Rockets in space, like all other objects, have to accelerate to change velocity. But space is a vacuum, so there is nothing to push against to create force. Instead, rockets accelerate by using the *conservation of momentum*. The momentum of an object is equal to the object’s mass multiplied by the object’s velocity: \vec{p} = m \vec{v}. In a closed system, the total momentum remains constant: \vec{p}_{0} = \vec{p}_{t}.

A rocket carries propellant that it expels at high velocities to accelerate. Imagine a rocket moving in space; at first the rocket is not expelling propellant and so its momentum does not change. Then it expels a part of its propellant. That propellant’s momentum is equal to its mass multiplied by its velocity. The rocket and propellant are part of a closed system, so the momentum of the rocket has to change such that the total momentum (that of the rocket plus that of the propellant) is equal to the momentum of the rocket before it expelled the propellant. As a result, the rocket gains velocity in direction opposite to that of the propellant.

Let’s find out how much velocity the rocket gains!

We define the mass of the propellant the rocket expels as \Delta m and the rocket’s mass after expelling the propellant as M. Before the propellant is expelled, the rocket has a mass of M + \Delta m and a velocity of v. The propellant is expelled in opposite direction with an exhaust velocity of v_e. For a stationary observer the propellant will have a velocity of v-v_e.

The initial momentum of the system consists only of the momentum of the rocket (before the propellant has been expelled).

p_i = (M + \Delta m) v

After the propellant is expelled, the momentum of the system, which we’ll call the final momentum, is equal to the momentum of the rocket and the momentum of the propellant. The rocket will have gained some velocity \Delta v.

p_f = \underbrace{M (v + \Delta v)}_{\text{Rocket}} + \underbrace{\Delta m (v-v_e)}_{\text{Propellant}}

Because of the conservation of momentum it holds that p_i = p_f.

\begin{aligned}

(M + \Delta m) v &= M (v + \Delta v) + \Delta m (v-v_e) \\

M v + \Delta m v &= M v + M \Delta v + \Delta m v – \Delta m v_e \\

0 &= M \Delta v – \Delta m v_e \\

\Delta m v_e &= M \Delta v \\

M \Delta v &= v_e \Delta m

\end{aligned}

We now look at the instantaneous rate of change of the mass and the velocity (i.e., we use an infinitesimally small time-frame).

M dv = v_e dm

The instantaneous change of propellant mass is equal to the negative change of mass of the rocket, because the propellant is carried by the rocket.

\begin{aligned}

M dv &= v_e dm \\

M dv &= v_e \cdot -dM \\

M dv &= -v_e dM \\

dv &= -v_e \frac{1}{M} dM

\end{aligned}

Now we take the definite integral at both sides, from the initial to the final situation.

\begin{aligned}

\int_{v_i}^{v_f} dv &= \int_{M_i}^{M_f} -v_e \frac{1}{M} dM \\

\int_{v_i}^{v_f} 1 dv &= -v_e \int_{M_i}^{M_f} \frac{1}{M} dM \\

v \Bigr\rvert_{v_i}^{v_f} &= -v_e \left( \ln{M} \Bigr\rvert_{M_i}^{M_f} \right) \\

v_f – v_i &= -v_e \left( \ln(M_f) – \ln(M_i) \right) \\

\Delta v &= v_e \left( \ln(M_i) – \ln(M_f) \right) \\

\Delta v &= v_e \ln\left( \frac{M_i}{M_f} \right)

\end{aligned}

Note that the definite integral of \frac{1}{x} from a to b is \ln(x)|_a^b and that \ln(m)-\ln(n) = \ln(m/n).

“Delta-v” refers to the amount of change in velocity a rocket can perform. If the rocket’s initial total mass is m_0 and the mass of the propellant is m_p, we can calculate the maximum delta-v.

\Delta v_{\text{max}} = v_e \ln\left( \frac{m_0}{m_0-m_p} \right)

We have a few options to increase the maximum delta-v. The rocket can be filled with more propellant to increase the ratio \frac{m_0}{m_0-m_p}. The rocket can be made lighter to increase this same ratio. An interesting option is to use staging, where parts of the rocket that are not needed anymore are thrown out. The last option is to increase the exhaust velocity v_e. In practice, all three of these options are used. Initially, by far the largest part of the mass of the rocket is its propellant. Most rockets use staging: some rockets have successfully been launched using a total of five stages. Finally, propellant is often heated or otherwise handled to increase the exhaust speed. Note that in the case of staging the equation must be applied to each stage separately, otherwise it is assumed that the stages are expelled at the same velocity that the propellant is expelled with!

The equation we found is commonly known as the *Tsiolkovsky rocket equation* or the *ideal rocket equation*. It is convenient to use, as this single equation expresses the essentials of rocket flight.

## 3 thoughts on “The Rocket Equation”