# A rocket ejecting mass

• Rikudo
The basic idea is:In summary, if you assume that mass ##dm## is positive, the answer will be different from the one when you regard the ##dm## as negative.f

#### 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
$$m \,dv=-dm \,u$$
$$ln\frac {m'} {m} = - \frac {v'-v} {u}$$

What exactly happened here?

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

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

Last edited:
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

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?

Last edited:
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 ?

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

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

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"

##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.

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.

Last edited:
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.

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.

Last edited:
• Rikudo
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.

• Rikudo
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.

• Rikudo and Drakkith
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.

• Rikudo, vanhees71, Filip Larsen and 1 other person