= Solution
Assume a circular <orbit>, nadir viewing, negligible planetary rotation and a fixed camera relative to the orbital ground track. If $R_V$ is the planet's radius and $h$ the orbital altitude, the orbital angular speed is
$$
n=\sqrt{\frac{GM_V}{(R_V+h)^3}}=1.137\times10^{-3}\,\mathrm{s^{-1}}.
$$
The nadir footprint moves across the surface at $v_g=nR_V\simeq6.88\times10^3\,\mathrm{m\,s^{-1}}$, rather than the slightly larger satellite speed $n(R_V+h)$. Hence
$$
\boxed{\Delta t\simeq\frac{200\,\mathrm m}{nR_V}=2.91\times10^{-2}\,\mathrm s.}
$$
This is an upper integration time for approximately one $200\,\mathrm m$ along-track sample without motion compensation. Using satellite speed directly changes it by only about four per cent; the expression above keeps the projection onto the planetary surface.
Back to article page
