= Solution
Take the <rocket exhaust speed> $u$ relative to the hovering ship, with exhaust directed vertically downward. If $\alpha=-\dot M>0$ is the <mass flow rate>, the upward <rocket thrust> is $u\alpha$. For example, over a short interval $dt$, the expelled mass $\alpha dt$ acquires downward <momentum> $-u\alpha dt$, giving the opposite recoil to the ship. Balancing its weight requires
$$
u\alpha(t)=M(t)g,\qquad\frac{dM}{dt}=-\frac gu M.
$$
Solving this <differential equation> with $M(0)=M_0$ yields the <hovering rocket mass flow>
$$
\boxed{M(t)=M_0e^{-gt/u},\qquad
\alpha(t)=\frac{gM_0}{u}e^{-gt/u}.}
$$
The required <mass flow rate> decreases because the mass to support decreases. These expressions apply while fuel is available and $g,u$ remain constant. If the <rocket dry mass> is $M_d>0$, hovering ends at $t=(u/g)\log(M_0/M_d)$; the mathematical limit $M\to0$ is not a model of an indefinitely fuelled physical ship.
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