It has to do with the intersection of the available orbits. Apollo launched from florida, not the equator. So Apollo was limited to certain orbits around earth. Extend these out to intercept the moon and there are only so many moon orbits available without burning excessive amounts of fuel.
The only way to visualize it easily is with a physical model. It is all about the angle of the orbits (planes) around the two spheres. You want to minimize 'plane change maneuvers' as they cost a huge amount of fuel (ie they require lots of Delta-V). Think of the earth orbit (the parking orbit before going to the moon) as a big piece of paper around the earth. The available moon orbits are where that piece of paper also touches the sphere of the moon. Now rotate the paper around the earth (the limited number of orbits from florida, between 25 and 90). As the paper rotates around the earth it also rotates around the moon. Some parts of the moon, as some parts of the earth, are not available.
Then add to this the sunlight issues. Apollo wanted to land in very specific light conditions. They wanted nice easy shadows (45) to increase situational awareness during decent and landing. The pilots practiced with the sun being at a particular angle, using surveillance photos taken with the sun at a similar angle. If the sun was too low/high then they might think a rock is bigger/smaller because of its shadow.
Unlike the Earth, the Moon rotates very slowly - once every 27 days. It's equator is not fast-moving - it is moving at only 4 m/s. This delta-v gain is nearly negligible.
The free-return trajectory, where you keep the command module's orbit in the same plane as the Moon's orbit around the Earth is more important (Since both the Moon, and the command module orbits at ~1,000 m/s - which is the delta-v that you gain, if you eject from lunar orbit, in the opposite direction to the Moon's direction of travel.)
If you mean the free return trajectory for the Earth-Moon trip, that was done only up through Apollo 11. Apollo 12 and after did not (although Apollo 13 was put back onto one after the oxygen tank explosion).
I believe that it is about getting from the lander to the orbital vehicle.
If you go into an equatorial orbit, then every orbit of the orbital vehicle gives you a chance to get the lander back.
There is also a stable polar orbit where the orbiter stays lined up between the Earth and the Moon. Since the surface is tidally locked you again get the lander back easily. This orbit takes more energy to get into and out of for the orbital vehicle. But gives you mid latitudes with the Earth high in the sky.
What is hard is getting to mid latitudes on the side of the Moon. Now what happens is that the object on the surface turns away from the orbital vehicle making it hard to line up everything for the return trip. The lander therefore has to be much more capable to land and return at will, rather than waiting weeks for things to line up right.
> There is also a stable polar orbit where the orbiter stays lined up between the Earth and the Moon. Since the surface is tidally locked you again get the lander back easily.
Wait, what keeps the polar orbit lined up with the Earth? If the orbit just stayed constant, it would deviate from being lined up from the Earth, right?
Source: I have never been to the Moon, but I have played a lot of Kerbal Space Program.
The problem was not landing. The problem was getting back into space, to rendezvous with the command module.
The reason this is difficult is because orbital plane changes (changes in your orbit's inclination) are expensive, and because astronauts can't sit on the lunar surface for weeks at a time.
Launching to rendezvous with a space space station (Or the command module) is, if you want to avoid very expensive plane changes, requires waiting for it to be a few minutes from passing directly overhead. Then, you launch into the same orbital plane as the space station, and, in the process of accelerating to orbital speeds, the space station catches up to you, and you dock to it.
If the command module is in an equatorial orbit, and you land on the equator, then rendezvous is as easy as taking off from the lunar surface, a few minutes before it is scheduled to pass overhead. This happens every ~2 hours.
If the comand module is in a polar orbit, and you land on the poles, then the same thing happens. The orbiter will be overhead every ~2 hours, so, again, rendezvous is very easy.
But if you land in the mid-latitudes, you suddenly have a problem.
The moon rotates. And it rotates slowly.
Imagine if the command module is in a polar orbit, and you land in the mid-latitudes. You spend two days driving golf carts on the moon. You want to get back to space, to meet up with the command module.
Except that the moon has rotated in this time. But the polar orbit hasn't. Instead of passing overhead, the command module is now one-tenth of the way around the moon from you. You can't launch into its orbital plane anymore.
You now have two options.
1. You can wait for a total of 13.5 days, when due to the moon's rotation, you will, again, be within the command module's orbital plane.
2. You can launch into an orbital plane that is ~30 degrees inclined to that of the command module. And then spend a tonne and a half of rocket fuel, to change your orbital inclination.
Both of these options, for obvious reasons, suck.
Now, option #1 gets a bit easier, if the command module is in a non-polar orbit - the launch window, instead of being every 13.5 days, may be every few days, or once a day.
However, if you need to, for whatever reason, perform an emergency launch, or if you miss the launch window, you may be screwed.
If the Apollo mission did not involve a command module (If the lunar lander was the craft that would come back to Earth), then all of this becomes irrelevant. You could then land at whatever latitude, and return to Earth, via a wide range of possible trajectories. (Except that, as another poster pointed out, you miss out on a part of, or all of the free-return trajectory boost, that you get from exiting the Moon's orbit in the opposite direction of its travel - which is ~1,000 m/s of delta-v saved.)
