Rearranging the tsiolkovsky rocket equation.

[Cripion101]

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[boogabooga]

Ill give you a hint,

the exponential function is the inverse of the natural log.

 

[Cripion101]

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[boogabooga]

No, I think it will be very, very beneficial for you to learn some algebra. :yes:

Give it a try and we will tell you if you are correct. First, rearrange terms so that the natural log term is on one side of the equals sign.

 

[Artlav]

In simpler terms "the exponential function is the inverse of the natural log" means that if ln(x)=y then e^y=x.

From there on, it's fifth-grade algebra.

 

[boogabooga]

Haha. Ninth grade in the U.S.

 

[Cripion101]

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[Enjo]

ln (natural logaritm), is a common name of log_base_e

 

[Urwumpe]

General:

x=log_a --> y = a**x

Special:

x=ld --> y = 2**x
x=lb --> y = 2**x
x=ln --> y = e**x
x=lg --> y = 10**x
x=li --> y = i**x

 

[Hlynkacg]

Haha. Ninth grade in the U.S.

Err 6th when I was in school.

 

[PeriapsisPrograde]

I'll walk you through it.

deltav = g • Isp • ln(m1/m2).

[math]\Delta v = g * I_{sp} * ln(\frac{m_1}{m_2})[/math]

The tricky part of this is the ln part. I divide by g and ISP to get ln by itself.

[math]\frac{\Delta v}{g*I_{sp}} = ln(\frac{m_1}{m_2})[/math]

Here's the trick to getting rid of the ln: raise everything to the e power.

[math]e^{\frac{\Delta v}{g*I_{sp}}} = \frac{m_1}{m_2}[/math]

From there it is indeed simple algebra to solve for m_2

[math]\frac{e^{\frac{\Delta v}{g*I_{sp}}}}{m_1} = \frac{1}{m_2}[/math]

[math]\frac{m_1}{e^{\frac{\Delta v}{g*I_{sp}}}} = m_2[/math]

And there it is solved for m_2

 

[SolarLiner]

[math]\frac{m_1}{e^{\frac{\Delta v}{g*I_{sp}}}} = m_2[/math]

And there it is solved for m_2

I'm glad I *still* have my maths before school returns ! I got to the same conclusion ^^
When entered "coding style", it should be: m1/(e^(dV/(g*isp))) = m2.

Tried to be a little more useful than a simple "I'm glad I did it" answer :tiphat: