The Temperature Profile of a Spherical Cloud of Ideal Gas

In summary: 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. Then the force on the piece of the shell is the same as the force on the whole shell.
  • #1
Saitama
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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
[tex]PV=nRT[/tex]
[tex]P(dV)=\frac{dm}{\mu}RT=\frac{dV \rho(r)}{\mu}RT[/tex]
[tex]P=\frac{\rho_0}{\mu r^2}RT[/tex]
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. :confused:

Any help is appreciated. Thanks!
 

Attachments

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  • #2
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
phyzguy said:
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. :tongue2:

Any hints about how would I form the equations then?
 
  • #4
Pranav-Arora said:
Well, I thought that there is no gravity in intergalactic space. :tongue2:

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.
 
  • #5
phyzguy said:
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? :confused:

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
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:

[tex] M1 = \int_0^R \rho(r) 4\pi r^2 dr[/tex]
[tex]M2 = 4\pi R^2 \rho(R) dR[/tex]
d = R
 
  • #7
phyzguy said:
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:

[tex] M1 = \int_0^R \rho(r) 4\pi r^2 dr[/tex]
[tex]M2 = 4\pi R^2 \rho(R) dR[/tex]
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.
 
  • #8
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
phyzguy said:
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.
 
  • #10
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
phyzguy said:
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. :)

[tex]F+(P+dP)(4\pi r^2)=P(4\pi r^2)[/tex]
where F is the gravitational force.
[tex]F=(4\pi \rho_0)^2\frac{dr}{r}[/tex]
Solving the equations, I get
[tex]dP=-\frac{4\pi \rho_0^2 G dr}{r^3}[/tex]
From the ideal gas law.
[tex]P=\frac{\rho RT}{\mu}[/tex]
[tex]dP=\frac{d\rho RT}{\mu}=-\frac{2\rho_0 RT dr}{r^3 \mu}[/tex]
Equating the expressions for dP, T comes out to be constant. Is this correct?
 
  • #12
Not quite right, I think. What is the mass inside R, if the density goes like:
[tex]\rho(r) = \frac{\rho_0}{r^2}[/tex]

Edit: My mistake, I think you do have the gravitational force and differential equation for dP/dr correct.
 
  • #13
phyzguy said:
Not quite right, I think. What is the mass inside R, if the density goes like:
[tex]\rho(r) = \frac{\rho_0}{r^2}[/tex]

##4\pi \rho_0 R##?
 
  • #14
Correct. Ignore my earlier post. I think you've done it all correctly now.
 
  • #15
phyzguy said:
Correct. Ignore my earlier post. I think you've done it all correctly now.

Thanks a bunch phyzguy! :smile:
 
  • #16
Glad to help!
 

1. What is a spherical cloud of gas?

A spherical cloud of gas is a mass of gas particles that is roughly spherical in shape due to gravity pulling the particles towards the center.

2. How do spherical clouds of gas form?

Spherical clouds of gas can form through various processes such as the collapse of interstellar gas clouds, the explosion of a supernova, or the collapse of a dying star.

3. What types of gases can be found in a spherical cloud of gas?

Spherical clouds of gas can contain a variety of gases, including hydrogen, helium, and various trace elements such as carbon, oxygen, and nitrogen.

4. What role do spherical clouds of gas play in the formation of stars and galaxies?

Spherical clouds of gas are important in the formation of stars and galaxies as they provide the raw materials for these objects to form. As the gas collapses under its own gravity, it can form stars or be incorporated into larger structures such as galaxies.

5. Are spherical clouds of gas visible to the naked eye?

Most spherical clouds of gas are not visible to the naked eye as they are typically located in deep space and are too faint to be seen without the aid of telescopes. However, some spherical clouds of gas, such as the Orion Nebula, can be visible to the naked eye under dark sky conditions.

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