- #1
MattRob
- 211
- 29
Hello,
A little something I've been working on for fun today was trying to derive the rocket equation:
[itex]Δv=V_{e}ln(R)[/itex]
So first I start with
[itex]F=ma[/itex]
[itex]=m\frac{dv}{dt}[/itex]
[itex]\frac{F}{m}dt = dv[/itex]
[itex]∫_{0}^{Tco} \frac{F}{m}dt = ∫_{0}^{v} dv[/itex]
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:
[itex]m(t) = m_{i} - αt[/itex]
Where α is the burn rate of the propellant in kg/s.
[itex]∫_{0}^{Tco} \frac{F}{m_{i} - αt}dt = ∫_{0}^{v} dv[/itex]
Now here's the part where I think something is going wrong:
[itex]∫_{0}^{Tco} \frac{F}{m_{i} - αt}dt = ∫_{0}^{v} dv[/itex]
[itex]F∫_{0}^{Tco} \frac{1}{m_{i} - αt}dt = v = F(ln(m_{i}-αTco) - ln(m_{i}))[/itex]
Now with [itex]m_{i} - αTco = m_{f}[/itex] and [itex]\frac{m_{f}}{m_{i}} = R[/itex] ,
[itex]v = F ln(\frac{m_{i} - αTco}{m_{i}}) = F ln(R)[/itex]
It all works excellently for the natural logarithm term, but not for the [itex]V_{e}[/itex] term. I was thinking I could write
[itex]F = m\frac{dv}{dt}[/itex] and integrate [itex]∫_{0}^{Tco} \frac{m\frac{dv}{dt}}{m_{i} - αt}dt[/itex],
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?
A little something I've been working on for fun today was trying to derive the rocket equation:
[itex]Δv=V_{e}ln(R)[/itex]
So first I start with
[itex]F=ma[/itex]
[itex]=m\frac{dv}{dt}[/itex]
[itex]\frac{F}{m}dt = dv[/itex]
[itex]∫_{0}^{Tco} \frac{F}{m}dt = ∫_{0}^{v} dv[/itex]
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:
[itex]m(t) = m_{i} - αt[/itex]
Where α is the burn rate of the propellant in kg/s.
[itex]∫_{0}^{Tco} \frac{F}{m_{i} - αt}dt = ∫_{0}^{v} dv[/itex]
Now here's the part where I think something is going wrong:
[itex]∫_{0}^{Tco} \frac{F}{m_{i} - αt}dt = ∫_{0}^{v} dv[/itex]
[itex]F∫_{0}^{Tco} \frac{1}{m_{i} - αt}dt = v = F(ln(m_{i}-αTco) - ln(m_{i}))[/itex]
Now with [itex]m_{i} - αTco = m_{f}[/itex] and [itex]\frac{m_{f}}{m_{i}} = R[/itex] ,
[itex]v = F ln(\frac{m_{i} - αTco}{m_{i}}) = F ln(R)[/itex]
It all works excellently for the natural logarithm term, but not for the [itex]V_{e}[/itex] term. I was thinking I could write
[itex]F = m\frac{dv}{dt}[/itex] and integrate [itex]∫_{0}^{Tco} \frac{m\frac{dv}{dt}}{m_{i} - αt}dt[/itex],
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?
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