Did NASA use something more efficient than a Hohmann Transfer?

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Summary:

Hohmann Transfer to the Moon?
It's common knowledge that it takes about 3 days to get to the moon. With a Hohmann transfer, I get a transit time of 5 days, not 3. I see NASA used something called "trans-lunar injection". Is this distinct from a Hohmann transfer, and more time efficient? What makes this trajectory different? More burns?

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Filip Larsen
Gold Member
For a maneuver that in some situation may be more fuel efficient than Hohmann transfer you may want to take a look at Bi-elliptic transfer orbits.

Regarding the TLI "designation", I remember it as just the maneuver needed to insert the vehicle from Earth parking orbit into a "free return" orbit around the Moon.

etotheipi
Thanks. Are Bi-elliptic transfer orbits quicker than Hohmann transfers as well? Are they more akin to what the apollo missions used?

Filip Larsen
Gold Member
No, the bi-elliptic transfer are slower than Hohmann transfers and its "just" a general maneuver.

For Apollo, the orbits were (as far as I know) based on the concept of free-return trajectory and as such not "just" simple Hohmann.

I see. Free return trajectories are faster then? Since they have enough energy to return a craft with minimal/no burns? I’m just trying to account for the discrepancy in time between the Hohmann transfer time of 5 days and the actual 3 day time.

A.T.
With a Hohmann transfer, I get a transit time of 5 days, ...

berkeman
$$t = \pi \sqrt{\frac{(r_1+r_2)^3}{8GM}}$$

$$r_1 + r_2 = 4e8 m$$
$$M = 6e24 kg$$

$$t=4.9 days$$

Filip Larsen
Gold Member
The trajectory for a Hohmann transfer will only match well in situations that are approximately two-body. For initial interplanetary mission designs one can for instance use Hohmann transfer "between" planets where the primary mass is the Sun. The start and end of the Hohmann can then be patched into an escape and capture orbit using method of patched conics [1].

The Earth-Moon system, however, is so coupled that to my knowledge even patched conics are a poor approximation. I would think you find slightly better approximation by patching an Earth escape trajectory directly into a Moon capture trajectory, but even then the later part will still be influenced a lot by Earth gravity.

The only "reliable" way to calculate trajectories in the Earth-Moon system is by using numerical integration that includes enough effects to achive desired precision, e.g. just Earth and Moon in circular orbit for rough estimate, full elliptic Earth and Moon orbit and with Sun for better estimate. Bate et. al [2] has a chapter on calculating lunar orbits as it was done at the time Apollo missions was designed. Here it is mentioned that patched conics can be used as a first approximation for Moon capture orbits, but the method is only good for the capture part and not a good approximation for the free-return orbit actually used. They use a sphere of influence for the Moon (the distance from Moon center where the Earth escape trajectory is patched to the Moon arrival trajectory) of around 66300 km.

[2] "Fundamentals of Astrodynamics", Bate et. al., Dover, 1971.

berkeman
A.T.
$$t = \pi \sqrt{\frac{(r_1+r_2)^3}{8GM}}$$

$$r_1 + r_2 = 4e8 m$$
$$M = 6e24 kg$$

$$t=4.9 days$$
Keep in mind, that they didn't want to get into the same orbit as the Moon, but into an orbit around the Moon. Also the formula above neglects the gravity of the Moon itself.

sophiecentaur, DuckAmuck and etotheipi
I remember those flights and the trajectory they took. You may remember (or note) an interesting trajectory.

Draw the Earth and the Moon and the line through their centers (the centerline). The trajectories they took crossed the centerline once on the way to the moon and once on the way back making a figure 8.

A.T.