Tsiolokovsky Equation

1. Dec 29, 2014

James Brady

Hello! I'm doing a little derivation of the Tsiolokovsky equation where I'm trying to find velocity with respect to time, here's what I got so far:

F=ma, a = F/m

Here I replace the force term and the mass term, taking into account that the rocket is losing mass:

$a = \frac{v_e\cdot \dot{x}}{m_0 - \dot{x}\cdot t}$

where:
v = exhaust velocity
m dot = mass flow rate
m naught = initial mass

After this I take the integral of acceleration to get velocity, it's a pretty easy one since the the top two terms, mass flow rate and exhaust velocity, are both constants:

$\int \frac{v_e\cdot\dot{x}}{m_0 - \dot{x}\cdot t} = -v_e\cdot ln(m_0- \dot{x}\cdot t) + c$

Which is the velocity. My problem with this though is that when t = 0, velocity is $-v_e \cdot ln(m_0)$ which doesn't make any sense, right from the start there is an instantaneous velocity? Maybe the constant of integration is suppose to fix that? Any help would be appreciated.

2. Dec 30, 2014

D.Biswas

if i am correct then , yes , the constant of integration is there to fix that. i haven't checked your calculations as they look fine. c is supposed to be (ve * ln(mo ). if you evaluate the definite integral of the function between t = 0 and t = ti (for any ti) then you'll see that the velocity function
v = -ve * ln(mo - m'ti) + ve*ln(mo)

3. Jan 2, 2015

James Brady

Ah OK, I was just beginning to suspect that I as I was writing that post. Thanks for the help.

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