33.1: The Rocket Equation

Let's now derive the rocket equation. Given a rocket of mass $m$, we will wish to find an equation that tells us how much fuel (propellant) is required to change the rocket's speed by an amount $\Delta v$. The complication here is that the rocket loses mass as it expels propellant, so we need to allow for that.

Suppose that at an initial time $t=0$, a rocket has velocity $v$ and total mass $m$, including propellant mass. The total momentum of the rocket and propellant at time $t=0$ is therefore $m v$.

Now let's look at the situation an instant later, at time $t=d t$. Let $d m$ be the (negative) change in mass of the rocket due to the expulsion of propellant, and let $d v$ be the corresponding (positive) change in the velocity of the rocket. Then at time $t=d t$, a mass of propellant $-d m$ is expelled at velocity $v-v_{p}$. (The rocket is moving at velocity $v$ with respect to the Earth, the propellant is moving at speed $-v_{p}$ relative to the rocket, and so the velocity of the propellant relative to the Earth is $v-v_{p}$.) This expulsion of propellant will cause the rocket to then have mass $m+d m$ and velocity $v+d v$. The total momentum of the system at $t=d t$ is then the sum of the rocket and propellant momenta, $(m+d m)(v+d v)+\left(v-v_{p}\right)(-d m)$. By conservation of momentum, the momentum of the system at time $t=0$ must equal the momentum at time $t=d t$ :

$$m v =(m+d m)(v+d v)+\left(v-v_{p}\right)(-d m)$$
$$=m v+v d m+m d v+d m d v-v d m+v_{p} d m$$

Now the two $m v$ terms cancel, the two $v d m$ terms cancel, and the term $d m d v$ is a second-order differential, which can also be cancelled. We're then left with

$$0 =m d v+v_{p} d m$$
$$m d v =-v_{p} d m$$
$$d v =-v_{p} \frac{d m}{m}$$

Now let the rocket burn all its propellant. The rocket's velocity will change by a total amount $\Delta v$ and its mass will change from $m$ to its empty mass $m_{e}$. Integrating Eq. $\PageIndex{5}$ over the entire propellant burn, we find

$$\int_{v}^{v+\Delta v} d v=-v_{p} \int_{m}^{m_{e}} \frac{d m}{m}$$

Or, evaluating the integrals,

$$\Delta v=-v_{p} \ln \frac{m_{e}}{m}$$

or

$$\Delta v=v_{p} \ln \frac{m}{m_{e}}$$

Eq. $\PageIndex{8}$ is called the rocket equation. It relates the fueled and empty masses of the rocket and the velocity of the propellant to the total change in velocity of the rocket.