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Quote: Originally Posted by dumdidum http://www.mathhelpforum.com/math-he...1&d=1237099772 Thanks for your help :D ...I can tell from the looks of it that it's a fairly simple question, but haven't done high school maths in...well, 5 years Do I simply plug p=A/r back into dP/dr....it seems a little too simple This was my try: ρ = A/r dρ/dr = –A/r^2 dP/dr = – GMρ / r^2 = – (G x M x –A/r^2) / r^2 = (G x M x A/r^2) / r^2 = GMA Ps = 0 (Ps – Pc)(rs – rc) = (0 – Pc) / r = –Pc / r –Pc / r = GMA Pc / r = – GMA Pc = – GMAr You start with: $\displaystyle \frac{dP}{dr}=-\frac{GM \rho}{r^2}$ Then as $\displaystyle \rho=\frac{A}{r}$ you have: $\displaystyle \frac{dP}{dr}=-\frac{GMA}{r^3}$ Now just rework the original argument with the new right hand side. CB

Thanks CB. Guess it really was that simple (Nod)