For reference, a 10^40 liter sphere has a radius of 10 au.


closed as off-topic by RonJohn, Logan R. Kearsley, JBH, dot_Sp0T, Mołot Apr 23 '18 at 7:14

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    $\begingroup$ This might be better served in the Physics SE or Astronomy SE. Also, 10 AU is an absurdly huge number for a radius. For scale, the radius of the sun itself is roughly 1/215th of an AU. So it's not really "pouring it onto the sun" and more of "the sun being engulfed by an impossibly big sphere of water" $\endgroup$ – Sydney Sleeper Apr 23 '18 at 3:09
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    $\begingroup$ Depends on what poring means. Most likely. Boom... supernova. That ball is orders of magnatude heavier than the sun. $\endgroup$ – Jaden Travnik Apr 23 '18 at 3:12
  • $\begingroup$ Ask xkcd. I'm sure Randall Munroe has done a What If on something similar. $\endgroup$ – RonJohn Apr 23 '18 at 3:15
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    $\begingroup$ I'm voting to close this question as off-topic because it's a physics question answered by what-if.xkcd.com/14 $\endgroup$ – RonJohn Apr 23 '18 at 3:16
  • $\begingroup$ For one thing, the surface of the sun is something like 6000 kelvin. I think the pressure at the bottom of 10 au of water might be something like 10^11 atmospheres. What phase would water be in under those conditions? $\endgroup$ – thebuttsicles Apr 23 '18 at 3:21

A sphere of $10^{40}$ L of water at standard density has a mass of $10^{40}$ kg. The sun has a mass of only $2\times 10^{30}$ kg and the entire Milky Way galaxy has a mass of about $10^{12}$ solar masses.

So that water sphere represents about half a percent of the mass of the entire galaxy.

For reference, a 10^40 liter sphere has a radius of 10 au.

No it does not.

It would in fact collapse into a Black Hole. This is absolutely impossible to avoid, but it would also most likely produce a huge super nova as well in the process of collapse.

And long before you gathered even a small fraction of that $10^{40}$ L it would have enough mass to collapse into a star (with an Oxygen-Hydrogen fusion core) and as you added a bit more water it would eventually be large enough to collapse into a black hole. And at that point you'd still have almost all of the $10^{40}$ L left, because the amount of water required to form a black hole is only $4\times 10^{30}$ kg, twice the Sun's mass and one billionth of your total water supply.

Now in practice you'd need way more water to make a star with an active core collapse into a Black hole ( as opposed to just becoming a bigger star ). However there is a limit and it's something like 150 solar masses ( still way smaller than your sphere ).

But one way or another you get a black hole.

A black hole of mass $10^{40}$ kg has a radius of about $93$ AU. And yes, that is larger than a sphere of water of standard density of the same mass.

  • $\begingroup$ I'm curious as to just what would happen if you dumped a reasonably large amount of water on a star. Though of course the 10 AU sphere is not possible, say it's a mass comparable to what a close binary might steal from its neighbor. There's a fast oxygen-fusion reaction in old stars, but that happens when the hydrogen in the core is exhausted. Would H2O do something different? $\endgroup$ – jamesqf Apr 23 '18 at 5:10
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    $\begingroup$ @jamesqf under those temperatures and pressures, H2O would undergo fusion just as any other element or compound. The sun is not a wooden ball on fire. $\endgroup$ – B.fox Apr 23 '18 at 9:18
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    $\begingroup$ I was going to say the same. Any question that starts with an astronomically-large (pun intended) amount of stuff always has the same answer - a black hole. $\endgroup$ – Matt Bowyer Apr 23 '18 at 9:19
  • $\begingroup$ @B.fox: To start with, H2O wouldn't exist on a star like the sun, because it would disassociate into (highly ionized) H and O atoms. Then H and O undergo different fusion processes. Then too, fusion occurs only towards the core of the star, not at the surface. So you have a large mass of infalling H20, disassociating as it heats to extreme temperatures due to gravitational potential energy... Do you get fusion ignition just from that? $\endgroup$ – jamesqf Apr 24 '18 at 3:21

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