Simple Rocket Motion

Rocket motion. No, not that rocket that’s attached to your body. An actual space rocket that probably was built by your sketchy neighbor who parks their cars outside the garage. The grand idea here is that we’re dealing with a moving object, but with a mass that changes with time. Since the rocket propels itself using fuel, it loses mass from the fuel to gain speed forward. Therefore it’s not just an average kinematics problem, and we’ll have to apply a different method to solve for the motion of the rocket.

Firstly, we need to find an equation relating the rocket’s velocity at two different times, but includes the rocket’s mass as well. We’re in luck, because the equation for momentum, p = mv, has exactly both of those.

Rocket Momentum

Let’s have the rocket start with mass m and velocity v, and thus let its initial momentum, p_i, be $$ p_i= mv $$

At some point after the initial time, the rocket will have lost some mass we’ll call dm, and gained some velocity we’ll call dv. The rocket’s momentum at this moment in time will be $$ \color{blue}{p_{\text{rocket}}=(m-dm)\cdot(v+dv)} $$

At the same moment in time, the fuel propelled from the back of the rocket will have mass dm and velocity v - v_ex, where v_ex is the “exhaust velocity” of the rocket, or the speed of the stuff coming out the back.

[The exhaust velocity is calculated relative to us, i.e. the fuel is already moving inside the rocket at velocity v of the rocket, and thus when it’s propelled, it comes out in the opposite direction of v with magnitude v_ex. Then the total fuel speed will be v-v_ex.]

The momentum of this fuel coming from the rocket is then $$ \color{green}{p_{\text{fuel}}=dm\cdot(v-v_{ex})} $$

The total momentum of the rocket AND the fuel at this moment in time is then $$ p_{\text{total}}=\color{blue}{p_{\text{rocket}}}+\color{green}{p_{\text{fuel}}}=\color{blue}{(m-dm)(v+dv)}+\color{green}{dm(v-v_{ex})} $$

Since p_total is conserved, we can compare this total momentum to the rocket’s initial momentum: $$ mv=\color{blue}{(m-dm)(v+dv)}+\color{green}{dm(v-v_{ex})} $$

And now we can finally start expanding and cancelling: $$ \begin{align} \cancel{mv}&=\cancel{\color{blue}{mv}}\color{blue}{+m~dv\color{black}{\cancel{\color{blue}{-v~dm}}} -dv~dm} + \color{green}{\color{black}{\cancel{\color{green}{v~dm}}} - v_{ex}~dm}\newline 0&=\color{blue}{m~dv -dv~dm} -\color{green}{v_{ex}~dm} \end{align} $$

One other thing we can say about this equation is that dv dm = 0, because this term is “second order”, as in, it has two “d” terms. We only want to deal with first order when talking about momentum and infinitesimal terms, so now we have $$ \tag{☆}\color{green}{v_{ex}~dm}=\color{blue}{m~dv} $$

This is basically the “final form” of the rocket equation, and every solution we can draw about this system will come from the equation above.

Solutions

If we differentiate both sides with respect to time t, we get: $$ v_{ex}\frac{dm}{dt}=m\frac{dv}{dt}=ma $$

since the derivative of velocity with respect to time is just acceleration. We also know that F = ma, so then we can derive the “thrust force”, F_thrust, in terms of the rate of change of mass and the exhaust velocity: $$ \tag{☆}v_{ex}\frac{dm}{dt}=F_{\text{thrust}} $$

We can also finally calculate the velocity of the rocket using separation of variables: $$ v_{ex}~dm=m~dv\implies dv=\frac{v_{ex}}{m}~dm $$

Integrating, we get:

$$ \begin{align} \int_{v_0}^{v(t)}dv&=v_{ex}\int_{m_0}^{m(t)}\frac{1}{m}~dm\newline v(t)-v_0&=v_{ex}\ln|m|_{m_0}^{m(t)} \end{align} $$

Let’s say for the sake of this example that v₀ ≡ 0, meaning the rocket starts at rest with no velocity. We can also get rid of the absolute value on the natural log ln term since mass is always positive. Lastly, we should remember that $dm$ indicates a decrease in mass, thus we will add a negative sign. Putting all this together leaves $$ \tag{☆}v(t)=v_{ex}\ln(m_0-m(t))=\boxed{v_{ex}\ln\left(\frac{m_0}{m(t)}\right)} $$ and thus we have solved for v(t) in terms of the exhaust speed, the initial mass of the rocket, and the mass of the rocket at some point in time t.