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$$v(t) = u\ln\left({\frac{m_0}{m_0-\alpha t}}\right)-gt $$ Is the typical equation for the velocity of a rocket under gravity, with no air drag. Now I want to solve it, but I have no idea how to solve it for $v(t) = 0$. Clearly one solution is $t_1=0$. Plugging in the values numerically ($m_0 = 70000, u = 2500, g = 10, \alpha = 250$) Mathematica tells me the second root is $t_2 = 57.7766$

How can I solve this equation? I tried rewriting $t = \ln{e^t}$ but this didn't help me at all as I end up with an equation of the type $g(t,e^t) = 0$.

Any help would be appreciated!

Euler_Salter
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1 Answers1

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This is a transcendental equation, and there are probably no closed form solutions. However, you can use a series approximation: $$\ln\left(\frac{1}{1-x}\right)\approx x+ \frac{x^2}{2} + \frac{x^3}{3}+\ldots$$ To get: $$gt = u \ln \left(\frac{m}{m-\alpha t}\right)= u \ln \left(\frac{1}{1-(\alpha/m) t}\right)\approx u (\alpha/m) t + u\frac{(\alpha/m)^2 t^2}{2}$$ Leading to: $$\frac{u\alpha^2}{2m^2}t \approx(g - (u\alpha/m))$$ $$t\approx \frac{2m^2(g - (u\alpha/m))}{u\alpha^2}\approx 67 \ \text{sec}$$ Which isn't that far off from your $57.78$. If you keep the cubic term in the approximation, you'll get a quadratic equation which you can solve to get $t\approx 59\ \text{sec}$, which is even closer.

Nathaniel Bubis
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  • thank you! You said that probably there are no closed solutions. Is there a way to know if there are any? – Euler_Salter Dec 06 '16 at 21:36
  • @Euler_Salter - by a closed form solution I meant a solution expressible using elementary functions. You could always invent a new function that satisfies your solution, or in this case use the Lambert W function. – Nathaniel Bubis Dec 06 '16 at 21:47
  • can I just ask you how you went about approximating $\ln{\left(\frac{1}{1-x}\right)}$? Did you use the Taylor's expansion? Around which point? – Euler_Salter Dec 06 '16 at 23:58