Systems with Varying Mass: A Rocket — Question 9

Question 9

A varying exhaust speed belongs inside the mass integral

A force-free rocket burns from $m_0$ to $m_0/2$. Its relative exhaust speed depends on remaining mass as $u(m)=u_0(m/m_0)$ with fixed backward direction.

1. Find the exact velocity gain.

2. Compare with assuming constant exhaust speed $u_0$ and with using the arithmetic mean of the initial and final exhaust speeds in the logarithmic formula.

Question 9 - Solution

A varying exhaust speed belongs inside the mass integral

(a) Derivation and calculation

The mass-dependent exhaust speed remains inside the integral: $$\Delta v=\int_{m_0/2}^{m_0}\frac{u(m)}m\,dm=\frac{u_0}{m_0}\int_{m_0/2}^{m_0}dm.$$ $$\boxed{\Delta v=u_0/2}.$$

(b) Result and interpretation

The two proposed substitutions give $$\Delta v_{\mathrm{const}}=u_0\ln2=0.6931u_0,\qquad\Delta v_{\mathrm{mean}}=0.75u_0\ln2=0.5199u_0.$$ Both exceed the exact $0.5000u_0$. The correct effective exhaust speed is weighted by $dm/m$: $$u_{\mathrm{eff}}=\frac{\int u(m)\,dm/m}{\ln2}=\frac{u_0}{2\ln2}.$$

Physical check

The exhaust speed falls from $u_0$ to $u_0/2$ during the burn.