Sign problem in the rocket equation

[Leo Liu]

Homework Statement

A quiz question

Relevant Equations

$\vec p_1=\vec p_f$

Question:

Solution:

Issue:
I would like to know why $\Delta m=-\Delta M$ rather than $=\Delta M$ if minus signs have been used in the second equation. Thank you.

 

[PeroK]

I would like to know why $\Delta m=-\Delta M$ rather than $=\Delta M$ if minus signs have been used in the second equation. Thank you.

I assume that $\Delta m$ is the mass ejected and $\Delta M$ is the change in the mass of the rocket. They must be equal and opposite. If they were equal, then the rocket would be ejecting mass and gaining that mass - which is not right.

 

[TSny]

Note that positive momentum is taken to be in the direction of the forward motion of the rocket. Does the quantity $u - v - \Delta v$ represent the velocity of the ejected fuel in the forward direction or the rearward direction? Thus, should the final momentum of the ejected fuel (in the forward direction) be written as $+\Delta m (u - v - \Delta v)$ or as $-\Delta m (u - v - \Delta v)$? [As @PeroK noted, $\Delta m$ represents the mass of the ejected fuel. So, $\Delta m$ is positive.]

 

[Leo Liu]

Hello. Thanks for your reply.

Does the quantity u−v−Δv represent the velocity of the ejected fuel in the forward direction or the rearward direction?

It probably means backward.

Thus, should the final momentum of the ejected fuel (in the forward direction) be written as +Δm(u−v−Δv) or as −Δm(u−v−Δv)?

The second one I guess since the momentum of the fuel points "upward". Yet would the answer change if I wrote the velocities in the vector notation?

 

[Leo Liu]

I assume that $\Delta m$ is the mass ejected and $\Delta M$ is the change in the mass of the rocket. They must be equal and opposite. If they were equal, then the rocket would be ejecting mass and gaining that mass - which is not right.

Could you tell me why this substitution is valid even if it makes the rocket seem as if it is gaining mass $(2M(t)+2dM+M_0)(v+dv)$?

 

[PeroK]

Could you tell me why this substitution is valid even if it makes the rocket seem as if it is gaining mass $(2M(t)+2dM+M_0)(v+dv)$?

Why does the rocket seem to be gaining mass?

 

[Leo Liu]

Why does the rocket seem to be gaining mass?

The reason is that the second term 2dM seems to be positive. But I get it now.

I am afraid that I still don't quite understand why we cannot make $\Delta m=\Delta M$. Does the reason have something to do with $-\Delta m(u-v-\Delta v)$?

 

[PeroK]

The reason is that the second term 2dM seems to be positive. But I get it now.

I am afraid that I still don't quite understand why we cannot make $\Delta m=\Delta M$.

You can do that. That means that the rocket is gaining mass. That's a different problem, where the rocket is bombarded by particles. In this problem the rocket is losing mass.

 

[TSny]

I don't know if this analogy will help. Suppose we have two glasses of water. The mass of water in the first glass is denoted by $M$ and the mass in the second glass is denoted by $m$. Suppose I pour a little water from the first glass into the second glass.

The change in mass of water in the first glass would be denoted $\Delta M$. Since the mass of water in the first glass decreases, $\Delta M$ is a negative number.

The change in mass of water in the second glass would be denoted $\Delta m$. Since the mass of water in the second glass increases, $\Delta m$ is a positive number.

The equation $\Delta m = \Delta M$ cannot be correct since it says that a negative number equals a positive number. The correct equation is $\Delta m = -\Delta M$.

 

[Leo Liu]

The change in mass of water in the first glass would be denoted $\Delta M$. Since the mass of water in the first glass decreases, $\Delta M$ is a negative number.

The change in mass of water in the second glass would be denoted $\Delta m$. Since the mass of water in the second glass increases, $\Delta m$ is a positive number..

I find this analogy very helpful. Many thanks!

 

[TSny]

In my opinion, the diagram in the first post is incorrect in how it writes the mass of the ejected fuel. The mass of the ejected fuel is $dm$, which equals $-dM$.