 Solution

During a short time $dt$, the rocket loses mass $-dm>0$. Conservation of upward momentum, including gravity's impulse and exhaust velocity $v-U$, gives to first orderwhich is the stated rocket equation.

Here $dm/dt=-\alpha$ and $U=U_0{m}_0/m$, soLift-off from rest requires positive initial acceleration,With $v(0)=0$, integration givesThe dimensions are $[m_0]=[m]=M$, $[\alpha ]=M/T$, $[U_0]=[U]=[v]=L/T$, $[g]=L/T^2$, and $[t]=T$. Both displayed terms in $v$ therefore have dimension $L/T$.