# Deriving the Rocket Equation

1. May 12, 2015

### MattRob

Hello,
A little something I've been working on for fun today was trying to derive the rocket equation:
$Δv=V_{e}ln(R)$

$F=ma$

$=m\frac{dv}{dt}$

$\frac{F}{m}dt = dv$

$∫_{0}^{Tco} \frac{F}{m}dt = ∫_{0}^{v} dv$

Where Tco is the time of engine cutoff, and v is the velocity at that point in time. ie; what will the velocity be at the end of this thrusting period?
Now, since mass is time-dependent, we write mass as a function of time:

$m(t) = m_{i} - αt$

Where α is the burn rate of the propellant in kg/s.

$∫_{0}^{Tco} \frac{F}{m_{i} - αt}dt = ∫_{0}^{v} dv$

Now here's the part where I think something is going wrong:

$∫_{0}^{Tco} \frac{F}{m_{i} - αt}dt = ∫_{0}^{v} dv$

$F∫_{0}^{Tco} \frac{1}{m_{i} - αt}dt = v = F(ln(m_{i}-αTco) - ln(m_{i}))$

Now with $m_{i} - αTco = m_{f}$ and $\frac{m_{f}}{m_{i}} = R$ ,

$v = F ln(\frac{m_{i} - αTco}{m_{i}}) = F ln(R)$

It all works excellently for the natural logarithm term, but not for the $V_{e}$ term. I was thinking I could write
$F = m\frac{dv}{dt}$ and integrate $∫_{0}^{Tco} \frac{m\frac{dv}{dt}}{m_{i} - αt}dt$,
but then when it comes time to integrate I can't get rid of the mass term. Same problem whether or not I cancel the dt's before integrating (honestly, I don't really understand much on how to handle the dx differentials - I'm only up to linear algebra and integral calculus, no multivariable or diff Eq. yet).

It seems like I should be able to leave the F term alone, though, since, as with α, it's constant over the duration of the engine burn. But I guess not?

So what's going on?

Last edited: May 12, 2015
2. May 12, 2015

### nasu

The force on the rocket depends on the rate of ejection of gases. And the ejection speed as well.
You did not seem to use this.

3. May 12, 2015

### DrStupid

That works for systems with constant mass only. As the mass of the rocket and of the ejected reaction mass changes, you need to use F=dp/dt or conservation of momentum instead.

4. May 12, 2015

### AlephNumbers

I would start from the beginning with conservation of momentum. Consider the instant before the exhaust products are ejected, and, a small increment of time dt later, an instant after the exhaust products are ejected. After you get an equation for that, consider the velocities of the rocket relative an inertial reference frame, the exhaust products relative to the frame, and the rocket relative to the exhaust products. You can then make some substitutions. Then, with a little trickery, you should be able to relate the rate of fuel consumption and the velocity of the rocket relative to the products to the mass and acceleration of the rocket.

5. May 13, 2015

### BobG

I think responders are on the right track - especially DrStupid.

Force is equal to (exhaust velocity/g)*(Fuel rate * g) and it is based on conservation of momentum. You throw a certain amount of stuff out the back at a certain velocity, it has a certain momentum. The rocket has a change in momentum equal in magnitude, but in the opposite direction of the stuff you threw out the back. Obviously, force is slightly misleading since, in space, there will be no equal and opposite force - there's just conservation of momentum. The units work out to be newtons or pounds, though.

* The g's in the equation are acceleration due to gravity at the surface of the Earth and are there because this is an old equation created using pounds force and pounds mass (slugs weren't invented until sometime around 1910). They're a byproduct of unit conversion; not there for some physical reason. They can be extremely misleading, especially when using this equation (or variations of this equation) on the Moon or other planets. People think that they have to calculate acceleration due to gravity for the celestial object they're sitting on instead of using g (the constant) no matter where they are.

Last edited: May 13, 2015