So to put it in words, we must find the direction of thrust and the delta V required to take an orbit with an apoapsis reaching out to moon altitude, and turn it into an orbit with a much lower apoapsis, without ruining the periapsis. I hope I'm using the right language
Cheers
EDIT: I got a little bit confused with the angle conventions. The flight-path angles are measured from the local vertical, not the local horizontal as usual.
Yeah, that is about right.
I'll try to not make it too complicated, but I will need orbital mechanics and vectors for the explanation.
I have taken a quicksave of the Apollo 8 mission I flew with NASSP at about 12h GET (Ground Elapsed Time) into the flight and extracted the position and velocity vector relative to the Earth. Then I looked at the
Apollo Flight Journal and found the so called P37 Block Data. I decided to use the numbers for a return burn at 45h into the mission. That burn would have used a deltaV of 6208 ft/s.
Before I start with the maths, I'll talk about my notation a little bit. For the time indizes I will use:
0 = Time of my scenario (12h GET)
1 = Time of the direct return burn (45h GET)
2 = Time of reaching the Entry Interface (400,000 feet above the Earth)
The basic problem of calculating the return burn is only a few equations and can be almost solved analytically. What has to be done, is basically varying the flight path angle, until we get the desired reentry trajectory. The most important equation is:
[MATH]p = \frac{2 R_1 (\frac{R_1}{R_2}-1)}{(\frac{R_1}{R_2})^2(1+x(t_2)^2)-(1+x(t_1)^2)}[/MATH]
p = parameter of the orbit
R1 = Radius at time 1 (radius at the burn time)
R2 = Radius at time 2 (Entry Interface radius)
[MATH]x(t_1)=\cot(\gamma)[/MATH] Indepdent variable x. We have to selected the correct value, so that it meets our requirements, in this case the desired velocity change of 6208 ft/s.
[MATH]x(t_2)=\cot(\gamma_r)[/MATH]
gamma_r is the reentry angle. The AGC has a quartic function to calculate the angle depending on reentry velocity. A higher reentry velocity means a steeper reentry, so that the Crew Module doesn't skip out of the atmosphere. For simplicity I will just use an angle of -6.5° [from the local horizontal].
Now to the vector maths:
[MATH]\boldsymbol{V}_2(t_1)=\frac{\sqrt{\mu_E}}{R_1} \sqrt{p}(x(t_1) \boldsymbol{u}_{R1}+\boldsymbol{u}_H)[/MATH]
V2 = the velocity vector at the time 1 we have to achieve, to reach the direct return trajectory.
mu_E = the standard gravitational parameter of the Earth
U_R1 = the unit vector of R1
U_H = The the local horizontal unit vector
The deltaV vector is then
[MATH]\Delta \boldsymbol{V} = \boldsymbol{V}_2(t_1)-\boldsymbol{V}_1(t_1)[/MATH]
I have plotted the independent variable x(t1) against the required delta V:
This is almost a linear function and it can be easily seen, that we get the desired velocity change at about x= -3.2, which corresponds to a flight path angle of about -17° [EDIT: Actually -73° measured from the local horizontal].
With that I have plotted the trajectories in the Earth-Moon-System:
The blue line is the trajectory before the direct return burn and the green line is the trajectory after the burn. The red line is the trajectory of the moon.
Before the burn the orbit had a periapsis radius of 5111 km and a apoapsis radius of 556,230 km. After the burn the periapsis becomes 6408 km and the apoapsis 379,550 km.
I have made many simplifications with the equations. The AGC implemention also accounts for:
-reentry angle based on reentry velocity
-Earth is an ellipsoid, not a sphere with a constant radius and gravity field
-gravitational influence of the moon and sun
To incorporate this stuff, the AGC has to do an iterative process, where the values for x are limited, et cetera, et cetera.
As you can see, you can make this infinitely complicated :lol:
