Expression for Atmospheric Pressure with Altitude, including Tidal Forces Is this simple to obtain?  I've tried to modify the constant-gravity model to use the $GM/r^2$ form instead, but my result was shown to be wrong with some testing.
Constant Gravity Model
You can find this in many places online.  Here's one.  I'll repeat that work here, jumping over the calculus, and putting it in terms of Earth's radius.  r0 will be the surface radius.
$$ \frac{ dP }{ dr } \frac{ 1 }{ P(r) } = - \frac{ M_0 g }{ R T }  \\
H \equiv \frac{ R T }{ M_0 g } \approx 7.4 \text{ km} \\
P(r) = P_0 e^{ \frac{ r_0 - r }{ H } }  $$
Non-constant Gravity Model
So to generalize this (for a small moon with a diffuse atmosphere, for instance), just replace the gravity with its expression $g(r)=GM/r^2$.  M0 is still the average formula mass of air, but we add M, which is the mass of the planet.  This should work fine for Earth, I see no reason it shouldn't.
$$ \frac{ dP }{ dr } \frac{ 1 }{ P(r) } = - \frac{ M_0 G M }{ R T } \frac{1}{r^2}  \\
H' \equiv \frac{ M_0 G M  }{R T } \approx 4,739 \text{ km} \\
P(r) = P_0 e^{ H' \left( \frac{1}{r} - \frac{1}{r_0} \right) }  $$
It looks so nice and congruent.  But I plot my non-constant gravity model side-by-side with the constant model and it's obviously wrong.
So for my question, could somebody do ANY of:


*

*Find something I did wrong

*Derive or find a correct model for this

*Give a clear reason it can't be obtained in a simple closed form

 A: Wait, don't bother.  I found the embarrassing error.  It's the H' value.  It should have been:
$$ H' \equiv \frac{ M_0 G M  }{R T } \approx 4,739,000 \text{ km} $$
I just read that output wrong.  The plots now seem to fit.
Sorry to answer my own question so fast.  I suppose whether any of this is right remains an open question.
A: My solution has been challenged, and this comes from a separate source so I'm posting it as community wiki.  The reference is here:
http://web.ist.utl.pt/~berberan/data/43.pdf
The proposed equation is:
$$ P(z) = P(0) \exp{ \left( - \frac{ m g_0 R_0 }{ kT} \frac{ z }{ z+R_0 } \right) }$$
You could put this in terms of characteristic height if you wanted.  The question in my mind is rather this solves the same equation that I was aiming for.
Any insight as to what might be wrong with the differential equation I formulated would be appreciated.

Also, their equation references equations 5 and 7 in that paper for a justification of where it comes from.  I'll echo those here.  Equation 5:
$$ p(z) = p(H) + \frac{m}{k} \int_z^H \frac{g p(u) }{ T} du$$
Equation 7:
$$ \frac{dp}{dz} = - \frac{ m g }{ k T } p $$
The idea is that you substitute in $g(z)$ in place of $g$.  Honestly, I think this is what I had.  Their math looks a little different, but it should come out to the same thing.  I don't know what went wrong.

I think I've solved it now.  The problem is that the other source uses altitude (z), whereas I have used radius (r).  It also used surface gravity where I used mass and the gravitational constant.  If you just re-write my equations from the OP and make these substitutions, you can match the two expressions.  I have done that here:
$$ z \equiv r_0 + z \\
H' \equiv \frac{ M_0 G M  }{R T } \approx 4,739 \text{ km} \\
P(r) = P_0 e^{ H' \left( \frac{1}{r} - \frac{1}{r_0} \right) }  \\
P(z) = P_0 e^{ H' \left( \frac{1}{r_0 + z} - \frac{1}{r_0} \right) }  \\
P(z) = P_0 e^{ - \frac{ H'}{r_0} \frac{ z}{ r_0 + z} } \\
P(z) = P_0 e^{ - \frac{ M_0 G M  }{R T r_0}  \frac{ z}{ r_0 + z} } \\
g_0 \equiv \frac{ G M }{ r_0^2 } \\
P(z) = P_0 \exp{ \left( - \frac{ M_0 g_0 r_0 }{R T }  \frac{ z}{ r_0 + z} \right) } $$
These are, in fact, the same thing.
