Integration of Tsiolkovsky rocket equation


by LogicalTime
Tags: integration, integration property, rocket equation
LogicalTime
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#1
May24-11, 12:25 PM
P: 115
In the Tsiolkovsky rocket equation derivation there is a part that says:

[tex]\frac{dV}{dt} = -\upsilon_e \frac{1}{m} \frac{dm}{dt}[/tex]
"Assuming v_e, is constant, this may be integrated to yield:"

[tex]\Delta V\ = v_e \ln \frac {m_0} {m_1} [/tex]

How does this work? The differential is an operator and I am pretty sure you just can't cancel the dt. I wonder what assumptions are needed to be able to legally just remove the "dt"s though.

Thanks!
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Char. Limit
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#2
May24-11, 01:11 PM
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Quote Quote by LogicalTime View Post
In the Tsiolkovsky rocket equation derivation there is a part that says:

[tex]\frac{dV}{dt} = -\upsilon_e \frac{1}{m} \frac{dm}{dt}[/tex]
"Assuming v_e, is constant, this may be integrated to yield:"

[tex]\Delta V\ = v_e \ln \frac {m_0} {m_1} [/tex]

How does this work? The differential is an operator and I am pretty sure you just can't cancel the dt. I wonder what assumptions are needed to be able to legally just remove the "dt"s though.

Thanks!
Multiply both sides by dt, and then integrate both sides to get:

[tex]\int \frac{dV}{dt} dt = \int -\upsilon_e \frac{1}{m} \frac{dm}{dt} dt[/tex]

Now, the left side can be given the substitution u=V, du/dt= dV/dt or du = dV/dt dt. The right side can be given the substitution w=m, dw = dm/dt dt, thus turning your equation into...

[tex]\int du = \int -\upsilon_e \frac{1}{w} dw[/tex]

And both sides of those can obviously be integrated.
LogicalTime
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#3
May24-11, 01:35 PM
P: 115
Thank you for helping me to see the substitution. :-)

Mute
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#4
May24-11, 03:27 PM
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Integration of Tsiolkovsky rocket equation


You can also look at it this way:

[tex]\int_{t_0}^{t_f} dt~\frac{dV}{dt} = V(t_f) - V(t_0)[/tex]
by the fundamental theorem of calculus.

Similarly,

[tex]\int_{t_0}^{t_f}dt~\frac{1}{m}\frac{dm}{dt} = \int_{t_0}^{t_f}dt~\frac{d}{dt}\left(\ln m(t) \right) = \ln m(t_f) - \ln m(t_0) = \ln\left(\frac{m(t_f)}{m(t_0)}\right),[/tex]
again due to the fundamental theorem. We also recognized that [itex]d(\ln m(t))/dt = (1/m) dm/dt[/itex].
LogicalTime
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#5
May24-11, 04:43 PM
P: 115
Thanks! Another way to see it.

One further detail:
I think we have been leaving out the absolute value sign.
[tex] \int \frac{1}{w} dw = \ln |w| + c[/tex]

However wiki says this is true:

[tex] \frac{d}{dw} \ln(w) = \frac{1}{w} [/tex]

Assuming that is true then using the fundamental theorem of calc we would get an answer without the absolute value. I am not sure how to reconcile that. Do you guys understand this subtlety?
Mute
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#6
May24-11, 05:13 PM
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P: 1,391
We could have written the absolute value sign when we integrated, but m is presumably the mass, which is never negative, so we don't need it in this case. The absolute value sign appears in infinite integrals when it's not specified whether the variable is positive or negative, or in definite integrals when you know for sure the variable is negative (and don't feel like writing [itex]\ln (-x)[/itex] all the time).

If x is positive:

[tex]\frac{d}{dx}\ln x = \frac{1}{x}[/tex]

If x is negative, only ln(-x) makes sense, since the argument of the log must be positive. So:

[tex]\frac{d}{dx}\ln(-x) = \frac{1}{-x}(-1) = \frac{1}{x}[/tex]

Since the derivative of both ln(x) and ln(-x) gives 1/x, we can write this in compact form saying the derivative of ln|x| is 1/x.

(Note also that ln(ax) has a derivative of 1/x. So, if you differentiate ln(ax), to get it back when you integrate, you can write the arbitrary constant as ln(a), which gives back ln(ax)).
LogicalTime
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#7
May24-11, 07:36 PM
P: 115
Nice, that clears that up as well. Thank you!


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