GREPhysics.NET: GR8677 #61

GR8677 #61

Problem

GREPhysics.NET Official Solution

Mechanics $\Rightarrow$ }Conservation of Momentum

One can easily derive the formula for rocket motion,

$\begin{eqnarray} p_{initial}&=&p_{final}\\ mv &=& (m-dm')(v+dv)+(dm')(v-V)\\ mv &=& mv + mdv - dm'v - dm'dv + dm'v - dm'V\\ mdV &\approx& dm'V\\ mdV &=& -dm V,<br /> \end{eqnarray}$

where $m$ is the mass of the rocket and $m'$ is the mass of the fuel rejected. $v$ is the initial velocity of the rocket. $V$ is the velocity of the exhaust. The final line comes about from realizing that $dm=-dm'$ .

The derivation starts with the assumption that the final mass of the rocket is its original mass minus $dm'$ , the eject mass, and that its final velocity is a tiny bit faster than it was before $dv$ . The exhaust mass' velocity relative to the rocket is $V$ . (Higher order terms such as $dxdy$ have been thrown out to arrive at the final differential equation.)

$V$ is equivalent to the velocity of the exhaust mass relative to the rocket, the inertial reference system. Thus $u=V$ and the answer is choice (E).

(Note that firing the exhaust backwards generates a forward thrust for the rocket---motion in space is quintessentially dependent on the phenomenon of farting. So, if one is well-equipped and wanna get going in space, just let out a bit of gas, and one's good to go.)

Alternate Solutions

Herminso 2009-09-21 19:28:34	Consider the system rocket of mass $m$ plus a fraction $\triangle m$ of combustible traveling at velocity $v$ just before the $\triangle m$ mass is exhausted. From a frame that moves at velocity $v$ we see the system instantaneously at rest, and then when the mass $\triangle m$ is exhausted, we see that the mass $m$ moves front at velocity $\triangle v$ , while $\triangle m$ moves backward at $u$ that is the speed of rocket's exhaust combustible. By momentum conservation $0=m\triangle v-\triangle m u$ $\Rightarrow$ $m(\triangle v/\triangle t)-(\triangle m/\triangle t) u=0$ or $m(\partial v/\partial t)-(\partial m/\partial t) u=0$ just the equation given since $u$ is negative for the test makers. Thus $u$ is truly the speed of rocket's exhaust combustible Reply to this comment
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. $\Rightarrow$ u = -V mdv/dt + udm/dt = 0: Here is the answer. Reply to this comment

Comments

subwoofer911
2009-11-03 23:43:34

v is the speed of the rocket, so A, B and C are automatically wrong. And there IS an acceleration $\partial$ v/ $\partial$ t, so the constant u cannot be an instantaneous velocity, D. Therefore it is E.

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abacus
2009-10-06 23:39:42

Rocket equation must remain correct under gallilean symmetry
v -> v+v0

this leaves E

Reply to this comment

Herminso
2009-09-21 19:28:34

Consider the system rocket of mass $m$ plus a fraction $\triangle m$ of combustible traveling at velocity $v$ just before the $\triangle m$ mass is exhausted. From a frame that moves at velocity $v$ we see the system instantaneously at rest, and then when the mass $\triangle m$ is exhausted, we see that the mass $m$ moves front at velocity $\triangle v$ , while $\triangle m$ moves backward at $u$ that is the speed of rocket's exhaust combustible. By momentum conservation

$0=m\triangle v-\triangle m u$ $\Rightarrow$ $m(\triangle v/\triangle t)-(\triangle m/\triangle t) u=0$ or $m(\partial v/\partial t)-(\partial m/\partial t) u=0$ just the equation given since $u$ is negative for the test makers.

Thus $u$ is truly the speed of rocket's exhaust combustible

Reply to this comment

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.
$\Rightarrow$ u = -V

mdv/dt + udm/dt = 0: Here is the answer.

Reply to this comment

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

Reply to this comment

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 $dm'dv$ 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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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

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