3 ms·

> The moon is not so far away in terms of distance but it is very far away in terms of Δv because, not least, you have to land propulsively because there is no

by kqr 2y ago

> The moon is not so far away in terms of distance but it is very far away in terms of Δv because, not least, you have to land propulsively because there is no atmosphere to slow you down. Not least, but certainly the requirement to brake before you land must be on the small order compared to achieving escape velocity from the much bigger rock I'm on?

PaulHoule 2y ago
You gotta get off Earth no matter where you go in space. It's almost free to come home from LEO, you get a huge amount of free velocity change returning from the moon. (At the cost of rejecting the heat) In the rocket equation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation the required mass ratio is an exponential function of the velocity change so adding another 2.5 km/sec for this and another 2.5 km/sec for that you are making the mission much more difficult. It's bad enough that it takes two stages to get to LEO comfortably but going beyond that adds cost and complexity pretty quick, for instance the large number of Starship launches required to get a Moon mission into the right orbit. I like to think about what interstellar travellers would do if they wanted to land on the Earth on the assumption that they are accustomed to life in deep space and have spent 1,000 to 10,000 years "living off the land" off comets and rouge planets and are used to a lifestyle like cutting up a planet like Pluto and building a number of small ringworlds powered by D-D fusion. I'd conjecture that despite having advanced technology they would still find the "reverse space shuttle" problem where you land with a full load of fuel and then take off from the ground to be difficult. It's not like they are going to haul a space shuttle along with them and would probably find it non-trivial to 3-d print one from plans that old. My take is that it would probably take them a decade to figure it out and that they might well come up with an alternative answer like https://en.wikipedia.org/wiki/Skyhook_(structure) https://en.wikipedia.org/wiki/Skyhook_(structure) which depends on in-space infrastructure that they'd be experience with although it could work together with an air-breathing aircraft which would be something new for them.
ryandrake 2y ago
Everything is on the small order of magnitude when compared with getting into Earth orbit. As the quote goes, "Once you get to earth orbit, you're halfway to anywhere in the solar system."
- mlyle 2y ago
  > As the quote goes, "Once you get to earth orbit, you're halfway to anywhere in the solar system." In terms of delta-v, yes. Going to the moon only takes 50% more delta-v than going to LEO. But the rocket equation makes getting that extra 50% a lot harder. Time for more staging and complexity and still having terrible payload fraction.
- PaulHoule 2y ago
  That might be Heinlein's most annoying quote ever and boy does it have competition. It is very expensive to change orbits. If you had two space stations like the ISS with ascending nodes 180 degrees from each other it would be about as expensive to transit between them as it is to launch a rocket from the Earth to begin with. See https://caseyhandmer.wordpress.com/2019/10/02/there-are-no-gas-stations-in-space/ https://caseyhandmer.wordpress.com/2019/10/02/there-are-no-g... You've got the advantage in space that you can use electric rockets with a high specific impulse. Back in the 1950s folks like von Braun imagined that manned space flight might use electric rockets but after they discovered the Van Allen belts they discovered this is much too slow to make it through the magnetosphere.
  - no-s 2y ago
    > If you had two space stations like the ISS with ascending nodes 180 degrees from each other it would be about as expensive to transit between them as it is to launch a rocket from the Earth to begin with. 1+1 = 2 2/2 = 1; technically halfway. To interpret “halfway to anywhere” that way is missing the point. Going from LEO to LEO wasn’t the point of “halfway to anywhere”. Yes a bit exuberant but not far off.
alexey-salmin 2y ago
> Not least, but certainly the requirement to brake before you land must be on the small order compared to achieving escape velocity from the much bigger rock I'm on? The problem is all the fuel you use to break before landing also has to achieve earth escape velocity at first. And it makes the original problem much harder because the total mass that needs acceleration grow exponentially with delta speed.

by kqr 2y ago

PaulHoule 2y ago

You gotta get off Earth no matter where you go in space. It's almost free to come home from LEO, you get a huge amount of free velocity change returning from the moon. (At the cost of rejecting the heat) In the rocket equation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation the required mass ratio is an exponential function of the velocity change so adding another 2.5 km/sec for this and another 2.5 km/sec for that you are making the mission much more difficult. It's bad enough that it takes two stages to get to LEO comfortably but going beyond that adds cost and complexity pretty quick, for instance the large number of Starship launches required to get a Moon mission into the right orbit. I like to think about what interstellar travellers would do if they wanted to land on the Earth on the assumption that they are accustomed to life in deep space and have spent 1,000 to 10,000 years "living off the land" off comets and rouge planets and are used to a lifestyle like cutting up a planet like Pluto and building a number of small ringworlds powered by D-D fusion. I'd conjecture that despite having advanced technology they would still find the "reverse space shuttle" problem where you land with a full load of fuel and then take off from the ground to be difficult. It's not like they are going to haul a space shuttle along with them and would probably find it non-trivial to 3-d print one from plans that old. My take is that it would probably take them a decade to figure it out and that they might well come up with an alternative answer like https://en.wikipedia.org/wiki/Skyhook_(structure) https://en.wikipedia.org/wiki/Skyhook_(structure) which depends on in-space infrastructure that they'd be experience with although it could work together with an air-breathing aircraft which would be something new for them.

ryandrake 2y ago

Everything is on the small order of magnitude when compared with getting into Earth orbit. As the quote goes, "Once you get to earth orbit, you're halfway to anywhere in the solar system."

mlyle 2y ago
> As the quote goes, "Once you get to earth orbit, you're halfway to anywhere in the solar system." In terms of delta-v, yes. Going to the moon only takes 50% more delta-v than going to LEO. But the rocket equation makes getting that extra 50% a lot harder. Time for more staging and complexity and still having terrible payload fraction.
PaulHoule 2y ago
That might be Heinlein's most annoying quote ever and boy does it have competition. It is very expensive to change orbits. If you had two space stations like the ISS with ascending nodes 180 degrees from each other it would be about as expensive to transit between them as it is to launch a rocket from the Earth to begin with. See https://caseyhandmer.wordpress.com/2019/10/02/there-are-no-gas-stations-in-space/ https://caseyhandmer.wordpress.com/2019/10/02/there-are-no-g... You've got the advantage in space that you can use electric rockets with a high specific impulse. Back in the 1950s folks like von Braun imagined that manned space flight might use electric rockets but after they discovered the Van Allen belts they discovered this is much too slow to make it through the magnetosphere.
- no-s 2y ago
  > If you had two space stations like the ISS with ascending nodes 180 degrees from each other it would be about as expensive to transit between them as it is to launch a rocket from the Earth to begin with. 1+1 = 2 2/2 = 1; technically halfway. To interpret “halfway to anywhere” that way is missing the point. Going from LEO to LEO wasn’t the point of “halfway to anywhere”. Yes a bit exuberant but not far off.

alexey-salmin 2y ago

> Not least, but certainly the requirement to brake before you land must be on the small order compared to achieving escape velocity from the much bigger rock I'm on? The problem is all the fuel you use to break before landing also has to achieve earth escape velocity at first. And it makes the original problem much harder because the total mass that needs acceleration grow exponentially with delta speed.