Calculate Density of Star w/ Mass & Radius

[Ayame17]

[SOLVED] Density of a star

Homework Statement

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

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

where $$\rho_{c}$$ 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 $$0.4\rho_{c}$$.

Homework Equations

See above and below.

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 $$M_{r}=\frac{4\pi}{3}\rho_{c}r^3(1-\frac{3r}{4R})$$, which I know to be right.
Part b I've done most of, ending up with $$\rho_{c}=\frac{3M}{\pi(R^3)}$$, which I also know to be right. The bit I'm having trouble with is getting to the $$0.4\rho_{c}$$. Can anyone offer any assistance?

 

[malawi_glenn]

How do you define average density? Find a formula for that first of all.

 

[Ayame17]

Ah, I assumed that the $$\rho$$ 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!

 

[malawi_glenn]

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

try both =)

 

[malawi_glenn]

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

otherwise:

$$M_{r}=\frac{4\pi}{3}\rho_{c}r^3(1-\frac{3r}{4R})$$

cant be right.

 

[Ayame17]

Yeah, I figured that must be right.
Been trying with the regular density equations - was looking good. Using $$\rho=\frac{M}{V}$$ and $$V=\frac{4\pi}{3}r^3$$, and the equation worked out at the start of part b (after rearranging to get M), I end up with $$\rho=\frac{\rho_{c}R^3}{4r^3}$$, which certainly looks nicer. However, $$\frac{R^3}{4r^3}$$ 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!

 

[Ayame17]

The worrying thing is, if I use the M calculated by the equation given that I think is wrong, I actually end up with $$\rho=0.4\rho_{c}$$...

 

[malawi_glenn]

The worrying thing is, if I use the M calculated by the equation given that I think is wrong, I actually end up with $$\rho=0.4\rho_{c}$$...

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

then$$\rho=0.4\rho_{c}$$?

But then the other two things that you said "i know these are right" are wrong? =)

 

[Ayame17]

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...!

 

[malawi_glenn]

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...!

Go and talk to her.

It doesent matter here if $$\rho = \rho_{c}[1-(\frac{r}{R})^2]$$ 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.

 

[dynamicsolo]

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

I get a mass function

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

yielding an average density of

$$0.1\rho_{c}$$ .

[In fact, if you "normalize" the mass functions by removing the $$\frac{4\pi}{3}\rho_{c}$$ 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 $$\rho_{c}$$, 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.)

 

[Ayame17]

Ah, it's the quadratic density model...I didn't know that. For that model, I got the mass function:

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

which gives an average density of $$0.4\rho_{c}$$ which is what she asked for...

 

[malawi_glenn]

Ah, it's the quadratic density model...I didn't know that. For that model, I got the mass function:

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

which gives an average density of $$0.4\rho_{c}$$ which is what she asked for...

Great, I only did the linear density model =)

 

[dynamicsolo]

Ah, it's the quadratic density model...I didn't know that. For that model, I got the mass function:

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

which gives an average density of $$0.4\rho_{c}$$ which is what she asked for...

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.)