A rocket of initial mass M_0 moves in a straight line under…
A rocket of initial mass \(M_0\) moves in a straight line under the action of a force produced by the continuous ejection of a jet of gases expelled by the rocket. The ejection velocity \(u_0\) of the gases (relative to the rocket) is constant and oriented opposite to the initial velocity \(v_0\) of the rocket. Find the law of motion of the rocket, neglecting gravity and air resistance (Tsiolkovsky's problem).
Explanation of Method
Conservation of momentum for the rocket+ejected-gas system over an infinitesimal time \(dt\), dropping the second-order term \(dM\,dv\), yields the classical relation \(M\,dv=-u_0\,dM\), which integrates to the Tsiolkovsky rocket equation \(v=v_0+u_0\ln(M_0/M)\). To obtain an explicit law of motion \(x(t)\) one needs a mass-loss law \(M(t)\); the standard, simplest physically reasonable choice — and the one used here — is a constant mass-ejection rate \(dM/dt=-\mu\), giving \(M(t)=M_0-\mu t\). Substituting into the velocity relation and integrating \(dx/dt=v(t)\) (verified term-by-term with Sympy, including the logarithmic integral) gives the explicit position law.
1. Momentum conservation \(\Rightarrow\) the Tsiolkovsky relation
At time \(t\) the system (rocket + not-yet-ejected fuel) has momentum \(Mv\). At \(t+dt\) the rocket has momentum \((M+dM)(v+dv)\) and the ejected gas (mass \(-dM>0\), lab-frame velocity \(v-u_0\)) has momentum \((-dM)(v-u_0)\). With no external forces, total momentum is conserved: \[ Mv=(M+dM)(v+dv)+(-dM)(v-u_0). \] Expanding and dropping the second-order term \(dM\,dv\) gives \(M\,dv=-u_0\,dM\), i.e. \(dv=-u_0\,dM/M\).
2. Integrate to get \(v(M)\)
\[ \int_{v_0}^{v}dv=-u_0\int_{M_0}^{M}\frac{dM}{M}\ \Longrightarrow\ v-v_0=-u_0\ln\frac{M}{M_0} \ \Longrightarrow\ v=v_0+u_0\ln\frac{M_0}{M}. \] This is the Tsiolkovsky rocket equation, independent of any assumption on how fast mass is ejected.
3. Adopt a constant mass-ejection rate to get an explicit time law
The problem's own data (\(u_0\) constant) does not by itself fix \(M(t)\); the simplest standard
choice, adopted here to produce an explicit law of motion, is a constant mass flow rate \(dM/dt=-\mu\)
(\(\mu>0\)), so \(M(t)=M_0-\mu t\) for \(0\le t \[
x(t)=\int_0^t v(\tau)\,d\tau=v_0t-u_0\int_0^t\ln\!\left(1-\frac{\mu\tau}{M_0}\right)d\tau.
\]
With \(u=1-\mu\tau/M_0\) and using \(\int\ln u\,du=u\ln u-u\), this evaluates (checked term-by-term with Sympy)
to
\[
x(t)=v_0t+\frac{u_0M_0}{\mu}\left[1-\frac{M(t)}{M_0}+\frac{M(t)}{M_0}\ln\frac{M(t)}{M_0}\right],
\]
which satisfies \(x(0)=0\). Differentiating the boxed \(x(t)\) with Sympy reproduces \(v(t)=v_0-u_0\ln(1-\mu t/M_0)\)
exactly (residual \(0\)), and \(x(0)=0\) by direct substitution — confirming the closed form is a genuine
antiderivative satisfying the initial condition, not just a plausible guess. Final answer: Velocity (Tsiolkovsky's equation, valid for any mass-loss law):
\[
\boxed{v(t)=v_0+u_0\ln\frac{M_0}{M(t)}}
\]
For the standard assumption of a constant mass-ejection rate \(dM/dt=-\mu\), so \(M(t)=M_0-\mu t\):
\[
\boxed{v(t)=v_0-u_0\ln\!\left(1-\frac{\mu t}{M_0}\right)},\qquad
\boxed{x(t)=v_0t+\frac{u_0M_0}{\mu}\left[1-\frac{M(t)}{M_0}+\frac{M(t)}{M_0}\ln\frac{M(t)}{M_0}\right]}.
\]4. Integrate \(v(t)\) to get \(x(t)\)
5. Verification
Result