How do pressure and temperature in a star depend on its overall mass and radius?

[AStaunton]

******* I attempted to post this already today but I don't think I'm not sure if it worked...I apologise if this is a dublicate!**********

the problem is:
suppose that a star is modeled as having a density that decreases linearly from the center to the surface,

rho(r)=rho_c(1-r/R)

where rho_c is the central density and R is the stellar radius.

(1)Find m, the mass enclosed at radius r and thus derive expression for (2)the central pressure and (3)temperature that depend only on the overall mass M and the radius of the star R and physical constants.

part (1) is quite straightforward:

because we are dealing with a sphere the volume is:

integral[0-r] (4*pi*r^2) dr

=> m(r) = integral[0-r] (rho_c(1-r/R)(4*pi*r^2) dr

=> m(r) = 4*pi*rho_c(r^3/3 - r^4/4R)

so to get M, the total mass of star set r=R, where R is star radius

=> M = 1/3*pi*rho_c*R^3


I am having trouble with part (2) (deriving expression for the central pressure the depends only on the overall mass M and radius of the star R), my attempt is below and any advice on where I'm going wrong is appreciated:

from equation of hydrostatic equilibrium:

dP = rho*g*dr

where P = pressure of star rho=density g=is grav accel of star
r=distance into the star from the surface

=> P = integral[0-r] rho(r)*g(r) dr {A}

equation for rho is given in the question:

rho(r) = rho_c(1-r/R)

to find g for this star we use Newton's law of gravitation:

g = GM/r^2

solved for M in terms of r in part (1):

=> g=G(1/3*pi*rho_c*R^3)/r^2

so going back to equation {A}:

=> P = integral[0-R] (rho_c(1-r/R))(G(1/3*pi*rho_c*R^3)/r^2) dr

=> P = 1/3*pi*rho_c^2*G[-1/r - Rlnr] <---solved from [0-R] as implied by above integral

the above equation for P seems problematic here, as we are solving between the limits [0-R]... often the 0 means that that term in the expression goes to 0 and so we only have the R part left...however in the above equation I have a 1/r term and a ln(r) and I don't think either of these works when you try to stick in a 0 for r...

Any tips and advice are welcome.

 

[gneill]

You said that for dP = rho*g*r, "r=distance into the star from the surface".

The r in the equations you developed for rho(r) and M(r) assume r is measured from the center. You'll have to do a little parameter matching, something like d = R - r, where d is the depth below the surface.

 

[AStaunton]

thanks..I have a quick follow up question if you don't mind!:

an example given in class with a constant density rho(r)=rho_c and using the two equations:

dm(r)/dr = 4*pi*r^2*rho(r) <------------- conservation of mass

and

dP(r)/dr= - G*rho(r)*m(r)/r^2 <----------hydrostatic equilibrium

we should be able to show that the central pressure P_c is given by:

P_c = 3/8*pi*(G*M^2/R4)

however I am not able to show this...

after attempting to solve this in much the same way as I did with the last problem I ended up with:

P_c = 2/3G*pi*rho_c^2*R^2

According to my teacher's final anser the rho_c's should cancel out by I seem to be squating them (you get one rho_c within the mass expression and also as the hydrostat eq. has a rho_c in it, makes two...so I end up with rho_c^2) its becoming aparant I don't understand this as well as I thought I did also the fact that the hydrostat eq. has a minus sign doesn't make sense either!

also...back to what you said about the way I chose my r from the surface...it felt odd to me to, but I seem to have solved the problem with the dividing by zero and ln0 that I mentioned and the resulting is equation:

P_c = 0.035*pi*G*rho_c^2*R^2 where P_c is the pressure at center

works quite well to describe the pressure at center of sun when I plug in values I found for rho_c (density at center) and R (the radius of sun)...however once again this is in terms of the wrong quantities, as the question I posted states, I should be getting my answer in terms of M and R...but rho_c^2 is all I can get..

 

[gneill]

Regarding P_c, you have an expression for the total mass of the star, M = (π/3)*rho_c*R³ which can be used to simplify it.

