Assume that a star has a uniform mass density. Then how does the temperature scale as a function of the distance to the center?
Note: I am not looking for an answer that involves derivations or integrations.

Here is what I did:
The star is in equilibrium so the pressure at distance $r$ must equal the gravity force at distance $r$. Pressure is given by $P=nKT$, and so $P\times Area=F_g$.
Then $nKT \times 4\pi r^2=F_G$. The mass of the part of the star contained within radius $r$ from the center is $\frac{4}{3}\pi r^3\times m_sn$ where $m_s$ is the mass of each particle and $n$ is the number of particles per volume.
Then we can rewrite $nKT \times 4\pi r^2=m(\frac{G\frac{4}{3}\pi r^3\times m_snF}{r^2})$ and simplify to $KTr=\frac{m}{3}Gm_s$. Isolating $T$ we get that $T=\frac{mGm_s}{3Kr}$.
This seems to imply that the temperature decreases the further we are from the core, however I am not sure what $m$ should stand for in $F_g=mg$, should $m=m_s$, i.e be the mass of a single particle? Or does $m$ itself depend on the radius in which case the answer I got is invalid.
Can someone verify that the answer I got is right?

  • $\begingroup$ m is the mass of the material "above" that area. Top of the atmosphere, low pressure. Bottom of the atmosphere, high pressure. $\endgroup$
    – BowlOfRed
    Oct 30 '17 at 7:10
  • $\begingroup$ Think about what you're trying to balance. You say that the force due to thermal pressure is $P\times Area$, but what is that the area of? If the pressure is applied to the area of something, then that must necessarily equate to the force of gravity on that same something. You have a disconnect on what the pressure is applied to and what the force of gravity is pulling on. $\endgroup$
    – zephyr
    Oct 30 '17 at 13:53
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    $\begingroup$ Assuming such a star requires breaking some laws of physics, so I find it hard to see how one could come up with a reasonable estimation of anything. $\endgroup$ Oct 31 '17 at 18:29

I am not sure what you mean by looking for an answer without derivations? Of course you have to solve the maths to get a mathematical answer.

The handwaving answer is that the star has to be hotter in the middle, because at constant density the pressure depends on temperature and you need a negative pressure gradient to support the weight of the gas above.

The answer is:

The equation of hydrostatic equilibrium is $$\frac{dP}{dr} = -\rho g,$$ where $\rho$ is the mass density and $g$ is the gravity at radius $r$, which by Newton's shell theorem is due to the total mass interior to $r$.

This equation is easily derived by considering the force equilibrium on a slab of gas in the star (your error above was in not recognising that it is the difference in pressure across the slab that provides the upward force).

If the density (and composition and ionisation state) are constant, then this equation can be written as $$ \frac{\rho k_B }{m} \frac{dT}{dr} = -\rho \frac{4\pi G r^3 \rho}{3r^2},$$ where $m$ is the mass per particle and a perfect gas has been assumed.

Integrating with respect to radius and assuming $T(r=0)=T_0$, we obtain $$T = T_0 - \frac{2\pi G\rho m}{3 k_B}r^2$$

Finally, if the star has radius $R$, then this will be radius at $T=0$, since nothing could be supported beyond that. In which case $$T = \frac{2\pi G\rho m}{3 k_B}(R^2-r^2)$$

Or in terms of mass $M$ $$T = \frac{GM m}{2 k_B R^3}(R^2-r^2)$$

Constant density is of course a poor approximation. The above treatment gives a central temperature of 7 million Kelvin for the Sun, assuming $m\simeq 10^{-27}$ kg.


Based on info from CalTech, a constant density is unrealistic. I wouldn't place much faith in any equations derived from such an assumption.

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  • 2
    $\begingroup$ This completely skates over the point of the problem. This is clearly a homework question that the OP is trying to solve and the assumption of constant density is simply there to make the problem easier. Saying such a thing doesn't exist is like saying you can't do physics 101 because frictionless systems don't exist. $\endgroup$
    – zephyr
    Oct 31 '17 at 19:41
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    $\begingroup$ You do this type of calculation because it gives you an answer that is correct to within a factor of 2-3. Furthermore, you can say whether your answer is going to be high or low by physical reasoning. You could then bracket said answer by making a different assumption. $\endgroup$
    – ProfRob
    Oct 31 '17 at 21:12

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