How Does the Sign of ##dm## Affect Rocket Propulsion Calculations?

[Rikudo]

Homework Statement

A rocket is in a space. The rocket's initial mass and velocity is m and v. After that, mass dm is ejected backwards with constant speed u relative to the rocket. Find the equation for the velocity of the rocket

Relevant Equations

momentum conservation

I have a question. If we assume that $dm$ is positive, is the answer supposed to be different from the one when we regard the $dm$ as negative?

1. If I assume that $dm$ is positive:
By using momentum conservation, we will get
$$mv=(m-dm)(v+dv)+dm (v-u)$$
simplify the equation
$$m \,dv=dm \,u$$
Integrate the equation and we will get
$$ln\frac {m'} {m} = \frac {v'-v} {u}$$

2. if I assume that mass $dm$ is negative:
$$mv=(m+dm)(v+dv)-dm (v-u)$$
Simplify
$$m \,dv=-dm \,u$$
The answer will be
$$ln\frac {m'} {m} = - \frac {v'-v} {u}$$

What exactly happened here?

 

[erobz]

It looks like if you assume $dm$ is negitive you have the rocket gaining mass. The momentum before in that case is not $mv$:

$$F_{ext}dt= ( m +dm)(v+dv) - [mv + dm u]$$

 

[Rikudo]

It looks like if you assume $dm$ is negitive you have the rocket gaining mass. The momentum before in that case is not $mv$ .

I copied it from my textbook. I doubt that its answer is wrong

 

[Frabjous]

You are orienting the axis differently.
What is ∫dx from 0 to 1 for dx positive?
What is ∫dx from 1 to 0 for dx positive?
What is ∫dx from 0 to 1 for dx negative?
Think about Riemann sums.

 

[Rikudo]

You are orienting the axis differently.

ah...

What is ∫dx from 0 to 1 for dx negative?

Um... imo, the dx need to be positive in order to solve it like usual. What should I do ?

 

[Frabjous]

Riemann sums assume you are adding up rectangles. What does this imply for the sign of the integral when dx is negative?

 

[Drakkith]

Forgive me, but how can $dm$ be negative if it represents the mass ejected from the rocket?

 

[Rikudo]

Forgive me, but how can $dm$ be negative if it represents the mass ejected from the rocket?

"during a small time dt, a negative mass dm gets added to the rocket, and a positive mass (−dm) gets shot out the back"

 

[Drakkith]

$mv=(m-dm)(v+dv)+dm (v-u)$

Note that this equation represents: $InitialRocketMomentum = FinalRocketMomentum + ExhaustMomentum$

It seems to be a physical impossibility for $dm$ to be a negative quantity, as that no longer gives us the correct momentum of the rocket or its exhaust.

"during a small time dt, a negative mass dm gets added to the rocket, and a positive mass (−dm) gets shot out the back"

That just switches the sign of $dm$ back to what it should be to make the equations and terms make sense.

I'm uncertain about what you're actually asking.

 

[Rikudo]

Riemann sums assume you are adding up rectangles. What does this imply for the sign of the integral when dx is negative?

If I solve it like usual, the result (or, we can also say the area under the curve) is negative.
So, does that mean I need to multiple the result by (-1) so that it become positive?

EDIT: nevermind. This is clearly incorrect.

 

[Rikudo]

I'm uncertain about what you're actually asking.

The second method in my first post is what is written in the book.(By the way the sentence that I just quoted is also taken from it). Then, I tried a slightly different way by regarding $dm$ as a positive mass, but the result is different from the book's. That is what I'm confused right now; why they are different.

 

[Frabjous]

I must apologize. My reply was a little misleading and there are some subtleties that I do not feel confident explaining (or am sure that I completely understand)
The basic idea is:
For a rocket, dm = m(t+Δt)-m(t) which is negative (or m(t+Δt) = m(t)+dm). This matters as one turns things into differentials. So if you want to define dm as positive, it has a sign inconsistency with m, so one needs to add a negative sign or reverse the limits of integration.

 

[Drakkith]

The second method in my first post is what is written in the book.(By the way the sentence that I just quoted is also taken from it). Then, I tried a slightly different way by regarding $dm$ as a positive mass, but the result is different from the book's. That is what I'm confused right now; why they are different.

Ah, I see. I was referencing this page, which has the original equation listed with the same signs as your first example in your first post does, as well as $mdv=-dmu$ (the integrated equation in 14.1 without the gravity term). Check the link and see if your steps match up with theirs.

 

[haruspex]

dm is defined to be the before-to-after gain in m. By taking it as positive, you have reversed the order of integration, making it $\int_{m'}^m$. Switching that back will make the signs match.

 

[Orodruin]

This is a very common issue. It ultimately boils down to using $dm$ to represent both the mass ejected (ie, positive if mass is ejected) and as the change in the mass when doing the integration. The two differ by a sign and you will get the wrong answer if you do not account for this. The notationally least confusing way in my opinion is to use $dm$ as the change in the mass $m$ because then you will not confuse yourself with the integration limits.