Invoking Newton's third law on this problem is in a sense more ad-hoc and more abstract than using conservation of momentum with regard to this problem. The concept of force (Newton's second law) is very abstract; anything change to an object's momentum involves some force. Applying conservation of momentum is no less abstract than applying Newton's third law and yields a deeper answer.
Put the spaceship in deep space, far from any massive object, so that there are no measurable external forces acting on the vehicle. Now, what exactly is the force that is making the spaceship accelerate? By assumption, there are no external forces. You can say posit some ad-hoc force F that results from burning the fuel and get an answer via Newton's second law. Note well: Since the rocket's mass is changing, the simpler form F=ma is not valid here. We have to use F=dp/dt instead.
F<br />
= \frac{dp_r} {dt}<br />
= \frac{d}{dt} ( m_r \, v_r )<br />
= \dot m_r v_r}<br />
+ m_r \, a_r<br />
Solving for the rocket's acceleration,
a_r =<br />
\frac {F} {m_r}<br />
- \frac {\dot m_r} {m_r} v_r}<br />
This is not very satisying. What exactly is this force? It's purely ad-hoc for one thing. Moreover, it will turn out that the force is frame-dependent.
Here is how things turn out from conservation of momentum point of view. For this derivation, I will use some math that mathematicians don't particular like but physicists use willy-nilly -- things \Delta v. Things can be done more formally using continuum physics (classical treatment of gases), but that is a lot messier.
Nomenclature:
\vec v_r(t) Rocket velocity at time t, inertial observer
\vec v_e(t) Rocket exhaust velocity at time t, relative to vehicle velocity
m_r(t) Rocket mass at time t
\dot m_f(t) Rate at which rocket consumes fuel att
Over a small time interval \Delta t, the rocket will eject a small mass of fuel \dot m_f(t)\,\Delta t. At the start of the interval, the rocket has mass m_r(t), velocity \vec v_r(t), and momentum m_r(t) \, \vec v_r(t). At the end of the interval, the rocket has mass, velocity, and linear momentum
m_r(t+\Delta t) = m_r(t)-\dot m_f(t)\,\Delta t
\vec v_r(t+\Delta t) = \vec v_r(t)+\Delta \vec v_r(t)
\vec p_r(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t))
The bit of ejected fuel carries some momentum from the vehicle. The mass, inertial observer velocity, and momentum of the exhaust are
\Delta m_e(t) = \dot m_f(t)\,\Delta t
\vec v_{e_{\text{inertial}}}) = \vec v_r(t)+\vec v_e(t)
\Delta \vec p_e(t+\Delta t) =<br />
\dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))[/tex]<br />
<br />
The momentum of the rocket+exhaust at the end of the time interval is thus<br />
\vec p_{r+e}(t+\Delta t) =<br />
(m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t)) +<br />
\dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))<br /><br />
<br />
Dropping the second-order term \Delta t \Delta \vec v_r(t) and simplifying,<br />
<br />
\vec p_{r+e}(t+\Delta t) =<br />
m_r(t)\,\vec v_r(t)+<br />
m_r(t) \, \Delta \vec v_r(t) +<br />
\dot m_f(t)\,\Delta t\, \vec v_e(t)<br /><br />
<br />
Assuming no external forces act on the rocket and the ejected fuel during this time interval, the rocket and the ejected fuel form a closed system. Momentum is conserved in a closed system, so \vec p_{r+e}(t+\Delta t)=\vec p_r(t)[/tex]:<br />
<br />
&lt;br /&gt;
m_r(t)\,\vec v_r(t)+&lt;br /&gt;
m_r(t) \, \Delta \vec v_r(t) +&lt;br /&gt;
\dot m_f(t)\,\Delta t\, \vec v_e(t) = m_r(t)\,\vec v_r(t)&lt;br /&gt;<br />
<br />
or<br />
<br />
&lt;br /&gt;
m_r(t) \, \Delta \vec v_r(t) +&lt;br /&gt;
\dot m_f(t)\,\Delta t\, \vec v_e(t) = 0&lt;br /&gt;<br />
<br />
Dividing by \Delta t and taking the limit \Delta t \to 0,<br />
<br />
&lt;br /&gt;
\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_f(t)} {m_r(t)} \, \vec v_e(t)&lt;br /&gt;<br />
<br />
This is the equation for the acceleration of the rocket at time t.<br />
<br />
By conservation of mass, the time derivative of the rocket&#039;s mass is just the additive inverse of the fuel consumption rate: \dot m_r(t) = -\dot m_f(t). If the relative exhaust velocity is a constant vector, both the left and right hand sides of the above acceleration equation are integrable. Integrating from some initial time t_0 to some final time t_1 yields<br />
<br />
&lt;br /&gt;
\vec v_r(t_1) - \vec v_r(t_0) = \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right) \vec v_e&lt;br /&gt;<br />
<br />
This is the Tsiolokovsky rocket equation. Since the final mass is smaller than the initial mass, the logarithm will be negative. The change in velocity is directed against the exhaust direction.