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Phobos

  • Radius: 11.267 km
  • Orbital period: 8 hours
  • Equatorial rotation velocity: 11.0 km/h (6.8 mph) (at longest axis)

With an 8 hour orbit around Mars it must be going at a fair clip already which got me to wondering, why not use it as a generation ship, if we can kick it out of orbit we might even be able to do it in a way where we slingshot it around a couple of other planets on the way out .. that might give us a respectable speed .. & there's plenty of room inside for some really big sealed habitats.

So assuming it's optimally positioned on the surface for the purpose.

How big a nuke would we need to break it out of its orbit of Mars?

Presumably a large concave pit in the appropriate spot is needed to guide the blast for best effect.

Pelinore
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    Recommend closing due to the inappropriateness of the concept. I recommend asking for a calculation of the total energy needed to be applied along a vector to move a satellite of this mass. A "nuke" dumps energy in a rather uniform 4*pi steradians (excepting some relatively small "directed energy" items which will be useless against anything bigger than a very small asteroid), so most of the energy won't even change the kinetic energy Phobos has. – Carl Witthoft Dec 06 '19 at 14:14
  • This is a hellishly difficult question to ask, due to the range of different ways a chunk of rock might respond to a really energetic dose of x-rays. Also, if I'm not misunderstanding, as you lift it out of Mars' gravity well, it'll slow right down (because Kepler hates you) so it isn't immediately obvious that there's any benefit over phobos vs some other rock potatoid in the asteroid belt. – Starfish Prime Dec 06 '19 at 14:16
  • @CarlWitthoft you could always bury it, of course. Doesn't make modelling what happens next any easier, but there'll be less wastage (in exchange for more mess). – Starfish Prime Dec 06 '19 at 14:19
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    Although an object in a close circular orbit will be moving faster than one in a distant circular orbit around the same body, it still takes more energy to get it out of orbit and away to infinity, so Phobos is basically the worst possible choice for this. A main belt asteroid, a distant Moon of Neptune or a KBO like the one photographed by New Frontiers would all be better choices. – Steve Linton Dec 06 '19 at 14:19

2 Answers2

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Simply to lift get Phobos out of Mars orbit you would need to increase its orbital velocity by a factor of $\sqrt{2}$ (this is generally true for any object in circular orbit). Phobos orbits at about 2.1 km/s (Wikipedia) relative to Mars, so this is a delta-V of $2.1 \times (\sqrt{2} -1)$ which is about $0.9 km/s$. It's mass, same source is about $10^{16}kg$ so the energy needed is $0.5\times 10^{16}\times 900^2 J$ which is $4\times 10^{21}J$. So, assuming perfect efficiency and no wastage at all (which you will certainly not get with a nuke) you must release at least this much energy. A ton of TNT equivalent is just over $4GJ$ so this is just about $10^{12}$ tons TNT equivalent, or a million megatons.

Steve Linton
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  • There was me trying to find a comparison with the size of the Stickney impactor and showing that no practical nuke would be big enough. This seems like a much simpler explanation ;-) – Starfish Prime Dec 06 '19 at 14:31
  • Eeek! Hiroshima was 20 kilotons (0.02 megatons) so that's 500 Hiroshimas per megaton which means we need 500 million Hiroshimas to get the job done .. but it's such a little rock! – Pelinore Dec 06 '19 at 14:35
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    @Pelinore little rock? I feel like you need to recalibrate your sense of size ;-) – Starfish Prime Dec 06 '19 at 14:41
  • @StarfishPrime : 11 an'a quarter kilometers around, I could walk that in less than a day, so to me little, relative to other big rocks whizzing through space of course ;) – Pelinore Dec 06 '19 at 14:44
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    @Pelinore and everest is less than 9km high... you could be up and back in time for tea and medals. – Starfish Prime Dec 06 '19 at 14:46
  • Using $GM_{Mars} =$ 4.3E+13 m^3/s^2 and $m_{Phobos} =$ 1.1E+16 kg and $r = 9,376,000$ meters and use $U = -GMm/a$ to get potential energy, I get 4.9E+22 Joules, about a factor of 10 larger. – uhoh Dec 06 '19 at 14:46
  • @StarfishPrime : if they lie it flat for me first sure. – Pelinore Dec 06 '19 at 14:47
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    @uhoh you're missing the kinetic energy of its current orbit. – Steve Linton Dec 06 '19 at 14:48
  • @Pelinore and that is why your idea has been foiled; you can't always ignore gravity. – Starfish Prime Dec 06 '19 at 14:48
  • @StarfishPrime : Ah now that's where you're wrong, of course you can, the problem is it so rarely returns the favour ;-) – Pelinore Dec 06 '19 at 14:49
  • @SteveLinton but wouldn't that would be a factor of 2 type effect, not a factor of 12? I don't think that it's right to use $E = \frac{1}{2} m \Delta v^2$ like that, but it's late and I can't prove it right now. The vis-viva equation has kinetic and potential energies on equal footings, so I think they always have have the same order of magnitude, at least for circular orbits. – uhoh Dec 06 '19 at 14:54
  • @uhoh you've got a point there; the energy required to effect a change in velocity depends on your initial velocity, after all... – Starfish Prime Dec 06 '19 at 14:56
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    @uhoh fair point. The KE of Phobos on an escape trajectory as measured by an observer left behing in Phobos old circular orbit is the number I computed. The change in KE of Phobos as observed by an observer at rest at the centre of mass of Mars is half the number you calculated. I think to reconcile these we have to think about what happened to the reaction mass and how much energy each observer thinks it had to start with, and thinks it has at the end. – Steve Linton Dec 06 '19 at 14:56
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    I'm still pondering this and fyi I've just asked When is it okay to use energy arguments in orbital mechanics? – uhoh Dec 07 '19 at 02:19
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It can't be done. Unless it's really, really solid (and that's not likely!) trying to shove it that hard will break it up instead. You would need to use a bunch of bombs to move it more gently.

Loren Pechtel
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  • Yes I noticed that a few hours back, my attention is shifting to Deimos now, nothing I've found so far suggests that's an accretion of loose rocks too but I'm still looking? the questions still good for the orbital mechanics of moving that much mass from that orbit, but your right, a single blast big enough likely breaks Phobos up. – Pelinore Dec 08 '19 at 03:05
  • @Pelinore I would be very surprised if any moon of any planet anywhere can be shoved out of orbit with one boom. You move such things around with a bunch of smaller bombs, not with one big boom. – Loren Pechtel Dec 08 '19 at 03:16