Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to do this question and getting stuck at interpreting what the question is asking for.

$$dP/dR = -g(R)\rho(R) = -[GM(<R)/R^2]\rho(R).$$
Thus when we descend a distance $\Delta R$ from the surface $R_s$ where $P=0$, the pressure will be approximately:
$$P = 0-\Delta P = GM(<R)/R^2\rho(R)\Delta R.$$

By taking $\Delta R = R = R_s$, the radius of the star, and $\rho(R) = \bar{\rho} = 3M/(4\pi R_s^3)$, the mean density of the star, derive a rough estimate for the central pressure of the star and show that the central pressure is independent of mass for the scalings given above.

I am substituting the mean density equation instead of $\rho(R)$ into the second equation on the image and simplifying it. I am not sure that $<R$ mathematically represents and when I simplify it I get $P = (3M)/(4\pi R_s)$. I'm not sure if this is correct because my answer still has mass.

share|cite|improve this question
$M(<R)$ means "the mass that has radius less than R". So if the density is $\rho(r)$, then $M(<R) = \int_0^R4\pi\rho(r)dr$. – Carl Brannen Mar 12 '11 at 3:56
Welcome to You'll note that we have latex formatting courtesy of MathJax, so it is possible to transcribe formulas as Carl has done for us rather than linking to images. – dmckee Mar 12 '11 at 4:11
Uh, I left off a factor of $r^2$ in the above comment. And hey, I edited in the complete equations but they didn't make it. :( – Carl Brannen Mar 13 '11 at 6:13
up vote 1 down vote accepted

That's some terrible notation in the question, and I think that may be what's tripping you up. Try the more conventional notation for hydrostatic equilibrium $$\frac{dP}{dr}=-\frac{GM(r)\rho(r)}{r^{2}}$$

They are saying

"Make the approximations $dp\approx\Delta P$ and $dr\approx\Delta r$, so instead of a differential pressure between two infinitesimally separated bits of the star, we are comparing broad regions. Then, evaluate this for $\Delta r=R_{s}$, the total radius of the star (what is $\Delta P$ in this case?) and use $\rho(r)=\frac{3M_{total}}{4\pi R_{s}^{3}}$ (the average density)."

They drop a hint that you should get an expression which includes the pressure at $r=0$ and which doesn't include a mass term.

Hope that this restatement of the problem makes more sense.

share|cite|improve this answer
Hello Spencer, Thanks for the clarification. It did make it a bit clear but Im still not completely sure what to do. How do I evaluate this ? From what I understood from your answer is that the equation now becomes $\frac{\Delta P}{\Delta r} = -\frac{GM(R_s)(\frac{3M_total}{4\pi R_s^2})}{R_s^2} $ ? and then I solve for P?? – Kartik Mar 12 '11 at 23:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.