Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 3C / Solution

Take upward velocity as positive, and consider the closed material system consisting of the rocket and the fuel it ejects during a short interval $dt$. Initially its mass and momentum are $m$ and $mv$. Let $\delta m>0$ be the ejected mass, so the rocket's mass change is $dm=-\delta m$. At the end its velocity is $v+dv$, while the ejected gas has inertial velocity $v-u$ to first order; its downward relative velocity need not be downward in the inertial frame.

The final total momentum, to first order, isNeglect air resistance and take the gravitational acceleration $g$ constant. The external gravitational impulse on this material system is $-mgdt$ to first order. Newton's second law therefore gives $mdv-u\delta m=-mgdt$, orThe positive thrust is $-u\dot{m}$ because fuel loss makes $\dot{m}<0$. Applying $d(mv)/dt$ to the rocket alone without accounting for the escaping momentum would give the wrong equation.

For constant $u$ and positive mass, divide by $m$ and integrate from release at $t=0$:Thus the gravity-corrected rocket equation is equivalent to