= Solution
{c}
Ignoring <refraction>, the line of sight is perpendicular to the radius at its tangent point. For a surface tangent, <Pythagorean theorem> gives the satellite-to-tangent distance
$$
\boxed{D=\sqrt{(R+H)^2-R^2}
=\sqrt{7171^2-6371^2}\,\mathrm{km}
=3291\,\mathrm{km}\simeq3300\,\mathrm{km}.}
$$
For tangent altitude $z_t$, replace $R^2$ by $(R+z_t)^2$; lower <atmospheric scale heights> change the estimate little. The required angular resolution is
$$
\theta\simeq\frac{8.5\,\mathrm{km}}{3291\,\mathrm{km}}
=2.58\times10^{-3}\,\mathrm{rad}.
$$
Using the <Rayleigh criterion> for a circular telescope aperture, $\theta=1.22\lambda/d$. The longest <wavelength> imposes the strongest condition, so
$$
\boxed{d_{\rm diffraction}\gtrsim
\frac{1.22(20\times10^{-6}\,\mathrm m)}{2.58\times10^{-3}}
=9.45\times10^{-3}\,\mathrm m\simeq9.5\,\mathrm{mm}.}
$$
The corresponding requirement at $7\,\mu\mathrm m$ is $3.3$ mm. This is a <diffraction> limit, without allowance for scan sampling, aberrations or instrumental field of view.
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