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Forum » SpaceEngine » Science and Astronomy Discussions » Government Human Spaceflight Thread (Anything related to manned spaceflight by governments)
Government Human Spaceflight Thread
HarbingerDawnDate: Tuesday, 09.04.2013, 22:35 | Message # 61
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Quote (neutronium76)
Correct me if I am wrong but wouldn't this be more risky and at a similar cost to just sending an Apollo-style mission to the moon?

No. This would be a lot cheaper than a Moon landing program. Also, not risky.

Quote (neutronium76)
I mean with all the research that has been done and is being done, wouldn't be wiser to carefully plan a mission for a permanent base on the moon with a 3-5 human crue doing research and preparing for the next step to Mars?

Obviously NASA doesn't think so, and I tend to agree. A lunar outpost is not a necessary prerequisite for a Mars program.

Quote (neutronium76)
If something goes wrong with that asteroid it may crash to earth, enter wrong orbit etc..

It can never hit Earth since they would never tow it into an Earth-crossing trajectory at any point when moving it. It would never be able to hit any body other than the Moon. And even if it did hit Earth it would be harmless. A 7-meter carbonaceous asteroid would cause no damage.

Quote (neutronium76)
Also I don't know if its orbit can remain stable around the moon which orbits the earth.

It only needs to be there for a matter of years, and in high lunar orbit it would definitely be stable for longer than that. It would probably crash to the Moon eventually, but not before we were done with it.





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WatsisnameDate: Tuesday, 09.04.2013, 22:47 | Message # 62
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Yeah, the asteroid would only be a few meters in size, so the worst thing that could happen would be it turning into a pretty meteor in the Earth's atmosphere.

Quote (Disasterpiece)
I'm not sure how long the asteroid would be stable, though.


This is a concern of mine as well; any stable orbit of the Moon also lies within its Roche limit. A rubble pile asteroid would be broken apart by the tidal forces if we tried to do this, but a dense and rigid asteroid might be okay.

Added (10.04.2013, 01:47)
---------------------------------------------
As for the stability of the orbit, the region of stable circular equatorial lunar orbits goes up to about 1200km. This region is smaller for inclined orbits, though there are also some special orbits that are higher and stable.




 
HarbingerDawnDate: Tuesday, 09.04.2013, 22:54 | Message # 63
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Here guys, knock yourselves out
http://www.kiss.caltech.edu/study/asteroid/asteroid_final_report.pdf





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DisasterpieceDate: Tuesday, 09.04.2013, 23:06 | Message # 64
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Quote (HarbingerDawn)
Here guys, knock yourselves out

Thank you! I was going to sit down with a pencil and paper and do the math myself at this point.

Also, if we moved an asteroid into the roche limit could we give the moon temporary rings.





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Edited by Disasterpiece - Tuesday, 09.04.2013, 23:41
 
NovaSiliskoDate: Wednesday, 10.04.2013, 00:14 | Message # 65
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Such an interesting possibility, for the first time that we know of, there would be a (mostly) natural moon-of-a-moon.
 
HarbingerDawnDate: Wednesday, 10.04.2013, 05:58 | Message # 66
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Also, if we moved an asteroid into the roche limit could we give the moon temporary rings.

Guys, guys! A 7-meter asteroid is not going to form rings. It's too small for the tidal forces to disrupt it. It would almost certainly stay intact all the way until impacting the surface. Keep in mind, this thing is about the same size as the Apollo spacecraft that were sent to the Moon. Even the rings of Saturn contain many intact pieces that are larger than this.





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WatsisnameDate: Wednesday, 10.04.2013, 08:24 | Message # 67
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Tidal disruption depends not on size but density, and whether or not the object is held together by forces other than gravity.

A rubble-pile asteroid of any size would be disrupted if it were brought into a close orbit around the moon, but an asteroid only a few meters across is certainly not going to be a rubble pile.







