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Simplification to first order

$\ddot{s} = \dot{s} \dfrac{d\dot{s}}{ds}$ hence Equation (??) becomes:

\begin{aligned} \dot{s} \dfrac{d\dot{s}}{ds} &= \dfrac{GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right] - \dfrac{\rho C_D A\dot{s}^2}{2m}. \end{aligned}

Letting $u=\dot{s}^2$ , then $\dfrac{du}{ds} = 2\dot{s}\dfrac{d\dot{s}}{ds}$ and so:

\begin{aligned} \dfrac{1}{2}\dfrac{du}{ds} &= \dfrac{GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right] - \dfrac{\rho C_D A u}{2m}. \end{aligned}

Multiplying both sides by 2 yields:

\begin{aligned} \dfrac{du}{ds} &= \dfrac{2GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right] - \dfrac{\rho C_D A u}{m}. \end{aligned}

Adding the $u$ term to the left-hand side:

\begin{aligned} \dfrac{du}{ds} + \dfrac{\rho C_D A}{m}u &= \dfrac{2GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right]. \end{aligned}

© Brenton Horne. I do not claim to be infallible, so feel free to open issues or pull requests on GitHub if you notice flaws or things that can be improved.
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Last modified: October 23, 2026.