Solve Mass & Mean Density of Star w/ Mass Density Function

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Homework Help Overview

The discussion revolves around a problem from astrophysics concerning the mass density of a star defined by a specific function of radius. The original poster seeks assistance in calculating the mass of the star in terms of the density parameter and its radius, as well as finding the mean density and central pressure of the star.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to integrate the density function over the volume to determine the total mass. There is mention of the symmetry of the problem simplifying the integration process. Some participants question the use of spherical coordinates and clarify the relationship between mass elements and the density function.

Discussion Status

The discussion is active with participants providing insights on the integration process and addressing potential confusion regarding the equations presented. There is a focus on clarifying the definitions and relationships between mass, density, and the integration method needed to solve the problem.

Contextual Notes

Participants note that the problem involves assumptions about spherical symmetry and the need for clarity in the notation used for density and mass. There is an emphasis on understanding the volume element in spherical coordinates, despite it not being explicitly stated in the problem.

jdw_
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Hi, I am wondering if somebody could explain how to answer these questions, it is from "Foundations of astrophysics" ch.15 q15.10

Suppose the mass density of a star as a function of radius is

p(r)=p0[1-(r/R)^2]

where R is the radius of the star

a) Find the mass M of the star in terms of p0 and R
b) Find the mean density of the star in terms of p0
c) Show that the central pressure of the star is:

Pc=15/16pi x GM^2/R^4

I have a good idea on how to do b) and c) but i need help with a)

useful equations :
P(r)= -GM(r)p(r)/r^2

M(r) = 4pi r^2 p(r)
 
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jdw,

It looks like you need to integrate the density over the volume in order to get the total mass. The problem is made simpler by the fact that there is symmetry (rho depends only on r, so that the density is constant everywhere on a sphere of radius r). Therefore, the integrals over the angular coordinates will just turn into 4pi, and you'll be integrating over successive spherical shells of surface area 4pi r^2, and thickness dr. To get this result, you need to know what a volume element is in terms of spherical coordinates.
 
cepheid said:
jdw,

It looks like you need to integrate the density over the volume in order to get the total mass. The problem is made simpler by the fact that there is symmetry (rho depends only on r, so that the density is constant everywhere on a sphere of radius r). Therefore, the integrals over the angular coordinates will just turn into 4pi, and you'll be integrating over successive spherical shells of surface area 4pi r^2, and thickness dr. To get this result, you need to know what a volume element is in terms of spherical coordinates.

Spherical coordinates weren't specified so why use them?

Mass is straight forward. First the mass element is...

dM = 4.pi.ρ(r).r2 dr

...sub in ρ(r) = ρo(1-(r/R)2) and expand out the brackets to make it easier to integrate.

From there it's straight forward. I think you might've been confused by what you've quoted as the mass equation:

M(r) = 4pi r^2 p(r) [sic]

...except that's the equation for dM(r) NOT M(r). Was that the problem?

Also 'p(r)' should've been 'ρ(r)' else you'd be thinking "pressure as a function of radius" instead of "density as a function of radius".
 
Last edited:
qraal said:
Spherical coordinates weren't specified so why use them?

I just meant that you implicity already *have* used them if you do the integration over "shells" having surface area 4pi r^2. It's just that the integral over the angular coordinates has already been done. Of course the spherical symmetry means that it reduces to a 1D integral -- I just thought it would be good if the OP could see how that comes about by starting with the full volume integral.
 

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