Force from changing mass; acceleration increases

Force due to change in mass

Let a system of mass $m$ moving with velocity $\vec v$ lose (eject) a small mass $\Delta m$ in time $\Delta t$, the ejected particles leaving with absolute velocity $\vec u$. An external force $\sum\vec F$ acts.

Momentum at $t$: $m\vec v$. At $t+\Delta t$: $(m-\Delta m)(\vec v+\Delta\vec v)+\Delta m\,\vec u$. Impulse-momentum:

$$m\vec v+\sum\vec F\,\Delta t=(m-\Delta m)(\vec v+\Delta\vec v)+\Delta m\,\vec u$$

Neglecting $\Delta m\,\Delta\vec v$ and dividing by $\Delta t$ (with $\dot m=-\Delta m/\Delta t$ the rate of change of the system mass, so $\dot m<0$):

$$\sum\vec F=m\frac{d\vec v}{dt}+\frac{\Delta m}{\Delta t}(\vec u-\vec v)$$

With $\vec u_{rel}=\vec u-\vec v$ (exhaust velocity relative to the body) and $\left|\dot m\right|=\Delta m/\Delta t$:

$$m\vec a=\sum\vec F-\left|\dot m\right|\vec u_{rel}$$

Here $\vec u_{rel}$ points backward, so $\vec T=-\left|\dot m\right|\vec u_{rel}$ is a forward thrust force. Hence $m\vec a=\sum\vec F+\vec T$ with $T=\left|\dot m\right|u_{rel}$.

Why the acceleration increases when mass is lost

Along the line of motion:

$$a=\frac{\sum F+T}{m(t)},\qquad m(t)=m_0-\left|\dot m\right|t$$

Take a constant burn rate and constant exhaust speed, so $T$ is constant, and assume the external force (weight, drag) is roughly constant. The numerator then stays the same while the denominator $m(t)$ decreases, so $a$ increases with time:

$$\frac{da}{dt}=\frac{(\sum F+T)\left|\dot m\right|}{m^{2}}>0$$

(when $\sum F+T>0$). The final acceleration is therefore larger than the initial one. For a rocket in free space ($\sum F=0$): $dv=u_{rel}\dfrac{-dm}{m}$, giving $v=v_0+u_{rel}\ln\dfrac{m_0}{m}$, so the velocity grows faster as $m$ becomes small.