# Solving the rocket differential equation

I'm trying to derive the rocket equation.

I'm pretty sure that the differential equation for the rocket equation is

$$v(t)\delta t =\frac{m(t)\delta t }{m(t)} V_e$$

where

• $$v(t)\delta t$$ is the rate of change in velocity of the rocket over time.
• $$m(t)\delta t$$ is the rate of change of the mass of the rocket over time.
• $$m(t)$$ is the mass of the rocket at a time $$t$$.
• $$V_e$$ is the exhaust velocity (a constant)

Now I want to solve for $$v(t)$$, so I integrate on both sides.

$$\int_0^t v(t)\delta t =\int_0^t \frac{m(t)\delta t }{m(t)} V_e$$

I belive I should get $$v(t) = ln\left(\frac{m(0)}{m(t)}\right)V_e$$

But whenever I try to actually solve the integral I come up with stuff that does not look remotely like that.

I tried a ton of videos and posts on the internet but most of the time there is some magic involved or some questionable not-quite-rigorous math going on.

So my question is:

Is this differential equation correctly formulated to get the rocket equation?

How can I go around solving this differential equation?

First of all, I believe part of the reason you're getting confused is that you're using confusing notation. Instead of $$v(t) \delta t$$, the usual way of writing the rate of change of the rocket's velocity is $$dv/dt$$; and for the rate of change of the mass, one usually writes $$dm/dt$$. That way we avoid using $$m(t)$$ in two different ways.

OK, let's now solve the problem. Starting from your correct differential equation (where I've made the right side negative to make it easier to remember that the exhaust velocity is in the opposite direction of the rocket velocity):

\begin{align} \frac{dv}{dt} &= -\frac{dm/dt}{m} V_e \\ \Rightarrow dv &= -V_e \frac{dm}{m} \\ \Rightarrow \int_0^t dv &= -V_e \int_0^t \frac{dm}{m} \end{align}

and we can now integrate both sides to obtain

\begin{align} v(t)-v(0) &= -V_e \left( \ln m(t) - \ln m(0) \right) \\ \Rightarrow v(t) &= v_0 + V_e \ln \frac{m(0)}{m(t)} \end{align}

where $$v_0$$ is the rocket's initial velocity (usually zero).

• Sorry if the notation was confusing. I grew into it. How is $\int_0^t \frac{dm/dt}{m(t)} dt= ln(m(t)-ln(m(0))$? I know you are using the fundamental theorem of calculus, but that's pretty much it. It feels like you are doing $\int_0^t \frac{1}{m(t)}$ for some reason – Joaquin Brandan Dec 18 '18 at 3:51
• @JoaquinBrandan yes, I'm using the fundamental theorem of calculus. The indefinite integral of $1/m$ is $\ln m$. Therefore, $\int_0^t \frac{1}{m} dm$ is $\ln(m(t)) - \ln(m(0))$. – Thorondor Dec 18 '18 at 3:59
• The step where I replaced $\frac{dm}{dt}dt$ with $dm$ might look a little fishy, but it can be justified using the reverse chain rule. – Thorondor Dec 18 '18 at 4:02
• but I'm not integrating by $\frac{dm}{dt}$, $\frac{dm}{dt}$ is the derivative of $m(t)$. my integral looks like this $\int_0^t \frac{1}{m}\frac{dm}{dt}dt$. – Joaquin Brandan Dec 18 '18 at 4:03
• Thanks. If this is the way to go I will crunch the numbers until I get it. – Joaquin Brandan Dec 18 '18 at 4:10