Staged rocket

Problem

Staged rocket. A two-stage rocket has total mass $1\,000\;\text{kg}$. It expels stage 1 ($600\;\text{kg}$) at relative velocity $1\;\text{km/s}$. Stage 2 (mass $400\;\text{kg}$) then expels more fuel ($200\;\text{kg}$), again at $1\;\text{km/s}$ relative. Calculate the final velocity of the residual payload ($200\;\text{kg}$) using the Tsiolkovsky formula $\Delta v = v_g \ln(m_0/m_f)$.

Solution

The Tsiolkovsky formula gives the velocity increment of each stage: $\Delta v = v_g \ln(m_0/m_f)$, with $v_g = 1\;\text{km/s} = 1\,000\;\text{m/s}$.

Stage 1: starts at $m_0 = 1\,000\;\text{kg}$; after expelling $600\;\text{kg}$, $m_f = 400\;\text{kg}$ remains. $\Delta v_1 = 1\,000 \cdot \ln\!\left(\frac{1\,000}{400}\right) = 1\,000 \cdot \ln(2{,}5) \approx 916\;\mathrm{m/s}$

Stage 2: starts at $m_0 = 400\;\text{kg}$; expels $200\;\text{kg}$ of fuel, leaving $m_f = 200\;\text{kg}$ (the payload). $\Delta v_2 = 1\,000 \cdot \ln\!\left(\frac{400}{200}\right) = 1\,000 \cdot \ln(2) \approx 693\;\mathrm{m/s}$

The increments add up because each stage starts from the velocity already gained: $v_\text{final} = \Delta v_1 + \Delta v_2 \approx 916 + 693 = 1\,609\;\mathrm{m/s}$ Result: $\ev{v_\text{final} \approx 1\,609\;\mathrm{m/s}}$

Links

Topics: Momentum and collisions Concepts: Momentum · Conservation of momentum Skills: Conservation of momentum Objects: Rocket