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Simplification to first order

s¨=ds˙dsdsdt=dsdtds˙ds=s˙ds˙ds.\begin{aligned} \ddot{s} &= \dfrac{d\dot{s}}{ds} \dfrac{ds}{dt} \\ &= \dfrac{ds}{dt} \dfrac{d\dot{s}}{ds} \\ &= \dot{s} \dfrac{d\dot{s}}{ds}. \end{aligned}

Hence Equation (??) becomes, after dividing by mm:

s˙ds˙ds=g(sin⁡θ(s)−Crrcos⁡θ(s))−ρCDAs˙22m.\begin{aligned} \dot{s} \dfrac{d\dot{s}}{ds} &= g(\sin{\theta(s)} - C_{rr} \cos{\theta(s)}) - \dfrac{\rho C_D A\dot{s}^2}{2m}. \end{aligned}

Let u=s˙2u=\dot{s}^2, then:

duds=d(s˙2)ds\begin{aligned} \dfrac{du}{ds} &= \dfrac{d(\dot{s}^2)}{ds} \end{aligned}

Hence by the chain rule:

duds=2s˙ds˙ds.\begin{aligned} \dfrac{du}{ds} &= 2\dot{s}\dfrac{d\dot{s}}{ds}. \end{aligned}

Consequently:

12duds=g(sin⁡θ(s)−Crrcos⁡θ(s))−ρCDAu2m.\begin{aligned} \dfrac{1}{2}\dfrac{du}{ds} &= g(\sin{\theta(s)} - C_{rr} \cos{\theta(s)}) - \dfrac{\rho C_D A u}{2m}. \end{aligned}

Multiplying both sides by 2 yields:

duds=2g(sin⁡θ(s)−Crrcos⁡θ(s))−ρCDAum.\begin{aligned} \dfrac{du}{ds} &= 2g(\sin{\theta(s)} - C_{rr} \cos{\theta(s)}) - \dfrac{\rho C_D A u}{m}. \end{aligned}

Adding the uu term to the left-hand side:

duds+ρCDAum=2g(sin⁡θ(s)−Crrcos⁡θ(s)).\begin{aligned} \dfrac{du}{ds} + \dfrac{\rho C_D A u}{m} &= 2g(\sin{\theta(s)} - C_{rr} \cos{\theta(s)}). \end{aligned}
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Last modified: October 23, 2026.