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Rocket equation 
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#1
Jul1508, 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]MdM[/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=(MdM)(\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??? 


#2
Jul1508, 11:29 PM

P: 56

latex problem seems to be fixed now...



#3
Jul1608, 11:39 PM

P: 7

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


#4
Jul1608, 11:45 PM

P: 7

Rocket equation
why won't the latex work



#5
Jul1708, 05:34 AM

P: 25

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


#6
Jul1708, 07:32 AM

Mentor
P: 15,202

[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. 


#7
Jul1708, 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 MdM i dont get why minus can be used in scalar, but cannot be used in vercot derivation? 


#8
Jul1708, 06:21 PM

Mentor
P: 15,202

Your error occurs much earlier than I stated earlier.
[tex]\overrightarrow{v}M=(M+dM)(\overrightarrow{v}+d\overrightarrow{v})\overrightarrow{u}dM[/tex] 


#9
Nov2710, 01:02 AM

P: 2

http://www.sciforums.com/showthread.php?t=74512
Go through this site 


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