [tex]M_{r}=\frac{4\pi}{3}\rho_{c}r^3(1-\frac{3r^2}{5R^2})[/tex]

    which gives an average density of [tex]0.4\rho_{c}[/tex] which is what she asked for...
     
    Last edited: Jan 20, 2008
  14. Jan 20, 2008 #13

    malawi_glenn

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    Great, I only did the linear density model =)
     
  15. Jan 21, 2008 #14

    dynamicsolo

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    Oops, I misinterpreted the density function as [1 - (r/R)]^2 when I set up the integral. >:/ Good, so that mystery's been resolved... (And the "normalized" mass function does indeed reach 0.4 at (r/R) = 1.)
     
    Last edited: Jan 21, 2008
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