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Rocket launches up starting from zero, conservation momentum

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ex81
#1
Nov3-11, 06:17 PM
P: 67
1. The problem statement, all variables and given/known data

A rocket ejects exhaust with an velocity of u. The rate at which the exhaust mass is used (mass per unit time) is b. We assume that the rocket accelerates in a straight line starting initial mass (fuel plus the body and payload) be mi, and mf be its final mass, after all the fuel is used up. (a) Find the rocket's final velocity, v, in terms of u, mi, mf. (b) A typical exhaust velocity for chemical rocket engines is 4000m/s. Estimate the initial mass of a rocket that could accelerate a one-ton payload to 10% of the speed of light, and show that this design won't work. (ignore the mass of the fuel tanks.)


2. Relevant equations

Kinetic Energy = 1/2 mv^2
momentum = mv
work = f*d
f=ma

3. The attempt at a solution

I started out with the conservation of energy, and then switched to the conservation of momentum.

The rocket starts out at rest. so mi*vi = 0

0 = mf*v +(mi-mf)*u
which makes sense since the rocket's momentum will be opposite to the exhaust velocity, and the two momentum will cancel.

Now is the part where I am stuck.
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Andrew Mason
#2
Nov3-11, 10:20 PM
Sci Advisor
HW Helper
P: 6,654
Look up the ideal rocket equation. Although it is basically a Newton's third law momentum problem, it is not that simple to develop it on your own.

AM
ex81
#3
Nov4-11, 12:27 AM
P: 67
It is in the conservation section of the book. Ergo I have to use the conservation law :-(
I do realise that I need to differentiate the equation but I am pretty sure I need to tweak my current equation. Due to the change in mass vs velocity.

Andrew Mason
#4
Nov4-11, 02:12 AM
Sci Advisor
HW Helper
P: 6,654
Rocket launches up starting from zero, conservation momentum

Quote Quote by ex81 View Post
It is in the conservation section of the book. Ergo I have to use the conservation law :-(
I do realise that I need to differentiate the equation but I am pretty sure I need to tweak my current equation. Due to the change in mass vs velocity.
Ok. You can do a rough estimate using conservation of momentum.

Use the centre of mass frame - the earth. Initial momentum is 0. In order for you to get 1000kg moving at .9c how much mass moving at 4000 m/sec in the opposite direction will you need so that the total momentum is 0?

AM
ex81
#5
Nov4-11, 04:40 PM
P: 67
I think I am on the right track just from looking at NASA's ideal rocket equation
0 = mf*v +(mi-mf)*u

now NASA has dm * u/dt= v * d(mass propellant)/dt. Cancel out their dt, and their equation is dm *u=v*d(mass propellant)

So based on the above my equation needs to change a little, and needs to be derived.
ex81
#6
Nov4-11, 04:41 PM
P: 67
P.S. my u, and v is switched from NASA's


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