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Lets say we shoot a rocket from Earth and assume it starts with a base velocity. For something close to Earth one would use this to calculate its height depending on time etc. $$ \frac{1}{2}at^2 + v_0t + h_0 = h(t) $$ If we try to calculate its height, we have to consider that the acceleration drops depending on its height. Thus the output of the function depends on itself (recursive). Normally I would solve this by programming, but I am interested if there is a "prettier" method. I know about using energies to calculate the height, but this method cannot be used if i have something like air friction or a continuous thrust.

Does somebody know about ways to calculate such trajectories or is it common to do it using programs with small timesteps.

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  • $\begingroup$ Do not forget that the mass of the rocket is variable, as per en.wikipedia.org/wiki/Variable-mass_system $\endgroup$ – ZeroTheHero Apr 14 '17 at 18:24
  • $\begingroup$ @ZeroTheHero Absolutely, thats a factor which would change the acceleration by far. Thanks for pointing that out! $\endgroup$ – RIJIK Apr 16 '17 at 12:33
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The moment acceleration becomes a function of time (burn characteristics of rocket, changing mass of rocket as fuel is spent), velocity (drag) and height (air density -> drag), it becomes very hard to give a general solution to this problem that isn't somehow numerical in nature.

Note - depending on the integration scheme that you use, the time steps don't have to be "very small". There are higher order methods (such as fourth-order Runge-Kutta) that are exact as long as the function is smooth and well-behaved. But you do have to use a "proper" integration scheme for these things to work reasonably well.

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  • $\begingroup$ I am not exactly sure what you mean by numerical. Does that mean that it can only done with computers and approximation algorythms? $\endgroup$ – RIJIK Apr 14 '17 at 19:49
  • $\begingroup$ Yes - "numerical" means you do it by number crunching (rather than elegant mathematical equations). Of course Runge-Kutta was developed around 1900, well before computers. But doing it by hand is laborious (which is why an efficient method that doesn't require tiny steps was such an important development. ) And the "approximation" of a well written numerical algorithm is really very good - quite likely, better than the approximation you are almost certainly making in your equations. $\endgroup$ – Floris Apr 14 '17 at 20:34
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For a start substitute in $acceleration=\frac{GM}{height(time)}$ and rearrange for height(time). Can you solve this?

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  • $\begingroup$ 1/2*("old acceleration")*time^2 + velocitytime - sqrt(GM/("new acceleration")); I don't understand the purpose of this, because its exactly what i don't need since it takes the accelerations in steps. $\endgroup$ – RIJIK Apr 14 '17 at 19:38
  • $\begingroup$ I did not refer to old and new acceleration only to acceleration. Do you know how to re-arrange an equation? $\endgroup$ – JMLCarter Apr 15 '17 at 17:18
  • $\begingroup$ I do, though i don't understand why it matters in this case. $\endgroup$ – RIJIK Apr 16 '17 at 9:26

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