> If the Apollo mission did not involve a command module (If the lunar lander was the craft that would come back to Earth), then all of this becomes irrelevant. You could then land at whatever latitude, and return to Earth, via a wide range of possible trajectories. (Except that, as another poster pointed out, you miss out on a part of, or all of the free-return trajectory boost, that you get from exiting the Moon's orbit in the opposite direction of its travel - which is ~1,000 m/s of delta-v saved.)
Yup. Another downside would be that the fuel needed to break lunar orbit and head back to Earth, enter Earth orbit, and then deorbit for Earth landing would have to be landed on the moon and lifted off from it again, requiring exponentially more fuel for lunar landing and liftoff.
Do you have any idea what orbit the linked story was talking about and what the launch timing of that would be?
It's astonishing to me that you can orbit a point in space between two planetary bodies. There's nothing on Earth that would give you that intuitive understanding.
I'm curious if over the next hundreds of years if the Lagrange points will become contentious regions of space, perhaps the subject of territorial disputes. It's perhaps cynical to project humanity's pettiness into the future but if the L* points are as valuable for lunar access as the article says, it's not unthinkable.
I'm picturing one of the swimming pools shaped like Ultima-Thule where a roller skater can circle around a highpoint in the middle. The high point itself is unstable (like L1 L2 L3) but can be orbited with minor corrections.
Yes, if coming from outside the 2-body system. But you aren't really getting an assist from the L point. You are being accelerated by the pair of bodies that create the point. Such an assist is very very slight, not worth the effort in comparison to using either of the bodies directly.
Unless you are an interstellar traveler looking for an assist from a pair of binary stars that you really do not want to get too close to. That is scifi territory atm.
Imagine you have two planetary bodies fixed in space. A free mass is attracted to both, but it's possible to find an unstable equilibrium somewhere between them. Put the mass here, and it'll stay still-- but it's like balancing a ball on the top of a sphere; it'll roll off towards one of the two planets unless you stabilize it.
Now imagine displacing the mass from the equilibrium, but in a direction perpendicular to the line between the planets-- it'll swing back and forth like one of those slingshot rides. There are two dimensions perpendicular to the line between the planets, so two dimensions with stable oscillations, therefore you can get circular orbits.
I guess theoretically a Lagrange point is a single point in space, but in practice, how large are they? Assuming a station with the mass of the ISS, how far from the ideal Lagrange point can you be while still maintaining a reasonably stable orbit?
We have twenty years of experience with space stations but no experience with an extra-terrestrial base. Yes, Antarctica or Biosphere come close in some ways, and yes there's still plenty to learn from space habitation outside our magnetosphere, but to me that pales in comparison to the skills we'd learn establishing an actual human outpost on another celestial body!
Cool orbit. The criticisms [1] of the project are pretty telling though. The most glaring of which IMO is why one would consider putting unshielded humans in a tin can rather than under a bunch of rock on the ground.
Probably a need for Solar power throughout the month. A lunar base would need a nuclear generator for 15 days a month or one heck of a battery system. Also probably easier to maintain a structure in microG.
A low earth orbit, like the current space station, has roughly half the sky blocked out by earth plus the magnetic field of earth to somewhat protect it. Getting to the lunar gateway's weird orbit, if you were already on the moon, could take several days versus less delta-v to just go to low earth orbit which can be done at any time.
> Sure, it’s rocket science, but so is playing Kerbal Space Program.
Best sentence in the article :D
On a serious note, I remember a university friend telling me about how they had been working on figuring out this orbit. Interesting to see it written up here now.
34 comments
[ 2.3 ms ] story [ 82.9 ms ] threadThe only way to visualize it easily is with a physical model. It is all about the angle of the orbits (planes) around the two spheres. You want to minimize 'plane change maneuvers' as they cost a huge amount of fuel (ie they require lots of Delta-V). Think of the earth orbit (the parking orbit before going to the moon) as a big piece of paper around the earth. The available moon orbits are where that piece of paper also touches the sphere of the moon. Now rotate the paper around the earth (the limited number of orbits from florida, between 25 and 90). As the paper rotates around the earth it also rotates around the moon. Some parts of the moon, as some parts of the earth, are not available.
Then add to this the sunlight issues. Apollo wanted to land in very specific light conditions. They wanted nice easy shadows (45) to increase situational awareness during decent and landing. The pilots practiced with the sun being at a particular angle, using surveillance photos taken with the sun at a similar angle. If the sun was too low/high then they might think a rock is bigger/smaller because of its shadow.
The free-return trajectory, where you keep the command module's orbit in the same plane as the Moon's orbit around the Earth is more important (Since both the Moon, and the command module orbits at ~1,000 m/s - which is the delta-v that you gain, if you eject from lunar orbit, in the opposite direction to the Moon's direction of travel.)
If you mean the free return trajectory for the Earth-Moon trip, that was done only up through Apollo 11. Apollo 12 and after did not (although Apollo 13 was put back onto one after the oxygen tank explosion).
If you go into an equatorial orbit, then every orbit of the orbital vehicle gives you a chance to get the lander back.
