Calculating a star's density profile

[Angela G]

Homework Statement

Hello!

I am trying to solve this problem but I'm struggling with it, could someone please help me?

The exercise is



A star with density profile: $$ \rho = \rho_c \left( 1- \frac{r^2}{R^2}\right)$$

a) What is the central pressure $P_c$ and the central temperature $T_c$ in terms of M and the $\rho_c$?

b) What is the pressure profile $P(r)$ and the temperature profile $T(r)$ in terms of $P_c$, $T_c$ in addition to r and R

c) What is the value of $\alpha$ in the expression for the total gravitational energy $$ \Omega = - \frac{\alpha GM^2}{R}$$

Relevant Equations

$$\frac{d P}{d r} = - \frac{\rho Gm}{r^2}$$
$$ P = \frac{\rho k T}{\mu m_H}$$
$$ \Omega = - \int_0^M \frac{G m dm}{r} $$

to solve a) I used The equation of hydrostatic equilibrium $$ \frac{d P}{d r} = - \rho \frac{GM}{r^2} \iff dP = - \rho \frac{GM}{r^2}dr \Longrightarrow \int_{P_c}^0 dP = - \int_0^R \rho \frac{GM}{r^2} dr $$
I replaced M as $\rho V$ and then
I integrated both the left and right-hand sides and got at the left-hand side $- P_c$, But I'm stuck on the right-hand side. I calculated it and got $$ - \frac{4 \cdot 8 \pi}{3 \cdot 15} G \rho_c^2 R $$
I was thinking to replace R with the definition of density $\rho = \frac{ 3M} { 4 \pi R^3}$ But I'm not sure how to proceed, some ideas?

 

[pasmith]

The density profile is given as <br /> \rho(r) = \rho_c\left(1 - \frac{r^2}{R^2}\right). This is not uniform.

You have to start from the equation of hydrostatic equilibrium in the form <br /> \frac{dp}{dr} = -\rho(r) \frac{d\chi}{dr} where the gravitational potential \chi satisfies <br /> \frac{1}{r}\frac{d}{dr}\left(r\frac{d \chi}{dr}\right) = 4 \pi G \rho. Multiplying the first equation by r/\rho and differentiating with respect to r yields <br /> \frac{1}{r}\frac{d}{dr}\left(\frac{r}{\rho}\frac{dp}{dr}\right) = -4\pi G \rho. As the density is not uniform, the mass of the star is given by <br /> M = 4\pi\int_0^R \rho(r) r^2\,dr rather than M= (4/3)\pi R^3, which holds only for a uniform density.

I assume you have some equation of state which relates pressure, density and temperature. It would have been helpful to include that as a relevant equation.

 

[Angela G]

The density profile is given as <br /> \rho(r) = \rho_c\left(1 - \frac{r^2}{R^2}\right). This is not uniform.

You have to start from the equation of hydrostatic equilibrium in the form <br /> \frac{dp}{dr} = -\rho(r) \frac{d\chi}{dr} where the gravitational potential \chi satisfies <br /> \frac{1}{r}\frac{d}{dr}\left(r\frac{d \chi}{dr}\right) = 4 \pi G \rho. Multiplying the first equation by r/\rho and differentiating with respect to r yields <br /> \frac{1}{r}\frac{d}{dr}\left(\frac{r}{\rho}\frac{dp}{dr}\right) = -4\pi G \rho. As the density is not uniform, the mass of the star is given by <br /> M = 4\pi\int_0^R \rho(r) r^2\,dr rather than M= (4/3)\pi R^3, which holds only for a uniform density.

I assume you have some equation of state which relates pressure, density and temperature. It would have been helpful to include that as a relevant equation.

yes, I'm sorry
For solving a) I have to do it with the hydrostatic equilibrium equation, to determine the pressure and to determine the temperature I have to use the expression I got from the pressure and use the ideal gas law $P = \frac{\rho k T}{\mu m_H}$, for solving c) I think I should use the $$ \Omega = - \int_0^M \frac{G m dm}{r}$$ and then replace m with $m = \rho V \Longrightarrow \frac{dm}{dr} = \frac{\rho}{dr} \frac{ dV}{dr}$. I think this is the way to solve it because I did so in the last exercise, but in that case, the central density was $\rho_c = \rho$, so the mass could be replaced by $\rho V(r)$. but I will try your way, but I think I need a little more guidance because I do not understand that with the gravitational potential, I'm sorry that was a new sign

edit: I think I understood, thanks

 

[TSny]

the gravitational potential \chi satisfies <br /> \frac{1}{r}\frac{d}{dr}\left(r\frac{d \chi}{dr}\right) = 4 \pi G \rho.

For spherical coordinates, I think the left-hand side should be
<br /> \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d \chi}{dr}\right) = 4 \pi G \rho.