Variable Mass & the Rocket Equation
- Analyze systems whose mass changes using conservation of momentum
- Relate the thrust on a rocket to its exhaust speed and burn rate
- Apply the ideal rocket equation Δv = v_ex ln(m_i/m_f)
A rocket pushes off its own exhaust
A rocket in empty space has nothing external to push on, so it throws mass backward and rides the recoil. Momentum conservation applied to the rocket plus a small ejected bit dm gives the equation of motion m dv = −v_ex dm, where v_ex is the exhaust speed relative to the rocket. The backward-flung exhaust produces a forward thrust on the rocket.
Integrating gives the rocket equation
Separate variables in m dv = −v_ex dm and integrate from the initial mass m_i (speed v_i) to the final mass m_f (speed v_f). The mass integral produces a natural logarithm, giving the ideal rocket equation Δv = v_ex ln(m_i/m_f). Because the velocity gain grows only with the logarithm of the mass ratio, doubling your speed change requires squaring the mass ratio — the tyranny of the rocket equation.
A rocket in deep space has an exhaust speed of 2000 m/s, an initial mass of 3000 kg, and a final mass of 1000 kg after the burn. Using ln 3 ≈ 1.10, find the change in the rocket's speed.
- 1.Use the ideal rocket equation: Δv = v_ex ln(m_i/m_f).
- 2.Compute the mass ratio: m_i/m_f = 3000/1000 = 3.
- 3.Take the logarithm: ln 3 ≈ 1.10.
- 4.Multiply: Δv = 2000 × 1.10 = 2200 m/s.
The ideal rocket equation is Δv = v_ex ln(m_i/m_f). To double the speed change gained (for the same exhaust speed), you must:
A rocket has exhaust speed 3000 m/s, initial mass 2000 kg, and final mass 1000 kg. Its change in speed is approximately (ln 2 ≈ 0.69):
The rocket equation comes from momentum conservation with changing mass, not from F = ma with constant mass. Watch the logarithm: the payoff for carrying more fuel diminishes, since Δv grows only as ln(m_i/m_f).
Answer the 2 checkpoints as you read.
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