# Math Help - Separable Equation Help

1. ## Separable Equation Help

I was assigned a problem that stated pressure, p, and density, ρ, are related by the equation dp = -gρ dy. It said it follow the adiabatic equation p=p0(ρ/ρ0)^γ, where γ is not equal to 0, and p0 is pressure at earths surface and ρ0 is density at earths surface. It then wants me to show that

p = p0(1-((γ-1)/γ)*((ρ0gy)/p0))^(γ/(γ-1))

I've tried for an hour or more, and cannot figure out this problem. If anyone has any insight or even knows where to start, it would be greatly appreciated.

2. hi there rickyticki, I have a suggestion.

Have a go at solving this

Originally Posted by rickyticki
I was assigned a problem that stated pressure, p, and density, ρ, are related by the equation dp = -gρ dy
After this you will have an equation for p, for this subtitute in

Originally Posted by rickyticki
p=p0(ρ/ρ0)^γ,
and if all goes well it might end up being

Originally Posted by rickyticki

p = p0(1-((γ-1)/γ)*((ρ0gy)/p0))^(γ/(γ-1))

3. Thanks for the suggestion.

When I do that, I end up with -gρy = p0(ρ/ρ0)^(γ).

Is that how it should be? The problem is I do not have a p variable to solve for now.

4. I see, I was confusing your $\rho$ and $p$

5. yes, and the $\gamma$ and $y$ are different too.

didnt know how to use math symbols

6. Originally Posted by rickyticki
I was assigned a problem that stated pressure, p, and density, ρ, are related by the equation dp = -gρ dy. It said it follow the adiabatic equation p=p0(ρ/ρ0)^γ, where γ is not equal to 0, and p0 is pressure at earths surface and ρ0 is density at earths surface. It then wants me to show that

p = p0(1-((γ-1)/γ)*((ρ0gy)/p0))^(γ/(γ-1))

I've tried for an hour or more, and cannot figure out this problem. If anyone has any insight or even knows where to start, it would be greatly appreciated.
This is a somewhat confusing question. The first thing to do is to obtain the derivative of the adiabatic equation and substitute this in the given DE. This is:

$dp=p_0 \gamma \left( \frac{\rho}{\rho_0}\right) ^{\gamma-1}\frac{d \rho}{\rho_0}$

and so:

$\frac{p_0 \gamma}{\rho_0^{\gamma}} \rho^{\gamma-2}d \rho = -gdy$

Integrating this gives:

$\frac{p_0}{\rho_0^{\gamma}}\frac{\gamma}{\gamma-1} \rho^{\gamma-1}= -gy + K$

The constant of integration can be found by the boundary condition that at y=0, the density [imath]\rho=\rho_0[/imath]. This gives:

$K=\frac{p_0}{\rho_0}\frac{\gamma}{\gamma-1}$

The solution is now -after some algebra- the one given in the post as:

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

coomast