Here is how I'd try to calculate at which distance to begin a braking burn for a rendesvouz using the distance, BurnTime and rocket equations (Vector intercepts the target).
1. Distance
[math]x[/math]=distance, [math]\Delta V[/math]=velocity relative to the target, [math]T_{burn}[/math]=BurnTime, [math]a[/math]= acceleration.
The distance is [math] x= \Delta V\cdot\ T_{burn} - {\frac{a \cdot\ T_{burn}^{2}}{2}}[/math] and since [math] a= \frac{\Delta V}{T_{burn}}[/math] we get: [math] x= \Delta V\cdot\ T_{burn} - {\frac{\frac{\Delta V}{T_{burn}} \cdot\ T_{burn}^{2}}{2}}= \frac{1}{2}\Delta V\cdot\ T_{burn}[/math]
It's simplification ( [math] x = \frac{1}{2}\Delta V\cdot\ T_{burn}[/math] ) .
To find the exact solution we need to solve few differential equations:
(1) [math]\frac{d \vec x(t)}{dt} = \vec v(t)[/math]
Reactive motion:
(2) [math]m(t)\frac{d \vec v(t)}{dt}-\vec u_e\frac{dm(t)}{dt}=0 \Rightarrow[/math]
(3) [math]\vec v(t)=- \vec u_e ln(\frac{m_0}{m(t)})+\vec v_0[/math]
Mass :
(4) [math] m(t) = m_0 - a t [/math]
a - fuel mass flow rate:
(5) [math] a = - \frac{\vec F_{thrust}}{\vec u_e} [/math]
Now we can get x(t) solving (1). I used Maple (math software ):
(6) [math] \vec x(t) = \vec u_e (t-\frac{m_0}{a}) ln(1-\frac{a t}{m_0}) - \vec u_e t+\vec v_0 t+\vec x_0 [/math]
Using (3) and (6) we can solve %subj% in more accurate way, finding where [math]v(t)=v_{target}[/math]:
(7) [math] t_{target} = \frac{m_0}{a} (1-e^{\frac{v_{target} - v_0}{u_e}})[/math]
and calculating [math]x(t_{target})[/math]
Lets's compare solutions. Situation: DG, empty mass 11000 kg, fuel 12900 kg, Isp 40000, Thrust 2*160000N, Initial velocity 10000 m/s, we need zero velocity:
[math] m_0=23900 kg[/math]
[math] x_0=0 m[/math]
[math] v_{target}=0 m/s[/math]
[math] v_0=10000 m/s[/math]
[math] u_e=40000 m/s [/math]
Pay attention, [math]\vec u_e[/math] - vector. If it's value in 1D has same sign with [math]\vec v_0[/math] then the ship are losing velocity.
[math] a=2*160000/40000=8 kg/s [/math]
[math] t_{target} = \frac{m_0}{a} (1-e^{\frac{v_{target} - v_0}{u_e}}) = 660.832660 s [/math]
[math] x(t_{target})= u_e (t_{target}-\frac{m_0}{a}) ln(1-\frac{a t_{target}}{m_0}) - u_e t_{target}+ v_0 t_{target}+ x_0 = 3441693.58 m [/math]
Using [math] x = \frac{1}{2}\Delta V\cdot\ T_{burn}[/math] simplification:
[math]\Delta V = 10000 m/s[/math]
[math]m_0=23900 kg[/math]
[math]u_e = 40000 m/s[/math]
[math]F_{thrust} = 2*160000 = 320000 N[/math]
[math]T_{burn}=\frac{{m_0} \cdot\ (1-e^{-\frac{\Delta V}{u_e}})\cdot u_e}{F_{thrust}} = 660.832660 s [/math] (correct)
[math]x= \frac{1}{2}\Delta V\cdot\ \frac{{m_0} \cdot\ (1-e^{-\frac{\Delta V}{u_e}})\cdot u_e}{F_{thrust}} = 3304163.30 m[/math] (wrong!!!)
Visual:
Green area is [math] x = \frac{1}{2}\Delta V\cdot\ T_{burn}[/math] approximation. Red - difference between approximation and exact value.
wned:
