Force Exerted on a Rocket

I am reading Spivak's "Physics for Mathematicians". He has the following setup up: We let $m(t)$ be the mass of a rocket and its fuel at time $t$. And, we let $q(t)$ be the velocity at which the fuel is being expelled. Then the amount of fuel being expelled at time $t$ can be approximated by $m(t)-m(t+h)$ for small $h$. So, the momentum of the fuel is approximately $[m(t)-m(t+h)]q(t)$. So far, so good. He next says that the force this fuel exerts must be $m'(t)q(t)$. That I don't get.

Until this point he has only treated systems with constant mass, where the force is the derivative of momentum. Reading on Wikipedia, it seems that this would not be the case in the above system, because mass varies. I also don't see why the force should not depend on the change in velocity.

Would someone be so kind to explain away my confusion?

• This problem is typically confusing. I suggest checking "The rocket equation" before wasting too much time. Dec 6, 2015 at 0:04

Maybe if you view it would another perspective. Lets follow the mass that gets expelled from the rocket from time $t$ to $t+h$, denoted by $m_h(t)=m(t)-m(t+h)$. I assume that all fuel initially is at rest relative to the rocket, so the expelled mass has to accelerate from zero to $q(t)$. The acceleration experienced by $m_h$ can be found with the change in velocity, $q(t)$, divided by the time step, $h$. This acceleration can also be expressed as the force acting on it divided my the mass $m_h(t)$ (using $F=ma$),

$$\frac{q(t)}{h} = \frac{F(t)}{m_h}.$$

By rewriting this to $F$ yields,

$$F(t) = \frac{q(t) m_h(t)}{h} = \frac{q(t) (m(t)-m(t+h))}{h}.$$

Now note that $\frac{m(t)-m(t+h)}{h}$ is equal to the derivative of $m(t)$ as $h\to0$.

This force acts on $m_h(t)$, however the same force, but in the opposite direction, acts on the rocket (Newton's third law of motion).

• I was missing the accelerating from $0$ to $q(t)$ part. Nice answer. I wish I could upvote it twice.
– J126
Dec 6, 2015 at 0:06