Rocket science

The momentum balance for the force-free rocket reads $\begin{aligned} \frac{dp}{dt} &= \left(\frac{dp}{dt}\right)_\text{rocket} + \left(\frac{dp}{dt}\right)_\text{gas} \\ &= (\dot m v + m \dot v) - \dot m (v - u_0) = \dot m u_0 + m \dot v \stackrel{!}{=} 0 \\ \Rightarrow \quad \dot v &= - u_0 \frac{\dot m}{m} = u_0 \frac{|\dot m|}{m_0 - |\dot m|(t - t_0)} \end{aligned}$ with the burning time $t_0 = m_\text{fuel}/|\dot m|$ of the rocket. Integration from $t = 0$ to $t_0$ results the final velocity $\begin{aligned} v_0 &= u_0 \int_0^{t_0} \frac{ dt}{m_0/|\dot m| + t_0 - t} = u_0 \ln \frac{m_0/|\dot m| + t_0}{m_0/|\dot m|} \\ &= u_0 \ln \frac{m_0 + m_\text{fuel}}{m_0} \end{aligned}$ The work integral over the gravitational force results the necessary energy to leave the earth: $\begin{aligned} & & W &= - \int_{R}^{\infty} F dr = \int_R^\infty G \frac{M m_0}{r^2} dr = \int_R^\infty m_0 g \frac{R^2}{r^2} dr = m_0 g R\\ & & &\stackrel{!}{=} \frac{1}{2} m_0 v_0^2 \\ \Rightarrow & & v_0 &= \sqrt{2 g R} = 11.167\,\frac{\text{km}}{s} \\ \Rightarrow & & m_\text{fuel} &= m_0 \left(\exp\left(\frac{v_0}{u_0}\right) - 1\right) \approx 40.37 \cdot m_0 \end{aligned}$