rocket equation

calculus_jy

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???

calculus_jy

latex problem seems to be fixed now...

Poincare1

i dont see how you get [tex]
M- dM
[tex] is is because of [tex]E=mc^2[tex]

Poincare1

why won't the latex work

luke

I hope this isn't too off topic but

i think you need [/tex] instead of [tex] at the end

D H

Good up to this point.
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 v_e 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

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

Your error occurs much earlier than I stated earlier.

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

http://www.sciforums.com/showthread.php?t=74512
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