# Rocket Motion

1. Jan 26, 2009

### Gogsey

A rocket is accelerating upwards from rest in a uniform gravitational field g. Notation: m(t) is the mass of the rocket plus remaining fuel, m0 is the initial total mass, vex is the exhaust speed (relative to the rocket), and k is the rate, in kg/s, at which fuel is consumed. By considering momentum changes in a short time dt, show that mdv/dt = kv(ext) - mg(upward direction is positive). Assuming k is constant, find the speed v as a function of the
remaining mass m. Show that, if k is very large, it agrees with the result derived in class (without gravity).

Ok, so i have already proved the first part.

P(t)=mv and P(t+dt) = (m+dm)(v+dv)-dM(v-v(ext))

=mv+vdm+dmv(ext)

The a couple of steps later we get: -mgdt=mdv+(ext)(dm/dt)

so we get -mg=mdv/dt -kv(ext), where -k=dm/dt

and therefore mdv/dt=kv(ext) - mg.

Ok, so I'm a little stuck on the second part. Not really sure where to start off. Any hints?

2. Jan 26, 2009

### chrisk

Assuming that k=dm/dt is constant isolate dv then integrate. You have to assume the exhaust velocity, v(ext) is constant.