Deriving the rocket equation

[Rusty6899]

I wasn't really sure where to post this (it's not a "gameplay" question so to speak), but I wanted to have a go at deriving the Tsiolkovsky rocket equation. The problem is, that I don't get the right answer.

My assumption is that the delta-V is the integral of the acceleration of the craft wrt time, so;

a = F/m

m = m0 - m_dot*t

a = F / (m0 - m_dot*t)

int(a)dt = (F / m_dot) * ln(m0 - m_dot*t)

F / m_dot = g0 * Isp

m_dot*t = m0-m1

so, I get

dV = g0*Isp*ln(m1)

instead of

dV = g0*Isp*ln(m0/m1)

I'd appreciate if anyone has any idea where I have gone wrong.

[foamyesque]

Do the integral w.r.t. mass, not time.

dV = dPc/Mc = dMe * Ve/Mc

integral dMe/Mc*Ve between m0 and m1 = Ve * integral dMe/Mc between m0 and m1 = [Ve * ln(m1)] - [Ve * ln(m0)] = Ve * ln(m1/m0) = deltaV.

[AbacusWizard]

I think your problem is that in the step where you integrate, you're treating it as an indefinite integral (and leaving out the +c); it would make more sense as a definite integral, with the limits of integration being initial and final time, leading to ∆v on the left side (from v_f - v_i ) and ln(m_f/m_i) on the right side (because subtracting leads to ln(m_f) - ln(m_i), and those can be combined into one logarithm with division).

- - - Updated - - -

Another elegant way to work it out is to start with the definition of I_sp and treat it as a separable differential equation.

- F / (dm/dt) = g * I_sp

(We need the negative sign because the rocket's dm/dt is negative--in other words, we're *losing* mass as we burn fuel--but we want I_sp to be a positive number.)

Multiply both sides by dm:

- F dt = g * I_sp dm

Rewrite F as m*a:

- m * a dt = g * I_sp dm

Divide both sides by -m, and rewrite a as dv/dt:

dv/dt dt = - g * I_sp / m dm

Cancel out the dt's, and integrate both sides:

∫ dv = - g * I_sp ∫ (1/m) dm

Evaluate the integrals:

v = - g * I_sp ln(m)

Plug in the limits of integration (v_f, v_i, m_f, m_i), and distribute the negative on the right side:

v_f - v_i = - g * I_sp * [ ln(m_f) - ln(m_i) ] = g * I_sp * [ ln(m_i) - ln(m_f) ]

Use the laws of logarithms to simplify the right side, and the definition of "delta" for the left side:

∆v = g * I_sp * ln( m_i / m_f ),

or ∆v = g * I_sp * ln( m_full / m_empty )

I think that ought to work.

[OhioBob]

It is derived here: http://www.braeunig.us/space/propuls.htm#impulse

[Vanamonde]

Moved to Science Labs, where space sciencey stuff is examined.

[K^2]

Abacus is right. You have to do a definite integral from t0 to t1, which gives you the ln(m0/m1) in the answer.

[N_las]

There are many different ways.

But going your way:

a = F/m

m = m0 - m_dot*t

(this only works because m_dot is a constant value)

a = F / (m0 - m_dot*t)

integral [from t=0 to t_now] a d(t) = (-F / m_dot) * ln(m0 - m_dot*t_now) - (-F / m_dot) * ln(m0 - m_dot*0)

=F/m_dot * ln(m0/(m0-m_dot*t_now))

F / m_dot = g0 * Isp

m0 - m_dot*t = m_now

dV = g0*Isp*ln(m0/m_now)

[Rusty6899]

Thanks to everyone who has provided an answer.

I have to say, it's been a fair while since I did integral calculus, and I don't think I was ever brilliant at it.

I think the main thing that cost me though was failing to consider that "ln(a) - ln( = ln(a/b)

I probaby would have twigged if I had thought to apply that reasoning.

Thanks again