Forest UCM Ch3 Rockets

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Rocket problem 1

Remember the hobo problem? This is how rockets are propelled by expending fuel (mass).

Lets first think about the inverse rocket where mass is added instead of expelled.

Consider a railroad car with frictionless wheels moving at speed v on level ground.

A hobo handing from a tree drops down onto the railroad car as it is moving under the tree.

What is the final velocity of a railroad car after the hobo jumps on.

Conservation of momentum
Mv=(M+m)vf
vf=MM+mv

The railroad car slows down after the hobo jumps on.

Rockets work in a similar fashion. They speed up after the fuel "jumps" off.

Rocket Problem 2

Consider a rocket of mass m moving at a speed v ejecting rocket fuel for propulsion.


Forest UCM Ch3 Rockets Fig.pngFile:Forest UCM Ch3 Rockets Fig.xfig.txt


m= mass of Fuel + Rocket

dm= mass of Fuel ejected over time interval dt ( dm = mass lost by rocket < 0)

u=vFR velocity of Fuel relative to the Rocket

v= velocity of rocket relative to the ground before ejecting fuel of mass (dm)

v+dv=vRG velocity of the rocket relative to the ground after ejecting fuel

VFG= Velocity of the fuel with respect to the ground

Apply Conservation of Momentum

mv=(dm)VFG+(m(dm))(v+dv)
dmVFG=mdv+dm(v+dv)

The velocity of the fuel with respect to the ground(VFG) may be written as the vector sum of the rocket's velocity with respect to the ground (VRG) and the velocity of the fuel with respect to the rocket (VFR) :Galilean transormation

VFGVFR+VRG

assuming 1-D motion and using the velocity variables defined above

VFG=u+(v+dv)

substituting

dm(u+(v+dv))=mdv+dm(v+dv)
udm=mdv

solving for the velocity

vv0=mm0udmm
vv0=uln(m0m)
v=v0+uln(m0m)

where m0 is the initial Rocket mass

In terms of Thrust

Defining

THRUST=˙mu

Then

m˙v= THRUST = Force on rocket as a result of expelling fuel

Forest_UCM_MnAM#Rockets