Editing Tsiolkovsky rocket equation (section)

===Other derivations===

====Impulse-based====
The equation can also be derived from the basic integral of acceleration in the form of force (thrust) over mass.
By representing the delta-v equation as the following:
<math display="block">\Delta v = \int^{t_f}_{t_0} \frac{|T|}{{m_0}-{t} \Delta{m}} ~ dt</math>

where T is thrust, <math>m_0</math> is the initial (wet) mass and <math>\Delta m</math> is the initial mass minus the final (dry) mass,

and realising that the integral of a resultant force over time is total impulse, assuming thrust is the only force involved,
<math display="block">\int^{t_f}_{t_0} F ~ dt = J</math>

The integral is found to be:
<math display="block">J ~ \frac{\ln({m_0}) - \ln({m_f})}{\Delta m}</math>

Realising that impulse over the change in mass is equivalent to force over propellant mass flow rate (p), which is itself equivalent to exhaust velocity,
<math display="block"> \frac{J}{\Delta m} = \frac{F}{p} = V_\text{exh}</math>
the integral can be equated to
<math display="block">\Delta v = V_\text{exh} ~ \ln\left({\frac{m_0}{m_f}}\right)</math>

====Acceleration-based====
Imagine a rocket at rest in space with no forces exerted on it ([[Newton's laws of motion|Newton's first law of motion]]). From the moment its engine is started (clock set to 0) the rocket expels gas mass at a ''constant mass flow rate R'' (kg/s) and at ''exhaust velocity relative to the rocket v<sub>e</sub>'' (m/s). This creates a constant force ''F'' propelling the rocket that is equal to ''R'' × ''v<sub>e</sub>''. The rocket is subject to a constant force, but its total mass is decreasing steadily because it is expelling gas. According to [[Newton's laws of motion|Newton's second law of motion]], its acceleration at any time ''t'' is its propelling force ''F'' divided by its current mass ''m'':
<math display="block"> ~ a = \frac{dv}{dt} = - \frac{F}{m(t)} = - \frac{R v_\text{e}}{m(t)}</math>

Now, the mass of fuel the rocket initially has on board is equal to ''m''<sub>0</sub> – ''m<sub>f</sub>''. For the constant mass flow rate ''R'' it will therefore take a time ''T'' = (''m''<sub>0</sub> – ''m<sub>f</sub>'')/''R'' to burn all this fuel. Integrating both sides of the equation with respect to time from ''0'' to ''T'' (and noting that ''R = dm/dt'' allows a substitution on the right) obtains: 
<math display="block"> ~ \Delta v = v_f - v_0 = - v_\text{e} \left[ \ln m_f - \ln m_0 \right] = ~ v_\text{e}  \ln\left(\frac{m_0}{m_f}\right).</math>

==== Limit of finite mass "pellet" expulsion ====
The rocket equation can also be derived as the limiting case of the speed change for a rocket that expels its fuel in the form of <math>N</math> pellets consecutively, as <math>N \to \infty</math>, with an effective exhaust speed <math>v_\text{eff}</math> such that the mechanical energy gained per unit fuel mass is given by <math display="inline">\tfrac{1}{2} v_\text{eff}^2 </math>.

In the rocket's center-of-mass frame, if a pellet of mass <math>m_p</math> is ejected at speed <math>u</math> and the remaining mass of the rocket is <math>m</math>, the amount of energy converted to increase the rocket's and pellet's kinetic energy is
<math display="block">\tfrac{1}{2} m_p v_\text{eff}^2 = \tfrac{1}{2}m_p u^2 + \tfrac{1}{2}m (\Delta v)^2. </math>

Using momentum conservation in the rocket's frame just prior to ejection, <math display="inline"> u = \Delta v \tfrac{m}{m_p}</math>, from which we find 
<math display="block">\Delta v = v_\text{eff} \frac{m_p}{\sqrt{m(m+m_p)}}.</math>

Let <math>\phi</math> be the initial fuel mass fraction on board and <math> m_0</math> the initial fueled-up mass of the rocket. Divide the total mass of fuel <math>\phi m_0</math> into <math>N</math> discrete pellets each of mass <math>m_p = \phi m_0/N</math>. The remaining mass of the rocket after ejecting <math>j</math> pellets is then <math>m = m_0(1 - j\phi/N)</math>. The overall speed change after ejecting <math>j</math> pellets is the sum
<ref name="discreteproof">{{cite journal |last1=Blanco |first1=Philip |title=A discrete, energetic approach to rocket propulsion|journal=Physics Education |date=November 2019 |volume=54|issue=6 |pages=065001 |doi=10.1088/1361-6552/ab315b|bibcode=2019PhyEd..54f5001B |s2cid=202130640 }}</ref>
<math display="block"> \Delta v = v_\text{eff} \sum ^{j=N}_{j=1} \frac{\phi/N}{\sqrt{(1-j\phi/N)(1-j\phi/N+\phi/N)}} </math>

Notice that for large <math>N</math> the last term in the denominator <math>\phi/N\ll 1</math> and can be neglected to give
<math display="block"> \Delta v \approx v_\text{eff} \sum^{j=N}_{j=1}\frac{\phi/N}{1-j\phi/N} = v_\text{eff} \sum ^{j=N}_{j=1} \frac{\Delta x}{1-x_j} </math> where <math display="inline"> \Delta x = \frac{\phi}{N}</math> and <math display="inline"> x_j = \frac{j\phi}{N} </math>.

As <math> N\rightarrow \infty</math> this [[Riemann sum]] becomes the definite integral
<math display="block"> \lim_{N\to\infty}\Delta v = v_\text{eff} \int_{0}^{\phi} \frac{dx}{1-x} = v_\text{eff}\ln \frac{1}{1-\phi} = v_\text{eff} \ln \frac{m_0}{m_f} ,</math> since the final remaining mass of the rocket is <math> m_f = m_0(1-\phi)</math>.