Edited by Watsisname - Wednesday, 10.04.2013, 08:25
 
HarbingerDawnDate: Wednesday, 10.04.2013, 10:00 | Message # 68
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Quote (Watsisname)
but an asteroid only a few meters across is certainly not going to be a rubble pile.

That was largely what I meant. Also, you aren't going to notice such extreme tidal effects with a 7 meter object as you would with a 70 km object. The difference in gravity across 70 km is so much more than across 7 m. And it just seems like common sense to me that most objects have forces other than gravity holding them together to some degree, especially small objects, so I figured that didn't need to be specified.





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NovaSiliskoDate: Wednesday, 10.04.2013, 10:02 | Message # 69
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Also, not to mention a 7-meter asteroid wouldn't really enough material to produce notable rings.
 
HarbingerDawnDate: Wednesday, 10.04.2013, 10:20 | Message # 70
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Quote (NovaSilisko)
Also, not to mention a 7-meter asteroid wouldn't really enough material to produce notable rings.

Indeed.

If we assume that the object is spherical (it won't be), perfectly solid (it won't be) and forms a ring at an altitude of 100 km (it won't), and if it completely and uniformly broke in to grains 1 cubic millimeter in size (lolno), and assuming the ring was a disk 5 meters wide, then you would have about 600 grains per square meter in that ring, and that assumes that they stay in that perfect, concentrated state. Unless you came to within a few meters of the ring it would probably be invisible.

And I probably messed up those calculations somewhere so it very well could be wrong





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Edited by HarbingerDawn - Wednesday, 10.04.2013, 10:21
 
WatsisnameDate: Wednesday, 10.04.2013, 11:24 | Message # 71
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Why not assume the ring starts out the same width as the asteroid? smile But yeah, your numbers look good. The ring formed by disrupting an asteroid of this size in lunar orbit would be pretty insubstantial. I calculated it in terms of surface mass density rather than particle density, and it's only a few grams per square meter.




 
HarbingerDawnDate: Wednesday, 10.04.2013, 11:30 | Message # 72
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Quote (Watsisname)
Why not assume the ring starts out the same width as the asteroid?

I thought about that, but figured that for the sake of generosity I would make it a tad smaller, since all the other assumptions were already absurdly generous smile





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HarbingerDawnDate: Wednesday, 10.04.2013, 19:13 | Message # 73
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NASA has just released an animation depicting the mission. Warning, it's sans audio, so put some music on with it smile



There's a longer version of the video with sound on NASA's website, but the quality is much lower
http://www.nasa.gov/mission....ve.html





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Edited by HarbingerDawn - Wednesday, 10.04.2013, 19:39
 
NovaSiliskoDate: Wednesday, 10.04.2013, 19:13 | Message # 74
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The President's Fiscal Year 2014 budget ensures the United States will remain the world's leader in space exploration and scientific discovery for years to come, while making critical advances in aerospace and aeronautics to benefit the American people.


Uh-huh...

http://www.planetary.org/blogs/bill-nye/live-nasa-budget-webcast.html
 
DisasterpieceDate: Thursday, 11.04.2013, 02:00 | Message # 75
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Quote (HarbingerDawn)
Guys, guys! A 7-meter asteroid is not going to form rings. It's too small for the tidal forces to disrupt it. It would almost certainly stay intact all the way until impacting the surface. Keep in mind, this thing is about the same size as the Apollo spacecraft that were sent to the Moon. Even the rings of Saturn contain many intact pieces that are larger than this.

I know, I was think we could use solar power and hall effect thrusters on an M-type asteroid (10-15 km) get it into the roche limit and then make rings. It may sound like a hideous waste of money and resources, but hey, they made a second jurassic park.





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Edited by Disasterpiece - Thursday, 11.04.2013, 02:01
 
Forum » SpaceEngine » Science and Astronomy Discussions » Government Human Spaceflight Thread (Anything related to manned spaceflight by governments)
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