4 ms·

Agreed. In all likelihood, the seven parts would collide a bit, heat up a lot, and generally lump back together into a new moon. On that scale, the moon is real

Agreed. In all likelihood, the seven parts would collide a bit, heat up a lot, and generally lump back together into a new moon. On that scale, the moon is really soft. So much so that each of the seven individual lumps would probably flow into a roughly spherical shape before colliding with anything.

Florin_Andrei 9y ago
Right. What most people don't realize is that at planetary scale (or even at Moon scale) things are fluid. There are no solids at that scale. Gravity dominates all other interactions except gas pressure - and that only for things made mostly of gas. That being said, I would love to see an accurate, 3D, high resolution simulation of the scenario described in Seveneves.
- Symmetry 9y ago
  It's not exactly what you wanted but this is pretty close. If I'd run into it earlier I'd have put it in the blog post. https://www.youtube.com/watch?v=0XDZR_gENo8 https://www.youtube.com/watch?v=0XDZR_gENo8
raattgift 9y ago
Indeed, and it would be interesting to take a fuller numerical simulation that takes https://en.wikipedia.org/wiki/Hydrostatic_equilibrium https://en.wikipedia.org/wiki/Hydrostatic_equilibrium into account. Absent enough input energy to push one or more of the fragments onto a new geodesic, to zeroth-order the moon would very quickly recollapse into a hot spherical droplet (likely by way of seven increasingly droplet-like components), and it would take an enormous amount of time for surface features -- like seas and so forth -- and lunar mascons[1] to appear. There's some brief discussion of this sort of thing at https://gizmodo.com/what-if-the-earth-suddenly-turned-flat-1795819464 https://gizmodo.com/what-if-the-earth-suddenly-turned-flat-1... which while it's Gizmodo, is one of the better "What If?"s. Dave Stevenson (Caltech)'s answer is short and sweet. The undistorted body in the Earth-moon pair under either scenario is distant enough that the squashed Earth or the fragmented moon would still source a close approximation of the Schwarzschild metric[2], with an interface between the two as part of the boundary condition of each. Usually for simplicity we'd just describe the motion of the lighter body in the geodesics of the Schwarzschild solution of the heavier body, and the whole moon, its seven fragments, and the fragments' successor arrangements are very likely amenable to that. One could always kick part of the mass of the moon such that it escapes "to infinity" from the Earth-moon system. Unfortunately one would have to recalculate the stress-energy tensor around the residual moon (and around the Earth, and around the sun) and then work out the geodesics. This would make for an entertaining project in numerical relativity. - -- [1] https://www.nasa.gov/mission_pages/grail/news/grail20121205.html https://www.nasa.gov/mission_pages/grail/news/grail20121205.... from https://www.nasa.gov/mission_pages/grail/main/index.html https://www.nasa.gov/mission_pages/grail/main/index.html [2] The Schwarzschild solution is after all an exterior solution, and is fine for approximately spherically symmetric bodies with approximately zero angular momentum. Moreover, if we don't care about ~ microsecond accuracies (and we almost certainly don't, because the conditions at lunar fragmentation are not that finely specified), we can do everything with small corrections to Newtonian gravity (cf. sec 6.4 of http://onlinelibrary.wiley.com/doi/10.1002/9783527634569.ch6/summary http://onlinelibrary.wiley.com/doi/10.1002/9783527634569.ch6... for the post-Newtonian Kepler solution and ch 8 on relativistic geodesy).
- keksicus 9y ago
  Susan Hough in that Gizmodo article gave me autism. She says a sphere has maximum surface area for unit volume. No honey, it's minimum! You'd think she just misspoke, but give her a bit more rope and she hangs herself not long after by saying turning a sphere into a disc would decrease the ratio of surface area to volume. Again, no sweetie, it would increase! Mansplaining is such hard work!
SJetKaran 9y ago
Because the new body will heat up a lot due to collisions, how bright do you think will this new body be, for people on Earth?
- Terr_ 9y ago
  Wouldn't that, in turn, depend on how much energy was put into the system by the original blast/splitting?
PDoyle 9y ago
Yep. In fact, this is how it formed in the first place.

