GR8677 #61



Alternate Solutions 
Herminso 20090921 19:28:34  Consider the system rocket of mass plus a fraction of combustible traveling at velocity just before the mass is exhausted. From a frame that moves at velocity we see the system instantaneously at rest, and then when the mass is exhausted, we see that the mass moves front at velocity , while moves backward at that is the speed of rocket's exhaust combustible. By momentum conservation
or just the equation given since is negative for the test makers.
Thus is truly the speed of rocket's exhaust combustible   kimseonta 20090107 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.  

Comments 
GPRE 20130924 07:26:28  Would it be right to look at it like this: The second term suggests a change in mass and the only way the rocket's mass changes is by spending fuel.   Almno10 20101111 23:11:14  If you wanted to fart in space in order to propel yourself forward, you'd have to pull your pants down first so the fart particles wouldn't collide with the seat of your pants.
padkins 20110206 12:44:41 
I only fart neutrinos

  subwoofer911 20091103 23:43:34  v is the speed of the rocket, so A, B and C are automatically wrong. And there IS an acceleration v/t, so the constant u cannot be an instantaneous velocity, D. Therefore it is E.
varsha 20160220 08:30:26 
Common sense...Nice

  abacus 20091006 23:39:42  Rocket equation must remain correct under gallilean symmetry
v > v+v0
this leaves E   Herminso 20090921 19:28:34  Consider the system rocket of mass plus a fraction of combustible traveling at velocity just before the mass is exhausted. From a frame that moves at velocity we see the system instantaneously at rest, and then when the mass is exhausted, we see that the mass moves front at velocity , while moves backward at that is the speed of rocket's exhaust combustible. By momentum conservation
or just the equation given since is negative for the test makers.
Thus is truly the speed of rocket's exhaust combustible
Quynh_Strange_Beauty 20141013 23:26:50 
We can quickly see that the meaning of equation is momentum conservation (Delta (mv)_1  Delta (mv)_2 = 0). That's enough. no more math.
So u must be the speed of the exhaust. (D or E)
OK. What happen if u = 0 ? the equation says rocket gain no momentum. So u must be speed of exhaust relative to rocket. (IF I picked D, my rocket can gain momentum if I choose certain frame of reference, not true)

  kimseonta 20090107 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 20090106 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 20080924 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 20081031 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

  hefeweizen 20061130 12:10:53  is there a typo in the last line?
mdv = Vdm
blue_down_quark 20080827 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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