# Rocket motion in space

Father_Ing
Homework Statement:
-
Relevant Equations:
momentum conservation
Consider a rocket with mass ##m## in space is going to move forward. In order to do so, it needs to eject mass backwards. Let the mass that is ejected has velocity ##u## relative to the rocket. What is the equation for the final velocity?

It is said that after ##dt## second, the rocket will have mass ##m-dm##, and velocity ##v+dv##.But, isn't it also possible for the speed to increase in high sum after a very small amount of time, or even, not changing at all?

And I tried to find this out by using conservation of momentum.
Let ##v'## be the rocket's speed after ##dt## second, and ##v## is the initial speed.
$$mv = dm(v-u)+(m-dm)v'$$
$$v'=\frac {(m-dm)v +udm}{m-dm}$$

$$v'=v+\frac {udm}{m-dm}$$
Since ##dm## is small, we can take the limit of dm->0. Therefore,$$v'= v$$
It can be concluded that the velocity neither increase nor decrease.

But, I searched about this matter in the internet, and they said that ##v'## is ##v+dv##. Are there any mistakes in my method?

## Answers and Replies

Staff Emeritus
Science Advisor
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Homework Statement:: -
Relevant Equations:: momentum conservation

Since dm is small, we can take the limit of dm->0.
No you cannot. You are looking for dv, the small change in velocity when you eject mass dm. Obviously, if you do not eject any mass at all (ie, take the limit dm -> 0), then velocity does not change.

You want to find the change dv while ignoring higher order contributions (ie, ##dm^2## etc).