Can you derive the Tsiolkovsky rocket equation without calculus? Can you derive the Tsiolkovsky rocket equation without calculus?  
 A: No. You want to get to the total change in velocity. To do this, you have to add up (integrate) all the tiny contributions from each little lump of fuel as it is ejected. Each contribution is different because as each "lump" is ejected, the rocket gets lighter.
You could get an estimate if you work out the $\Delta{v}$ from the first ton of fuel, reduce the rocket mass by 1 ton, get the next $\Delta{v}$ from the next ton, reduce by another ton, and so on until you use up all the fuel, then add all the $\Delta{v}$s - but this would give only a crude estimate of the total. You could repeat the calculation and do it for each kg of fuel (1000 $\times$ as many steps). This would get you a better result, but still not precise.
Finally, you'd do it for an infinite number of infinitesimal quantities of fuel at each step - which is exactly how integral calculus works.
A: Tsiolkovsky rocket equation gives the velocity of rocket at any instant as $v=v_{0}+u \log _{e}\left(\frac{m_{0}}{m}\right)$, and if effect of gravity is taken into account, then $v=v_{0}+u \log _{e}\left(\frac{m_{0}}{m}\right)-g t$. I found a simple derivation with little calculus involved.


