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A Derivation of the Rocket Equation

· 2 min read
Josh Kaplan

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.

ΣF⃗=m⋅a⃗\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:

ΣF⃗=d(m⋅v⃗)dt\Sigma \vec{F} = \frac{d(m \cdot \vec{v})}{dt}

Using the product rule, we expand this to

ΣF⃗=m⋅dv⃗dt+v⃗⋅dmdt\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⋅dv⃗dt+v⃗⋅dmdt0 = m \cdot \frac{d\vec{v}}{dt} + \vec{v} \cdot \frac{dm}{dt}

or

dv⃗dt=−v⃗m⋅dmdt\frac{d\vec{v}}{dt} = \frac{-\vec{v}}{m} \cdot \frac{dm}{dt}

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

∫dv⃗=∫−v⃗mdm\int d\vec{v} = \int \frac{-\vec{v}}{m} dm

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

v2⃗−v1⃗=−ve⃗⋅ln⁡m1m2\vec{v_2} - \vec{v_1} = -\vec{v_e} \cdot \ln\frac{m_1}{m_2}

or simply

Δv⃗=−ve⃗⋅ln⁡m1m2\Delta \vec{v} = -\vec{v_e} \cdot \ln\frac{m_{1}}{m_{2}}

Where m1m_1 and m2m_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 −ve⃗-\vec{v_e} with Isp⋅g0I_{sp} \cdot g_0. Where IspI_{sp} is the rocket's specific impulse, a measure of a rocket engine's fuel efficiency, and g0g_0 is Earth's surface gravity (≈9.8ms \approx9.8 \frac{m}{s}). Therefore, our final form of the equation is

Δv⃗=Isp⋅g0⋅ln⁡mimf\Delta \vec{v} = I_{sp} \cdot g_0 \cdot \ln\frac{m_{i}}{m_{f}}