A Derivation of the Rocket Equation

Here, a derivation of the Tsiolkovsky rocket equation from Newton's Second Law is shown. As a disclaimer, this is not original work and is not intended to be. It is shown for reference purposes. In fact multiple derivations of the equation can be found here.

We begin with the Newton's Second Law.

$\Sigma \vec{F} = m \cdot \vec{a}$

However, this is a simplification of the Second Law where the mass is equal to zero. Newton's Second Law is in fact conservation of momentum and, as a reminder, momentum is defined by mass times velocity. Thus, Newton's Second Law in its more general form is expressed as:

$$\Sigma \vec{F} = \frac{d(m \cdot \vec{v})}{dt}$$

Using the product rule, we expand this to

$$\Sigma \vec{F} = m \cdot \frac{d\vec{v}}{dt} + \vec{v} \cdot \frac{dm}{dt}$$

In the case of a rocket, the we are analyzing the change in velocity of the rocket by the engine, which is part of the rocket. Therefore, thrust is not an external force and no external forces are acting on the rocket and the sum of forces is zero (consider a rocket in orbit rather than one flying through the atmosphere).

$$0 = m \cdot \frac{d\vec{v}}{dt} + \vec{v} \cdot \frac{dm}{dt}$$

or

$$\frac{d\vec{v}}{dt} = \frac{-\vec{v}}{m} \cdot \frac{dm}{dt}$$

The $\vec{v}$ term on the right hand side of the equation is the exhaust velocity of the rocket engine (denoted $\vec{v_e}$ from here on), which is assumed to be constant. We can now integrate both side of the equation.

$$\int d\vec{v} = \int \frac{-\vec{v}}{m} dm$$

This gives us the following equation for the velocity change of the rocket.

$$\vec{v_2} - \vec{v_1} = -\vec{v_e} \cdot \ln\frac{m_1}{m_2}$$

or simply

$$\Delta \vec{v} = -\vec{v_e} \cdot \ln\frac{m_{1}}{m_{2}}$$

Where $m_1$ and $m_2$ are the initial mass and final mass of the rocket, respectively. Note that the negative sign indicates the the velocity change of the rocket will be in the opposite direction of the rocket's exhaust (which makes sense).

Finally, we will rewrite this in its more common form, which replaces $-\vec{v_e}$ with $I_{sp} \cdot g_0$. Where $I_{sp}$ is the rocket's specific impulse, a measure of a rocket engine's fuel efficiency, and $g_0$ is Earth's surface gravity ($\approx9.8 \frac{m}{s}$). Therefore, our final form of the equation is

$$\Delta \vec{v} = I_{sp} \cdot g_0 \cdot \ln\frac{m_{i}}{m_{f}}$$