Tsiolkovsky Rocket Equation
Formulas1
The equation depicting the velocity of a rocket ejecting fuel in one-dimensional space without external forces is known as the Tsiolkovsky rocket equation.
$$ v = v_{0} + V \ln \dfrac{m_{0}}{m} $$
Here, $v$ represents the final velocity of the rocket, $v_{0}$ the initial velocity of the rocket, $V$ the relative ejection speed of the fuel to the rocket, $m$ the final mass of the rocket, and $m_{0}$ the initial mass of the rocket.
Since it’s an equation for one-dimensional space without external forces, it is also referred to as the ideal rocket equation.
Derivation
Let’s denote $m$ as the mass of the rocket, $\mathbf{v}$ as the velocity of the rocket, $\mathbf{V}$ as the relative ejection speed of fuel to the rocket, and $\mathbf{F}_{\text{ext}}$ as the external force. Then, the equation of motion for a variable mass system can be written as follows.
$$ \mathbf{F}_{\text{ext}} = m \dot{\mathbf{v}} - \mathbf{V}\dot{m} $$
Here, $\dot{}$ [dot] signifies the derivative with respect to time.
$$ \dot{\mathbf{v}} = \dfrac{d \mathbf{v}}{d t},\quad \dot{m} = \dfrac{d m}{d t} $$
Moreover, let’s assume that external forces such as the gravity of the Earth or other stars, and air resistance, are ignored, and the external force is $0$.
$$ m \dot{\mathbf{v}} = \mathbf{V}\dot{m} $$
In this equation, the right side, $\mathbf{V}\dot{m}$, is also referred to as the thrust of the rocket. For simplicity, let’s consider a one-dimensional space where $\mathbf{v} = v$, and since the fuel is ejected in the opposite direction of the rocket, $\mathbf{V} = -V$.
$$ \begin{align*} && m \dot{v} &= -V\dot{m} \\ \implies && m dv &= -V dm \end{align*} $$
Assuming the ejection speed $-V$ is constant, to find the velocity of the rocket, we integrate using the method of separation of variables,
$$ \begin{align*} && m dv &= -V dm \\ \implies && dv &= -V \dfrac{1}{m} d m \\ \implies && \int_{v_{0}}^{v} dv &= -V \int_{m_{0}}^{m} \dfrac{1}{m} d m \\ \implies && v - v_{0} &= V \ln \dfrac{m_{0}}{m} \\ \end{align*} $$
$$ \implies v = v_{0} + V \ln \dfrac{m_{0}}{m} $$
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See Also
Grant R. Fowles and George L. Cassiday, Analytical Mechanics (7th Edition, 2005), p312-314 ↩︎