Differential Equations Pressure and Density derivation

The pressure $p$, and the density, $\rho$, of the atmosphere at a height $y$ above the earth's surface are related by $dp = -g \rho\; dy$. Assuming that $p$ and $\rho$ satify the adiabatic equation of state $p = p_0(\rho/\rho_0)^\gamma$, where $\gamma \neq 1$ is a constant and $p_0$ and $\rho_0$ denote the pressure and density at the earth's surface, respectively, show that

$$p = p_0\left[1 - \frac{\gamma - 1}{\gamma} \left(\frac{\rho_0 g y}{p_0}\right)\right]^{(\gamma - 1)/\gamma}$$

~~~~~~~~~~~~~~

My work:

I separated variables and got $p = -g\rho y + c$. That's as far as I got.

• Sorry to tell you this, but since $\rho$ is a function of $y$, even the first step you took is incorrect. – David K Aug 28 '15 at 1:23
• If I remember correctly, the exponent of the correct equation for pressure is $(\gamma - 1)/\gamma$, not $(\gamma - 1/\gamma)$. Did you copy the formula incorrectly? That will make it harder to confirm when you get the right answer. – David K Aug 28 '15 at 1:26

to make typing easier, i will use $k, r$ for $\gamma$ and $\rho$ respectively.

we are given $$dp = -gr dy, \, p = p_0 (r/r_0)^k.$$ we will solve for $r$ first, and then use the second relation to get hold of $p.$

taking logarithm and differentiating the $p$-$r$ relation gives us $$k\frac{dr}r = \frac{dp}{p} = -\frac{gr\,dy}p = -\frac{gr_0^kr\, dy}{p_0r^k}$$ separating the variables $$kr^{k-2} \, dr = -\frac{gr_0^k}{p_0}\, dy$$ and on integration yields $$\frac k{k-1}\left(r/ r_0\right)^{k-1} = -gy+C$$ setting the initial condition $r =r_0$ at $y= y_0$ gives $$(r/r_0)^{k-1} = \frac{k-1}{k}\left( \frac k{k-1}-gy\right)=1-\frac{k-1}kgy$$

i hope you can complete the rest.

You are given two equations:

$$dp = -g \rho\; dy \tag1$$

and

$$p = p_0(\rho/\rho_0)^\gamma. \tag2$$

You cannot simply integrate both sides of $(1)$, treating $\rho$ as a constant, because in fact $\rho$ is less at high altitudes than at low altitudes; it is not constant. What you can do is use $(2)$ to eliminate $\rho$ from $(1)$, leaving you with an equation involving only $p$, $y$, and some constants. You can then do your separation of variables and integrate.