# Suborbital docking - Apollo Landings

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1. Jan 20, 2016

### sevenperforce

Suppose that one of the Apollo landings had developed a fuel leak in the lunar ascent module, such that even with a lightened load, the ascent module only had enough delta-v to reach about half of lunar orbital velocity.

One possible rescue solution would be to lower the command module's orbit much nearer to the lunar surface, then execute a retrograde inclined burn, placing it into a high-arching suborbital trajectory. At the same time, the ascent module would take off and aim for that same suborbital trajectory, so that the two craft would still be able to dock, albeit with a very limited window due to the free-fall condition. The command module would then re-ignite its engines and head for Earth.

Assuming the command module had enough extra fuel for such a maneuver, what might the optimal launch trajectory have been? Should the command module have descended to a very low parking orbit to minimize gravity drag on the ascent module, or remained at a higher orbit to give more free-fall time? A lower orbit is faster, which means a bigger gap has to be closed between the velocities of the two modules, but a higher orbit is going to add uncertainty and suck more delta-v away from the ascent module due to gravity drag.

2. Jan 20, 2016

### Staff: Mentor

The actual command module didn't have enough fuel, but as this is a hypothetical question: I don't have calculations backing it up, but I would expect docking close to the surface to be the most efficient way. Lower periapsis of the command module (CM), then accelerate ascent module (AM) as fast as possible while slowing the CM as fast as possible. Dock quickly, go to CM, throw away AM, accelerate again as fast as possible. This both minimizes gravity drag and maximizes the Oberth effect.

With more fuel, you could try to meet as far out as possible, but with just half the orbital velocity you don't get far away.

3. Jan 20, 2016

### sevenperforce

Would circularizing the CM's orbit improve the scenario (by reducing the orbital speed) or not? If as low an orbit as possible is ideal, what's the lowest you can get without running into problems and/or mountains?

I know the moon has pretty significant mass concentration anomalies, which thankfully have all been mapped. If a mission was being planned from the start with this kind of escape/ascent profile, would there be any ideal landing zones for using gravitational anomalies to help boost/compensate during the burns, or would you basically just want to stay as far away from mascons as possible?

Half of orbital velocity is about 841 m/s of delta-v, which will get you to an altitude of about 250 km, less gravity drag losses. Launching straight up from the lunar surface with a speed of 841 m/s would give you twenty minutes of "hang time".

If the ascent module had more than half the necessary dV but still not quite enough to make orbit, would this change the optimal ascent profile?

Last edited: Jan 20, 2016
4. Jan 20, 2016

### Nidum

I remember this being discussed at the time .

I believe that if the LEM could get up into any stable orbit not too far off the scheduled orbit then there was a reasonable chance that the command module could manouvre enough to align for docking .

Anything less than a stable orbit for the LEM and all was lost .

In any case think about the different time scales involved

(a) LEM in any stable orbit and there's time to plan and execute a recovery strategy .
(b) LEM not achieving stable orbit is coming down again - fast . No time to do anything .

.

5. Jan 20, 2016

### sevenperforce

Well, if they knew exactly what remaining fuel they had, a suborbital trajectory with half the required orbital dV is still 20 minutes of "hang time". Should be enough for suborbital docking.

6. Jan 20, 2016

### Staff: Mentor

The 20 minutes are the total time in the parabolic flight. Assuming you don't want to crash right after docking, the whole approach and docking procedure should be done in at most a few minutes. Compare that to the typical 30+ minutes a docking procedure needs, and you get a problem.

The whole problem is scale-invariant, so let's simplify numbers: the radius of moon is 1, the escape velocity is sqrt(2), the lowest orbital velocity is 1. That gives an orbital period of 2 pi.
To scale back: a distance of 1 is 1740 km, a speed of 1 is 1680 m/s, and 2 pi time is 1.8 hours.
Assume that acceleration and docking is instant, if necessary. The moon is also perfectly symmetrical. AM Fuel is given in terms of delta_v capability, so 1 fuel is sufficient to reach a low orbit from the ground. The CM has as much fuel as necessary.

Scenario 1: AM has 0.5 fuel, CM starts in a very low moon orbit.
With the maneuver described in post 2, CM needs 0.5 velocity change to slow down, capture the AM, and again 0.5 to reach a stable orbit. Sum: 1.
With the capture "at rest", the CM has to stop completely and accelerate again, for a delta_v budget of 2.

Scenario 2: AM has 0.5 as fuel, CM starts in a circular orbit at 0.143 above the surface, which is the maximal height AM can reach.
CM has a velocity of 0.935.
With the maneuver described in post 2, we have to slow to 0.904 to lower periapsis to the surface. We arrive there at a speed of 1.033. Slow down to 0.5, couple with AM, do the same things in reverse order - not optimal, but that way it is easier to compare scenarios. Total delta_v budget is 2*(0.029+0.533)=1.124.
With the capture "at rest", the CM has to stop completely and accelerate again, for a delta_v budget of 1.870.
If the final goal is to reach escape velocity, the first maneuver allows to directly accelerate from the lower orbit, which makes it even better.

7. Jan 20, 2016

### Nidum

The command module could come down from the scheduled orbit 69 miles above the lunar surface to a safe minimum of 10 miles . So if the LEM could reach an orbit more than 10 miles high a rendevouz and docking was possible .

The command module coming down from the high orbit to lower orbit took time and possibly several turns around the moon . So the LEM had to be in stable orbit or it would crash back to lunar surface before the command module could get to it .

This and much more is described in 'A man on the Moon' by Chaiken .

8. Jan 21, 2016

### sevenperforce

Would it have helped if the ascent module had started at a higher absolute altitude, like near the equator or on some prominence regions? The moon's equator is 2.1 km higher than its poles, and there are some prominence regions that exceed 7 km. The Apollo landings were all on the near side of the moon for obvious reasons, but if one had been on the far side, then it could take advantage of the highest point on the lunar surface, which is 10.8 km above the mean lunar radius.

If a lander took off from there, treating it as the apocynthion of an elliptical orbit, there could be significant dV savings.

9. Jan 21, 2016

### A.T.

Lack of direct radio contact with Earth (no live TV)?

10. Jan 21, 2016

### sevenperforce

Yes, that's the primary reason. Nowadays it's trivial (well, fairly trivial) to throw a comsat up and use that instead. In fact I think we have several up already.

11. Jan 21, 2016

### Staff: Mentor

Negligible. It is ~0.5% of the radius, so it saves ~0.5% of the energy for an escape, or 0.25% of the escape velocity. That is 12 m/s. Another even smaller contribution comes from reduced gravity drag.
On Earth the height matters due to the atmosphere. On moon: forget it.