The Sunrise

Let's consider an atmosphere with radius $R_E + h$ and a single refraction coefficient $n_A$ . In reality, the coefficient depends on the height of the atmosphere, so we should integrate the influence of it over the height (this would complicate the problem a lot).

We can use Snell's Law (https://brilliant.org/wiki/snells-law/) to relate the angle of incidence $\theta_1 = \theta_2 +\Delta \theta$ and the angle of refraction $\theta 2$ :

$\boxed{ n_{void} \sin{\theta_1} = n_{A} \sin{\theta_2} }$

( $n_{void} = 1$ )

Using some geometry, we can determine the sine of the angle of refraction: $\sin{\theta_2} = \frac{R_E}{R_E+h}$ . The expression simplifies to:

$\theta_1 = \sin^{-1} \left[ n_{A} \frac{R_E}{R_E+h} \right]$

...so the angle $\Delta \theta = \sin^{-1} \left[ n_{A} \frac{R_E}{R_E+h} \right] - \sin^{-1}\left[\frac{R_E}{R_E+h}\right]$

With this angle, we can now determine the distance of the sun under the horizon $d$ :

$d \approx (D_{ES}-h)\cdot \Delta \theta \approx D_{ES} \cdot \Delta \theta$ .

So: $\boxed{ d \approx D_{ES} \cdot \left( \sin^{-1} \left[ n_{A} \frac{R_E}{R_E+h} \right] - \sin^{-1}\left[\frac{R_E}{R_E+h}\right] \right)= 750 000 \text{km}}$

Let the angle of incidence be $α$ and the angle of refraction be $β$ . Then $\sin β=\dfrac{R}{R+h}\approx 0.998199\implies \cos β\approx 0.06129, \cot β\approx 0.0614$ . Here $R$ and $h$ are the radius of earth and height of atmosphere respectively. Also $\sin (α-β)\approx α-β\approx \dfrac{d}{S}=\dfrac{d}{150\times 10^6}$ . R.I. of atmosphere $=n=1.0003=\dfrac {\sin α}{\sin β}\approx 1+(α-β)\cot β\implies α-β=\dfrac{0.0003}{\cot β}\approx 0.00488$ . So $\dfrac{d}{150\times 10^6}\approx 0.00488\implies d\approx \boxed {732835.82}$ km.