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UY1: Rocket Propulsion

Derive and apply the rocket equation using momentum conservation, with clear sign conventions, thrust, and mass-ratio checks.

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  • University Physics Year 1
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Cluster role + pathway links

This page is the application page for variable-mass momentum methods in the UY1 mechanics pathway.

1) At a glance

  • Prerequisite: momentum conservation (Linear Momentum, Impulse & Collisions), integration + logarithms (Integration Techniques).
  • Outcomes: derive thrust and the ideal rocket equation, then apply them safely.
  • Key result (ideal, no external force):
    Δ v = vf-vᵢ = vₑₓ ln(mᵢ/mf)
  • Common trap: using the wrong mass ratio. Large Δ v needs large mᵢ/mf (equivalently small mf/mᵢ).

2) Setup (model, assumptions, coordinates/signs)

Two snapshots along +x. At time t the rocket has mass m and velocity v. At time t + dt the rocket has mass m + dm and velocity v + dv, and the ejected fuel of mass −dm moves with velocity v − v_ex.
Treat rocket and fuel as one system. Because dm is negative, the ejected fuel has positive mass −dm; all velocities are measured in the same inertial frame.
  • Model: rocket + fuel as one system.
  • Assumptions: deep space, no gravity/drag, non-relativistic, constant vₑₓ.
  • Sign convention: rocket moves in + x; exhaust leaves backward with speed vₑₓ > 0 relative to rocket.
  • Mass convention: rocket mass decreases, so dm < 0.
Why momentum is the right method here

A rocket is a variable-mass system: mass crosses your chosen system boundary. The impulse from the exhaust is easiest to track through momentum conservation over a short interval. Trying to use F = ma naively without accounting for the outflowing mass is a common source of wrong results.

Sign conventions (write this once, then don’t fight your algebra)

Take + x along the rocket’s motion.

  • Rocket velocity: v(t) (positive forward).
  • Exhaust speed relative to the rocket: vₑₓ > 0 backward, so in the rocket frame the exhaust velocity is -vₑₓ x hat.
  • Rocket mass change: dm < 0 because the rocket loses mass.

If you keep these conventions, the log naturally comes out as ln(mᵢ/mf) with mᵢ > mf.

3) Core method / derivation

At time t: rocket mass m, speed v.

At t + dt: rocket mass m + dm, speed v + dv; expelled fuel mass -dm has inertial-frame speed (v-vₑₓ).

Momentum conservation:

mv = (m + dm)(v + dv) + (-dm)(v-vₑₓ)

Expand and drop the second-order term dm dv:

m dv = -vₑₓ dm

Hence:

dv/dt = -(vₑₓ/m)dm/dt

Since dm/dt < 0, acceleration is positive.

Integrate from (mᵢ,vᵢ) to (mf,vf):

∫_vᵢ^vf dv = -vₑₓ∫_mᵢ^mfdm/m ⇒ vf-vᵢ = vₑₓ ln(mᵢ/mf)

Thrust magnitude:

T = |vₑₓdm/dt|

With external forces (one-line extension)

If external forces are not negligible (gravity, drag), the momentum balance over dt becomes: m dv = -vₑₓ dm + Fₑₓₜ dt. This is the same idea: “rocket gains momentum from ejecting mass”, plus whatever external impulse acts during the burn.

Checks:

  • Units: Δ v has units of vₑₓ, i.e. m s⁻¹; ln(mᵢ/mf) is dimensionless.
  • Sign: if dm/dt < 0, then dv/dt > 0 for forward-pointing rocket axis.
  • Limits: if mf → mᵢ, then Δ v → 0; if mf≪ mᵢ, Δ v increases.
  • Domain check: you must have mᵢ > mf so ln(mᵢ/mf) > 0 for a positive Δ v in the chosen axis.

4) Worked example

Worked example 1

Δv from the rocket equation

Problem

A probe has initial mass mᵢ = 2.0 × 10⁴ kg, final mass mf = 8.0 × 10³ kg and exhaust speed vₑₓ = 3.0 × 10³ m s⁻¹. Find its change in speed.

Show full solution

  1. Find the mass ratio

    Method

    Divide initial by final mass.

    Reason

    The rocket equation depends only on the mass ratio, not on the burn time.

    Working

    Answer

    mᵢ/mf = 2.5.

  2. Apply the rocket equation

    Method

    Use Δ v = vₑₓ ln(mᵢ/mf).

    Reason

    Each kilogram of propellant is expelled from a lighter rocket, so the gain grows only logarithmically.

    Working

    Answer

    Δ v = 3000 ln 2.5.

  3. State the answer

    Working

    Answer

    Δ v ≈ 2.75 × 10³ m s⁻¹.

Worked example 2

Mass ratio for a target Δv

Problem

With vₑₓ = 3.0 × 10³ m s⁻¹, what mass ratio is needed for Δ v = 6.0 × 10³ m s⁻¹?

Show full solution

  1. Rearrange the rocket equation

    Method

    Solve for mᵢ/mf.

    Reason

    Exponentiate both sides of Δ v = vₑₓ ln(mᵢ/mf).

    Working

    Answer

    mᵢ/mf = e^(Δ v/vₑₓ) = e².

  2. State the answer

    Working

    Answer

    mᵢ/mf ≈ 7.39.

5) Practice set

Check your understanding 1

A rocket has vₑₓ = 2500 m s⁻¹, mᵢ = 5000 kg and mf = 2500 kg. Find Δ v.

Show hint

Use Δ v = vₑₓ ln(mᵢ/mf).

Show answer

Δ v = vₑₓ ln(mᵢ/mf) = 2500 ln 2 ≈ 1.73 × 10³ m s⁻¹.

Check your understanding 2

Rockets A and B have the same exhaust speed and mass ratios of 3 and 4. Which gains more speed?

Show hint

Compare the logarithms of the mass ratios.

Show answer

B, since ln 4 > ln 3: Δ vB/Δ vA = ln 4/ln 3 ≈ 1.26.

Check your understanding 3

A rocket’s thrust is 1.2 × 10⁵ N with vₑₓ = 3000 m s⁻¹. Find the rate at which it burns propellant.

Show hint

Thrust is T = vₑₓ|dm/dt|.

Show answer

|dm/dt| = T/vₑₓ = 1.2 × 10⁵/3000 = 40 kg s⁻¹.

6) Summary + next steps

  • Rocket motion comes from momentum exchange with expelled mass, not from pushing against the ground.
  • The ideal rocket equation is logarithmic in mass ratio.
  • For larger Δ v: higher vₑₓ and larger mᵢ/mf (smaller mf/mᵢ).

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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