I get

$$P_c = \frac{5}{12} \rho_c \frac{G M}{R}$$

Regarding the rest, I'll have to give it some thought.

 

[gneill]

Okay, so you're given that ρ is constant, and the two equations:

$$\frac{dm(r)}{dr} = 4 \pi r^2 \rho$$

$$\frac{dP(r)}{dr} = -G \rho \frac{m(r)}{r^2}$$

Looking first at the mass equation, rearrange as:

$$dm(r) = 4 \pi r^2 \rho \; dr$$

Integrating yields

$$m(r) = \frac{4}{3} \pi r^3 \rho$$

Now turn to your pressure relationship. You now have an expression to use for m(r), so you get:

$$dP(r) = -G \rho \frac{4}{3} r \pi \rho \; dr$$

Again this can be integrated, but this time integrate from the outside in (you want to find the pressure at the center):

$$P(0) = - \frac{4}{3} \pi G \rho^2 \int_R^0 r \; dr$$

$$P(0) = \frac{2}{3} \pi G R^2 \rho^2$$

Now do like before and insinuate the expression for the total mass into the above. This time the mass is M = (4/3) π r³ ρ. Divide by the square of this and multiply by M². You have arrived at your destination.

 

[AStaunton]

very helpful thanks...

does the minus sign in the hydrostatic equation simply correct for the problem I was having with whether r should be measured from the surface into the center or the other way round?

 

[gneill]

very helpful thanks...

does the minus sign in the hydrostatic equation simply correct for the problem I was having with whether r should be measured from the surface into the center or the other way round?

You're welcome. Glad to help.

The minus sign is not sufficient to correct for the outside-in versus inside-out problem before. You really had to translate the coordinates.

Here it just means that as the radius increases, the pressure drops. So the rate of change of pressure with increasing radius is negative.

 

[AStaunton]

I have a related question to what we discussed in the above:

for a similar problem I didn't take rho to be constant but to be a function of r:

$$\rho(r)=\rho_{c}(1-\frac{r}{R})$$

and using the hydrostatic equil equation:

$$\frac{dP(r)}{dr}=-\frac{Gm\rho(r)}{r^{2}}$$

and the conserve of mass equation:

$$\frac{dm}{dr}=4\pi r^{2}\rho(r)$$

I found the central pressure...

$$P_{C}=\frac{5}{4\pi}\frac{GM^{2}}{R^{4}}$$

and from this and using the ideal gas equation of state:

$$P=\frac{\rho kT}{\mu m_{H}}$$

I found the expression for central temperature in terms of mass and radius..

Then plugging in the values of M and R that correspond to that of the sun, I found temperature for the sun to be around 6X10^6 Kelvin. The actual temperature for the sun is around 1.5X10^7.
What is the main reason that the calculated value by the above method differs from the actual value.
I suspect that it comes mainly from the ideal gas equation used...ie. in reality I think electron pressure may also be significant? I don't think radiation pressure is significant in the sun though..

 

[gneill]

Your derivation of the expression for the central pressure looks okay to my eye.

What are the ρ and μ variables in your last expression for pressure? Presumably the m_H is the mass of a hydrogen atom. Is ρ the overall density or just the density of hydrogen in the core?

 

[AStaunton]

thanks for your reply mh is indeed the mass of H, I found rho_c (central density) in terms of M and R to eliminate rho from my final expression for P_c. also mu is to do with the composition ie. how much of the star is made up of hydrogen. the value I was given was 0.6. I should say that I think everything I did was correct, ie. I found the correct answer considering the equation I used.
I was told by teacher that the answer should indeed be less than the actual values for the sun, when using the equation stated in my previous post. I think this must be because the equation I used assumes ideal gas.
so what I'm trying to figure out is - what is actually going on in the sun that makes T and P higher than in the above.
Is it because the nuclear reactions at the center cause the temperature and pressure increase?

 

[gneill]

I can certainly see the temperature increase being due to the exothermic nature of the nuclear reactions! Offhand I can't think of a reason for higher pressure, unless it has to do with the change in composition.