Timeline for How big a nuke would be needed to break Phobos out of orbit?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7, 2019 at 2:19 | comment | added | uhoh | I'm still pondering this and fyi I've just asked When is it okay to use energy arguments in orbital mechanics? | |
| Dec 6, 2019 at 14:56 | comment | added | Steve Linton | @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. | |
| Dec 6, 2019 at 14:56 | comment | added | Starfish Prime | @uhoh you've got a point there; the energy required to effect a change in velocity depends on your initial velocity, after all... | |
| Dec 6, 2019 at 14:54 | comment | added | uhoh | @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. | |
| Dec 6, 2019 at 14:49 | comment | added | Pelinore | @StarfishPrime : Ah now that's where you're wrong, of course you can, the problem is it so rarely returns the favour ;-) | |
| Dec 6, 2019 at 14:48 | comment | added | Starfish Prime | @Pelinore and that is why your idea has been foiled; you can't always ignore gravity. | |
| Dec 6, 2019 at 14:48 | comment | added | Steve Linton | @uhoh you're missing the kinetic energy of its current orbit. | |
| Dec 6, 2019 at 14:47 | comment | added | Pelinore | @StarfishPrime : if they lie it flat for me first sure. | |
| Dec 6, 2019 at 14:46 | comment | added | uhoh | 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. | |
| Dec 6, 2019 at 14:46 | comment | added | Starfish Prime | @Pelinore and everest is less than 9km high... you could be up and back in time for tea and medals. | |
| Dec 6, 2019 at 14:44 | comment | added | Pelinore | @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 ;) | |
| Dec 6, 2019 at 14:41 | comment | added | Starfish Prime | @Pelinore little rock? I feel like you need to recalibrate your sense of size ;-) | |
| Dec 6, 2019 at 14:35 | comment | added | Pelinore | 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! | |
| Dec 6, 2019 at 14:31 | comment | added | Starfish Prime | 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 ;-) | |
| Dec 6, 2019 at 14:27 | history | answered | Steve Linton | CC BY-SA 4.0 |