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

There are no Alternate Solutions for this problem. Be the first to post one!

Comments

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

Reply to this comment

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.

Reply to this comment

Post A Comment!

You are replying to:

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

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

Chat Archives | Delete left banner ad | Donate

(Click to view chat.)

Hate being Anonymous? Login or Register

Upgrade to Poser 7 Now

Huge Textbook Selection, Low Prices – Phat Campus.

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone