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GR8677 #61 |
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Alternate Solutions |
kimseonta
2009-01-07 01:18:00 | Sorry about the one that I posted below...
it is totally incorrect...so I am now going to post the solution again...
The rocket is in a free space -> No external forces
the momentum must be conserved.
Pi=Pf
mv=(m+delta m)(v+ delta v)+(V+v)( - delta m)
, where the delta m should be negative when considering a rocket propulsion. And V is the velocity [Not speed] of the rocket's exhaust from the rocket's frame.
[m + (delta m)](delta v) - V(delta m) = 0
take /(delta t) and limit[delta t -> 0]
Then,
mdv/dt - Vdm/dt = 0
Here, of course, V is negative[vector]. Let u be the speed[scalar] of the rocket's exhaust from the rocket's frame.
 u = -V
mdv/dt + udm/dt = 0: Here is the answer. |  |
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Comments |
kimseonta
2009-01-07 01:18:00 | Sorry about the one that I posted below...
it is totally incorrect...so I am now going to post the solution again...
The rocket is in a free space -> No external forces
the momentum must be conserved.
Pi=Pf
mv=(m+delta m)(v+ delta v)+(V+v)( - delta m)
, where the delta m should be negative when considering a rocket propulsion. And V is the velocity [Not speed] of the rocket's exhaust from the rocket's frame.
[m + (delta m)](delta v) - V(delta m) = 0
take /(delta t) and limit[delta t -> 0]
Then,
mdv/dt - Vdm/dt = 0
Here, of course, V is negative[vector]. Let u be the speed[scalar] of the rocket's exhaust from the rocket's frame.
u = -V
mdv/dt + udm/dt = 0: Here is the answer. | | kimseonta
2009-01-06 20:13:36 | This is my own way to make the typo (?) clear...
The same idea goes: the conservation of the momentum.
Pi = Pf
mv = (m + dm)(v+dv) + (dm)V, where V <0 of course.
= mv+mdv+vdm+dmdv+(dm)v
Now there exist two appoximations that one can make.
- V >> v : one can simply assume that the speed of the exaust from a rocket should be much faster than the one of the rocket.
-dmvm is almost zero relative to the other parts of the equation.
so the simplified version will be here...
mdv+Vdm=0
m(dv/dt) + V(dm/dt) = 0 |  | Poop Loops
2008-09-24 10:44:52 | I don't get the jump from Line 2 to Line 3. Where does mdV come from? And why is it approx. equal to dm'V?
mv = mv + mdv - dm'v - dm'dv + dm'v - dm'V
Subtract mv from both sides and -dm'v and +dm'v cancel, so you get:
0 = mdv - dm'dv - dm'V
(m - dm')dv = dm'V
So I have no idea how (m - dm')dv relates to mdV
realcomfy
2008-10-31 03:25:25 |
It was stated in the solution that higher order terms such as  were neglected. The solution thus becomes what is already there
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| | hefeweizen
2006-11-30 12:10:53 | is there a typo in the last line?
mdv = -Vdm
blue_down_quark
2008-08-27 23:44:48 |
I think V is assumed to be negative in the reference frame where v is positive. It's simply the exhaust velocity without its sign.
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