Florin_Andrei 9y ago

Right. What most people don't realize is that at planetary scale (or even at Moon scale) things are fluid. There are no solids at that scale. Gravity dominates all other interactions except gas pressure - and that only for things made mostly of gas. That being said, I would love to see an accurate, 3D, high resolution simulation of the scenario described in Seveneves.

Symmetry 9y ago
It's not exactly what you wanted but this is pretty close. If I'd run into it earlier I'd have put it in the blog post. https://www.youtube.com/watch?v=0XDZR_gENo8 https://www.youtube.com/watch?v=0XDZR_gENo8

raattgift 9y ago

Indeed, and it would be interesting to take a fuller numerical simulation that takes https://en.wikipedia.org/wiki/Hydrostatic_equilibrium https://en.wikipedia.org/wiki/Hydrostatic_equilibrium into account. Absent enough input energy to push one or more of the fragments onto a new geodesic, to zeroth-order the moon would very quickly recollapse into a hot spherical droplet (likely by way of seven increasingly droplet-like components), and it would take an enormous amount of time for surface features -- like seas and so forth -- and lunar mascons[1] to appear. There's some brief discussion of this sort of thing at https://gizmodo.com/what-if-the-earth-suddenly-turned-flat-1795819464 https://gizmodo.com/what-if-the-earth-suddenly-turned-flat-1... which while it's Gizmodo, is one of the better "What If?"s. Dave Stevenson (Caltech)'s answer is short and sweet. The undistorted body in the Earth-moon pair under either scenario is distant enough that the squashed Earth or the fragmented moon would still source a close approximation of the Schwarzschild metric[2], with an interface between the two as part of the boundary condition of each. Usually for simplicity we'd just describe the motion of the lighter body in the geodesics of the Schwarzschild solution of the heavier body, and the whole moon, its seven fragments, and the fragments' successor arrangements are very likely amenable to that. One could always kick part of the mass of the moon such that it escapes "to infinity" from the Earth-moon system. Unfortunately one would have to recalculate the stress-energy tensor around the residual moon (and around the Earth, and around the sun) and then work out the geodesics. This would make for an entertaining project in numerical relativity. - -- [1] https://www.nasa.gov/mission_pages/grail/news/grail20121205.html https://www.nasa.gov/mission_pages/grail/news/grail20121205.... from https://www.nasa.gov/mission_pages/grail/main/index.html https://www.nasa.gov/mission_pages/grail/main/index.html [2] The Schwarzschild solution is after all an exterior solution, and is fine for approximately spherically symmetric bodies with approximately zero angular momentum. Moreover, if we don't care about ~ microsecond accuracies (and we almost certainly don't, because the conditions at lunar fragmentation are not that finely specified), we can do everything with small corrections to Newtonian gravity (cf. sec 6.4 of http://onlinelibrary.wiley.com/doi/10.1002/9783527634569.ch6/summary http://onlinelibrary.wiley.com/doi/10.1002/9783527634569.ch6... for the post-Newtonian Kepler solution and ch 8 on relativistic geodesy).

keksicus 9y ago
Susan Hough in that Gizmodo article gave me autism. She says a sphere has maximum surface area for unit volume. No honey, it's minimum! You'd think she just misspoke, but give her a bit more rope and she hangs herself not long after by saying turning a sphere into a disc would decrease the ratio of surface area to volume. Again, no sweetie, it would increase! Mansplaining is such hard work!

SJetKaran 9y ago

Because the new body will heat up a lot due to collisions, how bright do you think will this new body be, for people on Earth?

Terr_ 9y ago
Wouldn't that, in turn, depend on how much energy was put into the system by the original blast/splitting?

PDoyle 9y ago

Yep. In fact, this is how it formed in the first place.