# Rocket Motion in Free Space

1. Oct 24, 2005

### Euclid

For a rocket in free space, it seems that
$$P = mv + (m_0 - m)(v-u)$$
where m is the mass of the rocket, v is its speed, m_0 is its initial mass, u is the speed of the propellent relative to the ship.
But in solving dP/dt=0, I get that the velocity is linear in m. There seems to be a mistake in what I have taken to be the total momentum. Can anyone point it out?

2. Oct 24, 2005

### Cyrus

$$P = (m+dm)(v+dv) + (-dm)(v-v_{ex})$$ Both dm and dv are functions of time.

3. Oct 24, 2005

### Euclid

The problem is I am trying to solve the problem without appealing to the use of infinitesimals. It seems to me I should be able to write down P, differentiate and solve, with having to use these bizarre little quantities. What you have written is mv, the momentum of the rocket, not what I am calling P, the total momentum of the system, which must be constant.

4. Oct 24, 2005

### pervect

Staff Emeritus
Hmmm, well then using your notation and variables (not what I would have used) we get

const = P =
$$\int -(v(t)-u)\frac{dm}{dt} dt + m(t)*v(t)[itex] for the same reason that cyrusabdollahi wrote - v is not a constant, it is a function of time. We want to add together all the exhaust masses dm moving at a velocity of v(t)-u. Note that dm/dt is <0, that's the reason for the minus sign in the first equation (easy to miss). If we differentiate this expression with respect to time, the derivative is zero becaue it's constant, and we get [tex] -(v(t)-u) \frac{dm}{dt} + m(t)\frac{dv}{dt} + v(t)\frac{dm}{dt} = 0$$

which reduces to the standard rocket equation

m dv = u dm or
$$v = u \int \frac{dm}{m}$$