KAHIBARO
Table of Contents
Variable mass and why rockets are special
Most mechanics problems use objects whose mass stays constant. A rocket is different. As it burns fuel, it throws part of its mass backward as exhaust, and its own mass decreases with time. Because of this, the usual form $F = ma$ must be used carefully. The rocket equation describes how the speed of a rocket changes because of expelled fuel.
The key idea is simple. A rocket moves forward by pushing mass backward. This is a direct consequence of conservation of momentum. No air is required for this process, which is why rockets work in space.
The basic rocket model
To derive the rocket equation, imagine a rocket moving in empty space, with no external forces acting on it. At one moment, the rocket has mass $m$ and velocity $v$. A short time later, it has ejected a small amount of fuel.
Let the rocket lose a small mass $\Delta m$. This expelled fuel moves backward relative to the rocket with exhaust speed $v_e$. The rocket mass becomes $m - \Delta m$, and its speed increases by a small amount $\Delta v$.
Because momentum is conserved, the total momentum before and after the fuel is expelled must be equal.
Before ejection,
$$
p_{\text{before}} = mv
$$
After ejection, the rocket has momentum
$$
(m - \Delta m)(v + \Delta v)
$$
The expelled fuel has mass $\Delta m$ and moves with speed
$$
v + \Delta v - v_e
$$
relative to the original frame, because $v_e$ is backward relative to the rocket.
So the fuel momentum is
$$
\Delta m (v + \Delta v - v_e)
$$
Setting total momentum before and after equal,
$$
mv = (m - \Delta m)(v + \Delta v) + \Delta m (v + \Delta v - v_e)
$$
Now expand and ignore the very small product $\Delta m \Delta v$. After simplification,
$$
m \Delta v = v_e \Delta m
$$
Since the rocket mass decreases, it is cleaner to write the rocket mass change as $dm < 0$. Then the speed change is
$$
dv = -v_e \frac{dm}{m}
$$
The ideal rocket equation
Integrate both sides from initial mass $m_0$ to final mass $m_f$, and from initial velocity $v_0$ to final velocity $v$:
$$
\int_{v_0}^{v} dv = -v_e \int_{m_0}^{m_f} \frac{dm}{m}
$$
This gives
$$
v - v_0 = v_e \ln\left(\frac{m_0}{m_f}\right)
$$
The change in speed is usually written as
$$
\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)
$$
This is the ideal rocket equation, also called the Tsiolkovsky rocket equation.
Important formula:
$$
\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)
$$
Here,
$m_0$ is the initial mass, including fuel,
$m_f$ is the final mass, after fuel is burned,
$v_e$ is the exhaust speed relative to the rocket.
Meaning of the formula
The rocket equation tells us that the speed increase depends on two things, the exhaust speed and the mass ratio.
The mass ratio is
$$
\frac{m_0}{m_f}
$$
A larger mass ratio means the rocket started with much more fuel compared with its final mass. A larger exhaust speed means the rocket throws fuel backward more effectively.
The logarithm is very important. It means that gaining more $\Delta v$ becomes harder and harder. Doubling the required $\Delta v$ does not mean doubling the fuel. It requires much more.
Interpreting the masses
The initial mass $m_0$ includes everything at launch for the burn being studied, structure, payload, and unburned fuel.
The final mass $m_f$ is the mass after the fuel for that burn is gone. It still includes structure and payload, and any fuel not yet used for later burns.
This is why rocket design is difficult. Fuel adds mass, but carrying more fuel also requires more fuel to accelerate that extra fuel.
Exhaust speed and specific impulse
The exhaust speed $v_e$ is one way to measure rocket engine performance. A related quantity often used in rocketry is the specific impulse, written $I_{sp}$.
They are related by
$$
v_e = I_{sp} g_0
$$
where $g_0 \approx 9.81 \, \text{m/s}^2$.
So the rocket equation can also be written as
$$
\Delta v = I_{sp} g_0 \ln\left(\frac{m_0}{m_f}\right)
$$
Equivalent form:
$$
\Delta v = I_{sp} g_0 \ln\left(\frac{m_0}{m_f}\right)
$$
A simple example
Suppose a rocket has exhaust speed
$$
v_e = 3000 \, \text{m/s}
$$
and mass ratio
$$
\frac{m_0}{m_f} = 4
$$
Then
$$
\Delta v = 3000 \ln(4)
$$
Using
$$
\ln(4) \approx 1.386
$$
we get
$$
\Delta v \approx 3000 \times 1.386 = 4158 \, \text{m/s}
$$
So the rocket can gain about
$$
4.16 \times 10^3 \, \text{m/s}
$$
Why multistage rockets are used
The rocket equation explains why staging is useful. After one stage uses its fuel, the empty tank and engines of that stage are dropped. Then the next stage does not need to accelerate that dead mass.
This makes the effective mass ratio for later stages much better. Even though each stage obeys the same rocket equation, combining stages allows a much larger total $\Delta v$.
If two stages provide speed changes $\Delta v_1$ and $\Delta v_2$, then the total is
$$
\Delta v_{\text{total}} = \Delta v_1 + \Delta v_2
$$
Ideal equation versus real rockets
The equation derived above is an ideal result. It assumes no external forces during the burn. Real rockets often experience gravity and air resistance, especially near Earth.
So in practice, the actual speed gain is less than the ideal value predicted by
$$
\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)
$$
Still, the equation is extremely important because it gives the fundamental limit set by momentum conservation.
The ideal rocket equation does not include gravity loss or drag loss. It gives the maximum speed change for a given exhaust speed and mass ratio in the ideal case.
Summary table
| Quantity | Meaning | Formula or note |
|---|---|---|
| $\Delta v$ | Change in rocket speed | $\Delta v = v_e \ln(m_0/m_f)$ |
| $v_e$ | Exhaust speed relative to rocket | Measures engine effectiveness |
| $m_0$ | Initial mass | Includes fuel before burn |
| $m_f$ | Final mass | Mass after burn |
| $I_{sp}$ | Specific impulse | $v_e = I_{sp} g_0$ |
| Mass ratio | Initial mass divided by final mass | $m_0/m_f$ |
Visual picture of rocket propulsion
Final idea
The rocket equation is one of the most important results in mechanics for variable mass systems. It shows that rocket motion comes from conservation of momentum, and it explains why high exhaust speed, large fuel fraction, and staging are so important in spaceflight.
Core result to remember:
$$
\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)
$$
A rocket gains speed by ejecting mass backward.