There is also a stable polar orbit where the orbiter stays lined up between the Earth and the Moon. Since the surface is tidally locked you again get the lander back easily. This orbit takes more energy to get into and out of for the orbital vehicle. But gives you mid latitudes with the Earth high in the sky.
What is hard is getting to mid latitudes on the side of the Moon. Now what happens is that the object on the surface turns away from the orbital vehicle making it hard to line up everything for the return trip. The lander therefore has to be much more capable to land and return at will, rather than waiting weeks for things to line up right.
Wait, what keeps the polar orbit lined up with the Earth? If the orbit just stayed constant, it would deviate from being lined up from the Earth, right?
https://en.wikipedia.org/wiki/Sun-synchronous_orbit
That is why in the article they offer a map of where the landing could have been and came up with https://hackadaycom.files.wordpress.com/2019/02/moonpossible...
You see the landings on the equator, and also the strip going over the pole.
https://space.stackexchange.com/questions/5312/why-would-sun...
The problem was not landing. The problem was getting back into space, to rendezvous with the command module.
The reason this is difficult is because orbital plane changes (changes in your orbit's inclination) are expensive, and because astronauts can't sit on the lunar surface for weeks at a time.
Launching to rendezvous with a space space station (Or the command module) is, if you want to avoid very expensive plane changes, requires waiting for it to be a few minutes from passing directly overhead. Then, you launch into the same orbital plane as the space station, and, in the process of accelerating to orbital speeds, the space station catches up to you, and you dock to it.
If the command module is in an equatorial orbit, and you land on the equator, then rendezvous is as easy as taking off from the lunar surface, a few minutes before it is scheduled to pass overhead. This happens every ~2 hours.
If the comand module is in a polar orbit, and you land on the poles, then the same thing happens. The orbiter will be overhead every ~2 hours, so, again, rendezvous is very easy.
But if you land in the mid-latitudes, you suddenly have a problem.
The moon rotates. And it rotates slowly.
Imagine if the command module is in a polar orbit, and you land in the mid-latitudes. You spend two days driving golf carts on the moon. You want to get back to space, to meet up with the command module.
Except that the moon has rotated in this time. But the polar orbit hasn't. Instead of passing overhead, the command module is now one-tenth of the way around the moon from you. You can't launch into its orbital plane anymore.
You now have two options.
1. You can wait for a total of 13.5 days, when due to the moon's rotation, you will, again, be within the command module's orbital plane.
2. You can launch into an orbital plane that is ~30 degrees inclined to that of the command module. And then spend a tonne and a half of rocket fuel, to change your orbital inclination.
Both of these options, for obvious reasons, suck.
Now, option #1 gets a bit easier, if the command module is in a non-polar orbit - the launch window, instead of being every 13.5 days, may be every few days, or once a day.
However, if you need to, for whatever reason, perform an emergency launch, or if you miss the launch window, you may be screwed.
If the Apollo mission did not involve a command module (If the lunar lander was the craft that would come back to Earth), then all of this becomes irrelevant. You could then land at whatever latitude, and return to Earth, via a wide range of possible trajectories. (Except that, as another poster pointed out, you miss out on a part of, or all of the free-return trajectory boost, that you get from exiting the Moon's orbit in the opposite direction of its travel - which is ~1,000 m/s of delta-v saved.)
Yup. Another downside would be that the fuel needed to break lunar orbit and head back to Earth, enter Earth orbit, and then deorbit for Earth landing would have to be landed on the moon and lifted off from it again, requiring exponentially more fuel for lunar landing and liftoff.
Do you have any idea what orbit the linked story was talking about and what the launch timing of that would be?
I'm curious if over the next hundreds of years if the Lagrange points will become contentious regions of space, perhaps the subject of territorial disputes. It's perhaps cynical to project humanity's pettiness into the future but if the L* points are as valuable for lunar access as the article says, it's not unthinkable.
On a side note, can somebody much smarted then me tell me if it's possible to "sling shot" (Gravity assist?) Off a Lagrange point?
Unless you are an interstellar traveler looking for an assist from a pair of binary stars that you really do not want to get too close to. That is scifi territory atm.
Imagine you have two planetary bodies fixed in space. A free mass is attracted to both, but it's possible to find an unstable equilibrium somewhere between them. Put the mass here, and it'll stay still-- but it's like balancing a ball on the top of a sphere; it'll roll off towards one of the two planets unless you stabilize it.
Now imagine displacing the mass from the equilibrium, but in a direction perpendicular to the line between the planets-- it'll swing back and forth like one of those slingshot rides. There are two dimensions perpendicular to the line between the planets, so two dimensions with stable oscillations, therefore you can get circular orbits.
We have twenty years of experience with space stations but no experience with an extra-terrestrial base. Yes, Antarctica or Biosphere come close in some ways, and yes there's still plenty to learn from space habitation outside our magnetosphere, but to me that pales in comparison to the skills we'd learn establishing an actual human outpost on another celestial body!
1. https://en.wikipedia.org/wiki/Lunar_Orbital_Platform-Gateway...
Best sentence in the article :D
On a serious note, I remember a university friend telling me about how they had been working on figuring out this orbit. Interesting to see it written up here now.