# Conservation of momentum for rockets [closed]

I'm reading Kleppner & Kolenkow Example 4.16 on page 140-141 of the second edition and I am confused by something.

So we have a rocket in free space. No friction and no gravity. The rocket has mass $M$ and the fuel has mass $m$. The rocket has initial velocity $v_0$ and the fuel is ejected out the back at velocity $u$ relative to the rocket.

My approach (which I understand to be wrong) is to do this. There is no external force on the system. So momentum should be conserved. So,

$$P_0 = v_0 (M + m)$$

$$P_1 = v_1 M + m (v_0 - u)$$

We set

$$P_0 = P_1$$ $$v_0 M + v_0 m = v_1 M + v_0 m - m u$$ $$M(v_1 - v_0) = mu$$ $$v_1 = v_0 + \frac{m}{M}u$$

K&K does something completely different and finds a very different answer of

$$v_1 = v_0 + u \ln \frac{m + M}{M}$$

I'm confused. Where did I go wrong?

• Consider writing $P_1=M(v_0+dv) + dm(v_0-u)$. The key is that dm is teh second variable. You've got a differential equation with separated variables: dm and dv. – FGSUZ Feb 12 '18 at 21:39
• – Kyle Kanos Feb 14 '18 at 11:05
• Possible duplicate of How does a rocket's exhaust gas' momentum propel a rocket? – Kyle Kanos Feb 14 '18 at 11:05
• Kyle, that question is indeed related, but a link to it would not have answered the question I was asking. My question was "here is a specific derivation I know to be wrong, please tell me how it is wrong." – azani Feb 14 '18 at 14:41

The fuel is taken to be a continuous medium, expelled at a steady rate, not all at once like a block thrown out the back (for which your result would be correct). Once some of the fuel has been expelled, the speed of the rocket is no longer what it was. So the final speed of each bit of fuel varies depending on when it was expelled. It does not all end up moving at the speed $v_0 - u$.

I've decided to complete my comment with a more detailed answer.

You're missing the fact that each tiny bit of fuel increments the speed of a different total mass. Let the variables be M=rocket mass; m=fuel mass; v=velocity/speed, and u=velocity of fuel expulsion assumed to be constant.

Your actual formula for momentum conservation is

$$(M+m)v=(M+m-dm)\cdot(v+dv) \ \ + \ \ dm\cdot(v-u)$$

If you expand both sides and simplify, you will have

$$M dv + m dv - dmu=0$$

where I have neglected the second order differential $dm dv$, which is a frequent technique.

If you rearrange all this, you get

$$dv= u \frac{dm}{M+m}$$

And this is what you integrate.

By the way, you can use that, since $M+m=constant=total\ mass$, differentiation gives

$dM+dm=0$, or $dM=-dm$, which is a good way to change the integration variable, so that you can find the limits of the integral more easily.

You are not following the book definition of $u$ (fuel speed becomes $v+u$) and you are assuming all of the fuel is expelled in 1 go.