rocket equation


by calculus_jy
Tags: equation, rocket
calculus_jy
calculus_jy is offline
#1
Jul15-08, 11:17 PM
P: 56
recenty i read the rocket equation, derivation of, however i think i have a slight confusion with signs
suppost initially a rocket has
mass= [tex]M[/tex]
velocity= [tex]\overrightarrow{v}[/tex]
then at a time dt later,
mass of rocket= [tex]M-dM[/tex]
velocity of rocket= [tex]\overrightarrow {v} +d\overrightarrow {v} [/tex]
mass of ejacted gas= [tex]dM[/tex]
velocity of gas= [tex]\overrightarrow{u}[/tex]

using conservation of momentum
[tex]\overrightarrow{v}M=(M-dM)(\overrightarrow{v}+d\overrightarrow{v})+\overrightarrow{u}dM[/tex]

[tex](\overrightarrow{u}-\overrightarrow{v})dM+Md\overrightarrow{v}=0[/tex]

but [tex](\overrightarrow{u}-\overrightarrow{v})[/tex]=velocity of gas relative to rocket
let [tex](\overrightarrow{u}-\overrightarrow{v})=\overrightarrow{U}[/tex]which is a constant

[tex]\overrightarrow{U}dM+Md\overrightarrow{v}=0[/tex]
[tex]-\int_{M_0}^{M}\frac{dM}{M}=\frac{1}{\overrightarrow{U}}\int_{\overright arrow{v}_0}^{\overrightarrow{v}}d\overrightarrow{v}[/tex]

now [tex]-ln\frac{M}{M_0}=\frac{\overrightarrow{v}-\overrightarrow{v_0}}{\overrightarrow{U}}[/tex]

the problem is , when taking the velocity in the direciton rocket is travelling
[tex]\overrightarrow{U}<0[/tex]
[tex]-ln\frac{M}{M_0}>0[/tex]since [tex]\frac{M}{M_0}<1[/tex]
then
[tex]\overrightarrow{v}-\overrightarrow{v_0}<0[/tex] which is impossibe as the rocket is accelerating???
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calculus_jy
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#2
Jul15-08, 11:29 PM
P: 56
latex problem seems to be fixed now...
Poincare1
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#3
Jul16-08, 11:39 PM
P: 7
i dont see how you get [tex]
M- dM
[tex] is is because of [tex]E=mc^2[tex]

Poincare1
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#4
Jul16-08, 11:45 PM
P: 7

rocket equation


why won't the latex work
luke
luke is offline
#5
Jul17-08, 05:34 AM
P: 25
Quote Quote by Poincare1 View Post
why won't the latex work
I hope this isn't too off topic but

i think you need [/tex] instead of [tex] at the end
D H
D H is offline
#6
Jul17-08, 07:32 AM
Mentor
P: 14,433
Quote Quote by calculus_jy View Post
[tex]\overrightarrow{U}dM+Md\overrightarrow{v}=0[/tex]
Good up to this point.
[tex]-\int_{M_0}^{M}\frac{dM}{M}
=\frac{1}{\overrightarrow{U}}
\int_{\overrightarrow{v}_0}^{\overrightarrow{v}}d\overrightarrow{v}[/tex]
This step is not valid. There is no such thing as the multiplicative inverse of a vector. Better is to define a unit vector v-hat directed along the rocket's delta-v vector. From the correct equation, this delta-v vector is directly opposed to the relative exhaust velocity vector. Thus

[tex]
\begin{aligned}
d\overrightarrow{v} &= dv \hat{v} \\
\overrightarrow{U} &= U \hat{v} & (U &\equiv \overrightarrow{U}\cdot \hat v)\\
&= -v_e \hat{v} & (v_e&\equiv -U)
\end{aligned}[/tex]

Note that ve is simply the magnitude of the relative velocity vector. With this, the vector differential equation becomes the scalar equation

[tex]-v_edM+Mdv=0[/tex]

from which

[tex]\int_{M_0}^{M}\frac {dM}{M} = \frac 1{v_e}\int_{v_0}^v dv[/tex]

or

[tex]\ln\frac{M}{M_0} = \frac{\Delta v}{v_e}[/tex]

You can use a vector formulation, but you can't divide by a vector like you did.
calculus_jy
calculus_jy is offline
#7
Jul17-08, 05:36 PM
P: 56
however is the equation
[tex]v=v_0+\overrightarrow{U}ln\frac{M}{M_0}[/tex] not right?using the notion in the first post
in the step with the integrals, to prevent multiplicative of inverse of vector, simply put [tex]\overrightarrow{U}[/tex]
on the same side of the equation as [tex]\frac{dM}{M}[/tex]
i have been told that its the problem associated with dM such that the mass of rocket after dt is M+dM not M-dM
i dont get why minus can be used in scalar, but cannot be used in vercot derivation?
D H
D H is offline
#8
Jul17-08, 06:21 PM
Mentor
P: 14,433
Your error occurs much earlier than I stated earlier.

Quote Quote by calculus_jy View Post
using conservation of momentum
[tex]\overrightarrow{v}M=(M-dM)(\overrightarrow{v}+d\overrightarrow{v})+\overrightarrow{u}dM[/tex]
Here your dM is the quantity of mass ejected by the spacecraft. With this definition, a positive dM means the spacecraft loses mass. Things would have worked properly if you had used dM as positive meaning the spacecraft gains mass. Then the conservation of momentum equation becomes

[tex]\overrightarrow{v}M=(M+dM)(\overrightarrow{v}+d\overrightarrow{v})-\overrightarrow{u}dM[/tex]
yathiraj
yathiraj is offline
#9
Nov27-10, 01:02 AM
P: 2
http://www.sciforums.com/showthread.php?t=74512
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