Variable-Mass Systems and the Rocket Equation
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You must be comfortable with Newton's Second Law for constant mass and the concept of linear momentum conservation for isolated systems, as covered in the previous lecture on Systems of Particles.
Why this matters
Standard Newtonian mechanics assumes a fixed mass, but rockets, falling rain, and leaking tanks require a different approach.
When a system changes its mass by ejecting or accreting matter, the simple form \(F = ma\) becomes ambiguous because the mass \(m\) is no longer constant. This lecture resolves that ambiguity by applying momentum conservation to a carefully defined system boundary. We will derive the fundamental relationship governing self-propelled bodies, known as the rocket equation, and clarify the common pitfalls associated with variable-mass dynamics.
The Core Idea: Defining the System
The primary source of confusion in variable-mass problems is the definition of the "system." In standard mechanics, we track a fixed set of particles. In a rocket, however, the fuel is part of the system initially but becomes external exhaust later. To apply Newton's laws correctly, we must consider the total momentum of the rocket plus the fuel it is about to eject. As the rocket pushes fuel backward, the fuel pushes the rocket forward. This interaction is an internal force within the combined system of rocket and fuel. Therefore, if no external forces act, the total momentum of this combined system remains constant. The rocket gains forward momentum exactly equal to the backward momentum gained by the ejected fuel. This exchange of momentum, driven by the relative speed of the exhaust, is the fundamental mechanism of propulsion.
Momentum Conservation in Variable-Mass Systems
Consider a rocket of instantaneous mass \(m\) moving with velocity \(v\). In a small time interval \(dt\), it ejects a small mass \(dm_{\text{ex}}\) (where \(dm_{\text{ex}} > 0\)) with an exhaust velocity \(v_{\text{ex}}\) relative to the rocket. The velocity of the exhaust in the inertial frame is \(v - v_{\text{ex}}\). The mass of the rocket decreases by \(dm_{\text{ex}}\), so its new mass is \(m - dm_{\text{ex}}\), and its new velocity is \(v + dv\).
According to the principle of momentum conservation for isolated systems, the change in total momentum is zero. The initial momentum is \(mv\). The final momentum is the sum of the rocket's momentum and the exhaust's momentum: \((m - dm_{\text{ex}})(v + dv) + dm_{\text{ex}}(v - v_{\text{ex}})\). Equating initial and final momentum and neglecting second-order infinitesimals (terms like \(dm_{\text{ex}} \, dv\)), we obtain the differential relationship governing the rocket's acceleration. This derivation relies on the fact that internal forces cancel out, leaving only the net external force to change the total momentum, as established in the conservation laws.
Since the rocket mass decreases as exhaust mass increases, \(dm_{\text{ex}} = -dm\), where \(dm\) is the change in the rocket's mass (a negative quantity). Substituting this yields the standard differential form of the rocket equation.
The Tsiolkovsky Rocket Equation
To find the total velocity change \(\Delta v\) over a finite burn, we integrate the differential equation. We assume the exhaust velocity \(v_{\text{ex}}\) is constant relative to the rocket. The integration limits for mass are from the initial mass \(m_0\) to the final mass \(m_f\). The integration limits for velocity are from \(v_0\) to \(v_f\).
Evaluating the integrals gives the logarithmic relationship between the velocity gain and the mass ratio. This result is profound because it shows that the velocity gain depends only on the exhaust velocity and the ratio of initial to final mass, not on the absolute mass or the burn time. This is the Tsiolkovsky Rocket Equation.
This equation highlights the exponential nature of mass requirements. To double the \(\Delta v\), you do not double the mass ratio; you square it. This is why multi-stage rockets are necessary for reaching orbital velocities.
External Forces and Generalization
In many practical scenarios, external forces such as gravity or air resistance are present. The general equation of motion for a variable-mass system is derived by applying Newton's Second Law to the system of particles. The net external force equals the rate of change of the total momentum of the system. When we isolate the rocket's momentum change, we find that the thrust force is given by \(F_{\text{thrust}} = v_{\text{ex}} \dot{m}_{\text{ex}}\), where \(\dot{m}_{\text{ex}}\) is the mass flow rate. The full equation of motion becomes \(m \frac{dv}{dt} = F_{\text{thrust}} + F_{\text{ext}}\). This form allows us to calculate the rocket's trajectory in a gravitational field by simply adding the gravitational force term to the thrust term.
The derivations
The Tsiolkovsky Rocket Equation
For a body that propels itself by ejecting mass at a constant speed \(v_{\mathrm{ex}}\) relative to itself, in the absence of external forces, momentum conservation yields the velocity gained as a logarithmic function of the ratio of initial to final mass: \(\Delta v = v_{\mathrm{ex}} \ln\!\left(\dfrac{m_0}{m_f}\right)\).
Momentum Conservation from Newton's Third Law
For an isolated system of \(N\) point particles interacting only through pairwise internal forces that obey Newton's third law, the total linear momentum \(\vec{P}=\sum_{i} m_i\vec{v}_i\) is constant in time, \(\frac{d\vec{P}}{dt}=\vec{0}\). The internal forces produce no net change in the system's momentum because they cancel in equal-and-opposite pairs, so \(\frac{d\vec{P}}{dt}\) equals the net external force alone.
A common error is to apply \(F = ma\) directly to the rocket mass \(m(t)\) without accounting for the momentum carried away by the exhaust. If one writes \(F_{\text{ext}} = \frac{d}{dt}(mv)\), one incorrectly assumes the exhaust leaves with zero velocity relative to the inertial frame. The correct application requires tracking the momentum of the ejected mass separately. The thrust arises from the relative velocity of the exhaust, not just the mass flow rate. C
Worked example
Problem. A rocket with an initial mass of \(2000\ \mathrm{kg}\) (including fuel) ejects exhaust at a constant speed of \(2500\ \mathrm{m/s}\) relative to the rocket. If the final mass of the rocket (after fuel is exhausted) is \(500\ \mathrm{kg}\), calculate the maximum velocity change \(\Delta v\) achievable in the absence of external forces.
- Identify the given values: \(m_0 = 2000\ \mathrm{kg}\), \(m_f = 500\ \mathrm{kg}\), \(v_{\text{ex}} = 2500\ \mathrm{m/s}\).
- Recall the Tsiolkovsky Rocket Equation: \(\Delta v = v_{\text{ex}} \ln\!\left(\dfrac{m_0}{m_f}\right)\).
- Calculate the mass ratio: \(\dfrac{m_0}{m_f} = \dfrac{2000}{500} = 4\).
- Compute the natural logarithm of the mass ratio: \(\ln(4) \approx 1.3863\).
- Multiply by the exhaust velocity: \(\Delta v = 2500 \times 1.3863\).
Units check. The term \(\ln(m_0/m_f)\) is dimensionless because the ratio of two masses has no units. The exhaust velocity \(v_{\text{ex}}\) has units of \(\mathrm{m/s}\). Therefore, the product has units of \(\mathrm{m/s}\), which is correct for velocity.
Three quick tests
- Does the rocket equation depend on the time over which the fuel is burned?
Answer
No, \(\Delta v\) depends only on \(v_{\text{ex}}\) and the mass ratio, assuming constant exhaust velocity. - If the exhaust velocity is doubled, how does the required mass ratio change to achieve the same \(\Delta v\)?
Answer
The required mass ratio decreases. Since \(\Delta v \propto v_{\text{ex}}\), a higher \(v_{\text{ex}}\) requires a smaller \(\ln(m_0/m_f)\), thus a smaller \(m_0/m_f\). - Why is it incorrect to say the rocket accelerates because "it pushes against the air"?
Answer
Rockets work in a vacuum. The acceleration is due to the conservation of momentum between the rocket and the ejected exhaust mass, not interaction with external air.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-034 · D-025 · Threads: force, energy, matter