The Temperature Profile of a Spherical Cloud of Ideal Gas

[Saitama]

Homework Statement

(see attachment 1)

Homework Equations

The Attempt at a Solution

(see attachment 2)
As the gas is ideal and there is no gravity, the pressure is same throughout the cloud. In the thin sphere shown, the mass of the gas is $dm=dV \cdot \rho(r)$. Let $\mu$ be the molar mass of gas.
From ideal gas law
PV=nRT
P(dV)=\frac{dm}{\mu}RT=\frac{dV \rho(r)}{\mu}RT
P=\frac{\rho_0}{\mu r^2}RT
Since $P, \rho_0$ and $\mu$ are constant, $T$ is directly proportional to $r^2$.
This gives $T_{r_0}/T_{2r_0}=0.25$, but this is wrong.

Any help is appreciated. Thanks!

 

[phyzguy]

Where does the problem statement say that there is no gravity? Without gravity, there would be nothing to hold it together and it would just dissipate into space.

 

[Saitama]

Where does the problem statement say that there is no gravity? Without gravity, there would be nothing to hold it together and it would just dissipate into space.

Well, I thought that there is no gravity in intergalactic space.

Any hints about how would I form the equations then?

 

[phyzguy]

Well, I thought that there is no gravity in intergalactic space.

There is no gravity from outside the cloud, but the cloud itself is self-gravitating.

Any hints about how would I form the equations then?

Consider a spherical shell, like you were doing. You should be able to calculate the mass inside the shell. This mass will attract the shell due to Newton's law of gravity and provide an inward force on the shell. Pressure provides an outward force on the shell. Write expressions for these two forces, and set them equal. Then the shell is in equilibrium.

 

[Saitama]

Consider a spherical shell, like you were doing. You should be able to calculate the mass inside the shell. This mass will attract the shell due to Newton's law of gravity and provide an inward force on the shell. Pressure provides an outward force on the shell. Write expressions for these two forces, and set them equal. Then the shell is in equilibrium.

I see your point but how would I calculate the gravitational force?

The mass inside the thin shell is $4\pi \rho_0 dr$. The mass enclosed in the sphere of radius r is $4\pi \rho_0 r$. The gravitational force is $F=\frac{GM_1M_2}{d^2}$ but what should I substitute for d?

 

[phyzguy]

The gravitational force of a spherically symmetric body acts as though the mass is concentrated at the center. This is a well-known fact that was first proved by Newton. So for a thin shell of radius R:

M1 = \int_0^R \rho(r) 4\pi r^2 dr
M2 = 4\pi R^2 \rho(R) dR
d = R

 

[Saitama]

The gravitational force of a spherically symmetric body acts as though the mass is concentrated at the center. This is a well-known fact that was first proved by Newton. So for a thin shell of radius R:

M1 = \int_0^R \rho(r) 4\pi r^2 dr
M2 = 4\pi R^2 \rho(R) dR
d = R

How d=R? Won't their CM overlap? The CM for both spherical shell and a solid sphere lies at their geometric centre.

 

[phyzguy]

Probably I explained it wrong. I should have said that the gravitational force of a spherically symmetric body on a point outside the sphere acts as though the mass is concentrated at the center. We're trying to calculate the force of the mass inside the shell on the shell itself, and the shell is outside the sphere. So the mass in the shell is a distance R away from the center of the shell inside it. If you want a more detailed proof, see this page.

 

[Saitama]

Probably I explained it wrong. I should have said that the gravitational force of a spherically symmetric body on a point outside the sphere acts as though the mass is concentrated at the center. We're trying to calculate the force of the mass inside the shell on the shell itself, and the shell is outside the sphere. So the mass in the shell is a distance R away from the center of the shell inside it. If you want a more detailed proof, see this page.

I am still confused. In that wiki proof, it is a point mass on which they are calculating the force but we have a spherical shell here.

 

[phyzguy]

Consider the spherical shell as being made up of an assembly of point masses, or, if you like, an assembly of wedges of width dθ. Each one is attracted to the sphere inside of it as though the sphere inside is at a distance R away. So the same is true for the shell as a whole.

In fact, if it makes it easier for you, you don't need to consider the shell as a whole for this analysis. Just consider a piece of the shell of width dθ, calculate the (inward)gravitational force on it, and the (outward)pressure force on it and set them equal.

 

[Saitama]

Consider the spherical shell as being made up of an assembly of point masses, or, if you like, an assembly of wedges of width dθ. Each one is attracted to the sphere inside of it as though the sphere inside is at a distance R away. So the same is true for the shell as a whole.

In fact, if it makes it easier for you, you don't need to consider the shell as a whole for this analysis. Just consider a piece of the shell of width dθ, calculate the (inward)gravitational force on it, and the (outward)pressure force on it and set them equal.

Thanks phyzguy, please check if I am making the equations right. :)

F+(P+dP)(4\pi r^2)=P(4\pi r^2)
where F is the gravitational force.
F=(4\pi \rho_0)^2\frac{dr}{r}
Solving the equations, I get
dP=-\frac{4\pi \rho_0^2 G dr}{r^3}
From the ideal gas law.
P=\frac{\rho RT}{\mu}
dP=\frac{d\rho RT}{\mu}=-\frac{2\rho_0 RT dr}{r^3 \mu}
Equating the expressions for dP, T comes out to be constant. Is this correct?

 

[phyzguy]

Not quite right, I think. What is the mass inside R, if the density goes like:
\rho(r) = \frac{\rho_0}{r^2}

Edit: My mistake, I think you do have the gravitational force and differential equation for dP/dr correct.

 

[Saitama]

Not quite right, I think. What is the mass inside R, if the density goes like:
\rho(r) = \frac{\rho_0}{r^2}

$4\pi \rho_0 R$?

 

[phyzguy]

Correct. Ignore my earlier post. I think you've done it all correctly now.

 

[Saitama]

Correct. Ignore my earlier post. I think you've done it all correctly now.

Thanks a bunch phyzguy!

 

[phyzguy]

Glad to help!