I have thought about this some more. The calculation you want can be done without detailed knowledge of the index-vs.-altitude dependence, you just need to know the pressure at sea level (or wherever the light beam is to be "observed") as well as at the satellite's altitude.
Note: fairly detailed solution follows; scroll down to the bottom for final result.
We want to know the time it takes for a light signal to reach a point on Earth at height
h0 from a satellite located at altitude
hs. The beam of light makes an angle
θ from the vertical. If the beam is directed straight downward, then
θ=0 and the cos
θ factor that appears in the derivation is simply equal to 1 and may be ignored.
From before, we had
<br />
\begin{flalign*}<br />
\text{time} & = & & \frac{1}{c} \int {n \ dx} \\<br />
\text{which may be written as} \\<br />
\text{time} & = & & \frac{1}{c} \int {(1+n-1) \ dx} \\<br />
\\<br />
& = & & \frac{1}{c} \int dx \ \ + \ \ \frac{1}{c} \int {(n-1) \ dx} \\<br />
\\<br />
& = & & \frac{x}{c} \ \ + \ \ \Delta t \\<br />
\text{where} \\<br />
\Delta t & \equiv & & \frac{1}{c} \int {(n-1) \ dx} \\<br />
\end{flalign*}<br />
In other words, the total time is simple the time x/c that light would take if traveling through a vacuum,
plus an additional time
Δt. We'll now derive an expression for
Δt which does not involve any integrals.
Using the (very good) approximation that (
n-1) is proportional to the density of air,
ρ, we have
<br />
\begin{flalign*}<br />
n-1 & = & & k \rho \ \ \text{ where } \ k \ \text{is a constant} \\<br />
\text{and so} \\<br />
\Delta t & \equiv & & \frac{1}{c} \int_{x_0}^{x_s} {(n-1) \ dx} \\<br />
\\<br />
& = & & \frac{k}{c} \int_{x_0}^{x_s} {\rho \ dx} \\<br />
\text{(Note } & \text{the} & & \text{ inclusion now of the integration limits, indicating the observation point and the satellite position.)} \\<br />
\text{We know } & \text{that} & & \text{ the pressure at an altitude} \ h_0 \ \text{is the weight of air mass above it in a unit-area, or} \\<br />
P_0 & = & & \int_{h_0}^{\inf} \rho \ g \ dh \\<br />
\text{and} \\<br />
P_0 - P_s & = & & \int_{h_0}^{h_s} \rho \ g \ dh \ \ \text{is the pressure difference between the two altitudes } h_0 \text{ and } h_s.<br />
\end{flalign*}<br />
Assume
g is constant over altitude changes small compared to the 6400 km radius of the Earth.
Also,
dh = dx cos
θ, where
θ is the angle of the light beam from vertical (i.e., "straight down" would mean that
θ=0, cos
θ=1, and
dh=dx.
Continuing with the derivation:
<br />
\begin{flalign*}<br />
P_0 - P_s & = & & \int_{h_0}^{h_s} \rho \ g \ dh \\<br />
& = & & g \ \cos \theta \int_{x_0}^{x_s} \rho \ dx \\<br />
\text{so that} \\<br />
\int_{x_0}^{x_s} \rho \ dx & = & & \frac{P_0 - P_s}{g \ \cos \theta} \\<br />
\text{and} \\<br />
\Delta t & = & & \frac{k}{c} \int_{x_0}^{x_s} {\rho \ dx} \\<br />
\\<br />
& = & & \frac{k}{c} \cdot \frac{P_0 - P_s}{g \ \cos \theta} <br />
\end{flalign*}<br />
We are getting there, but we need to know what the constant
k is. As long as we know that the refractive index is
nk at some
known pressure and temperature
Pk and
Tk, then
<br />
\begin{flalign*}<br />
k & = & & \frac{n_k-1}{\rho_k} \\ <br />
\text{where } \rho_k & = & & \text{ air density at } P_k \text{ and } T_k, \\<br />
\text{Since} \\<br />
P \ V & = & & N \ R \ T \ \text{, the (hopefully) familiar Ideal Gas Law,} \\<br />
\text{then} \\<br />
\rho_k & = & & \frac{\text{mass}}{\text{volume}} = \frac{\text{moles}}{\text{volume}} \cdot \frac{\text{mass}}{\text{mole}} \\<br />
\\<br />
& = & & (\frac{N}{V})_k \cdot m_m \ \text{, where } m_m = \text{ molar mass of air} \\<br />
\\<br />
& = & & \frac{P_k}{R \ T_k} \cdot m_m \\<br />
\text{and} \\<br />
k & = & & (n_k - 1) \cdot \frac{R \ T_k}{P_k \ m_m} \\<br />
\end{flalign*}<br />
Therefore
<br />
\begin{flalign*}<br />
\Delta t & = & & \frac{k (P_0-P_s)}{c \ g \ \cos \theta} \\<br />
\\<br />
& = & & \frac{1}{\cos \theta} \cdot \frac{R}{c \ g \ m_m} <br />
\cdot \frac{T_k \ (n_k-1)}{P_k} \cdot (P_0-P_s)<br />
\end{flalign*}<br />
And the total time is given by
time = x/c + Δt
Note: We are neglecting the deviation in
θ due to refraction. We also neglect effects due to the Earths' curvature, which is okay as long as
θ is not very close to 90
o, i.e. the satellite does not appear close to the horizon when viewed from the observation point.
If I have time tomorrow, I'll put some numbers on this. Note that if
R uses N/m
2 pressure units then
mm should be expressed in kg/mole, instead of the more familiar g/mole values given in the periodic table.
