# Homework Help: Spherical cloud of gas

1. Apr 15, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment 1)

2. Relevant equations

3. 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!

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2. Apr 15, 2013

### 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.

3. Apr 15, 2013

### Saitama

Well, I thought that there is no gravity in intergalactic space. :tongue2:

Any hints about how would I form the equations then?

4. Apr 15, 2013

### phyzguy

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

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.

5. Apr 15, 2013

### Saitama

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?

6. Apr 15, 2013

### 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

7. Apr 15, 2013

### Saitama

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

8. Apr 15, 2013

### 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.

9. Apr 15, 2013

### Saitama

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.

10. Apr 15, 2013

### 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.

11. Apr 15, 2013

### Saitama

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?

12. Apr 15, 2013

### 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.

13. Apr 15, 2013

### Saitama

$4\pi \rho_0 R$?

14. Apr 15, 2013

### phyzguy

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

15. Apr 15, 2013

### Saitama

Thanks a bunch phyzguy!

16. Apr 15, 2013