Timeline for Can you catch a spaceship with a train on the Moon?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 25, 2018 at 20:17 | vote | accept | Mike Nichols | ||
| Oct 29, 2018 at 16:46 | history | edited | user535733 | CC BY-SA 4.0 |
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| Oct 28, 2018 at 5:39 | comment | added | John Ray | @Kyle I remember seeing a system to transport cargo between two moon bases about 40 years ago in a kids book. Cargo pods were accelerated through a series of magnets (like square Quidditch hoops) which became larger and spaced farther apart as the cargo accelerated. The final magnet made sure the cargo had the proper speed/trajectory to travel to a duplicate set of magnets half way around the moon where the cargo was caught and decelerated. Unfortunately I don't remember the name of the book. | |
| Oct 27, 2018 at 23:10 | comment | added | Kyle | @KJO Where do you get that inbound velocity? I don't understand why you're multiplying it by 10. | |
| Oct 27, 2018 at 14:07 | history | edited | user535733 | CC BY-SA 4.0 |
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| Oct 27, 2018 at 12:47 | comment | added | RonJohn | @Kyle and the spacecraft would have to be made of steel instead of aluminum... | |
| Oct 27, 2018 at 11:19 | comment | added | dotancohen | @Kyle: Perhaps that braking plate could be connected to a generator, and the excess current generated could be stored in a battery. Thus, the incoming ships supply both material and energy to the moonbase. | |
| Oct 27, 2018 at 10:19 | history | edited | user535733 | CC BY-SA 4.0 |
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| Oct 27, 2018 at 8:31 | comment | added | Mołot | , their combined momentum will try to pull the train into orbit. — false/imprecise. If velocities are matched then train is anyway on orbital velocity and there is some mechanism preventing it from flying away. No need for combined momentum. Mass has nothing to do with it. | |
| Oct 27, 2018 at 7:51 | comment | added | Kyle | Of course you'd also generate an absurd amount of heat in the braking plate in the process since you're effectively dumping all your kinetic energy into heating that plate. It would definitely melt. | |
| Oct 27, 2018 at 7:46 | comment | added | Kyle | Almost makes it seem like it'd be easier to forego the train entirely and just have the spacecraft get close enough for some magnets to slow it down. I did a back of the envelope calculation assuming something the mass of the ISS and an acceleration of 3g and you'd only need a magnetic field of about 3.7 mT passing through a 1 m^2 braking plate. Large, but not unreasonable. The hard part would be accurately cruising down between the braking magnets and then how to safely touch the ground. At 3g the spacecraft would come to a stop in 81 seconds after covering 96 km. | |
| Oct 27, 2018 at 2:37 | history | edited | user535733 | CC BY-SA 4.0 |
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| Oct 27, 2018 at 0:17 | history | edited | user535733 | CC BY-SA 4.0 |
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| Oct 26, 2018 at 23:54 | history | answered | user535733 | CC BY-SA 4.0 |