Hydrostatic equilibrium of a star - simplification math error

Question

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.

Answer 1

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.