4 ms·

I would like to upvote this 1000 times to underline "pretty sturdy box." It can be a particularly amusing exercise to do the math for things like this, calculat

I would like to upvote this 1000 times to underline "pretty sturdy box." It can be a particularly amusing exercise to do the math for things like this, calculate the gravitational forces involved, and compare the results to e.g. the strength of steel.

lanna 5y ago
The net gravitational force is zero: https://en.wikipedia.org/wiki/Shell_theorem https://en.wikipedia.org/wiki/Shell_theorem
- BoiledCabbage 5y ago
  I don't see how Shell theory applies here. The box wouldn't be inside the blackhole, it'd be outside it. If the box were a perfect sphere then yes it would have zero net gravitational pull on the black hole, but the black hole would still have an extremely strong pull on it. And that's still assuming it's perfectly centered which would be almost impossible to maintain.
  - lanna 5y ago
    > but the black hole would still have an extremely strong pull on it No, it would not. Newton's third law: if an object has zero net pull on the black hole, the black hole also has to have a zero net pull on the object. The shell theorem is more general than you are assuming: it doesn’t matter if the heavier object is inside or outside, the surrounding object doesn't need to be spherical (it can be any shape as long as it fully envelopes the black hole) and the black hole doesn't need to be centered. I know it is a counter-intuitive result, but it can be easily proven with calculus. See for instance this sentence about Dyson shells: "Such a shell would have no net gravitational interaction with its englobed star (see shell theorem)" https://en.wikipedia.org/wiki/Dyson_sphere#Dyson_shell https://en.wikipedia.org/wiki/Dyson_sphere#Dyson_shell Edit: Related, what is bigger? The gravitational force the Moon exerts on Earth or the gravitational force the Earth exerts on the Moon? Veritasium has a nice video explaining the answer: https://www.youtube.com/watch?v=8bTdMmNZm2M https://www.youtube.com/watch?v=8bTdMmNZm2M
    - aj3 5y ago
      You're wrong despite your confidence. Cage will have zero net gravitational effect on the black hole (assuming the cage is perfectly symmetrical), but that just means it won't by itself exert gravitational force on the black hole (and vice versa) causing it to move in any direction. Cage's material will still get attracted to the black hole and close to the event horizon structural forces exerted by gravity will be hard / impossible to counteract with current materials: https://en.wikipedia.org/wiki/Strength_of_materials https://en.wikipedia.org/wiki/Strength_of_materials
    - hansvm 5y ago
      The net pull on the _center of mass_ of a shell would be zero. The pull on any local region would still exist though. All that the shell theorem says is that the sum of all those local directional forces is zero. Reading the exact passage you quoted, "the compressive strength of the material forming the sphere would have to be immense to prevent implosion due to the star's gravity."
    - MauranKilom 5y ago
      I also exert zero net pull when I tear open a package of snacks. Yet the package is destroyed in the process.

lanna 5y ago

The net gravitational force is zero: https://en.wikipedia.org/wiki/Shell_theorem https://en.wikipedia.org/wiki/Shell_theorem

BoiledCabbage 5y ago
I don't see how Shell theory applies here. The box wouldn't be inside the blackhole, it'd be outside it. If the box were a perfect sphere then yes it would have zero net gravitational pull on the black hole, but the black hole would still have an extremely strong pull on it. And that's still assuming it's perfectly centered which would be almost impossible to maintain.
- lanna 5y ago
  > but the black hole would still have an extremely strong pull on it No, it would not. Newton's third law: if an object has zero net pull on the black hole, the black hole also has to have a zero net pull on the object. The shell theorem is more general than you are assuming: it doesn’t matter if the heavier object is inside or outside, the surrounding object doesn't need to be spherical (it can be any shape as long as it fully envelopes the black hole) and the black hole doesn't need to be centered. I know it is a counter-intuitive result, but it can be easily proven with calculus. See for instance this sentence about Dyson shells: "Such a shell would have no net gravitational interaction with its englobed star (see shell theorem)" https://en.wikipedia.org/wiki/Dyson_sphere#Dyson_shell https://en.wikipedia.org/wiki/Dyson_sphere#Dyson_shell Edit: Related, what is bigger? The gravitational force the Moon exerts on Earth or the gravitational force the Earth exerts on the Moon? Veritasium has a nice video explaining the answer: https://www.youtube.com/watch?v=8bTdMmNZm2M https://www.youtube.com/watch?v=8bTdMmNZm2M
  - aj3 5y ago
    You're wrong despite your confidence. Cage will have zero net gravitational effect on the black hole (assuming the cage is perfectly symmetrical), but that just means it won't by itself exert gravitational force on the black hole (and vice versa) causing it to move in any direction. Cage's material will still get attracted to the black hole and close to the event horizon structural forces exerted by gravity will be hard / impossible to counteract with current materials: https://en.wikipedia.org/wiki/Strength_of_materials https://en.wikipedia.org/wiki/Strength_of_materials
  - hansvm 5y ago
    The net pull on the _center of mass_ of a shell would be zero. The pull on any local region would still exist though. All that the shell theorem says is that the sum of all those local directional forces is zero. Reading the exact passage you quoted, "the compressive strength of the material forming the sphere would have to be immense to prevent implosion due to the star's gravity."
  - MauranKilom 5y ago
    I also exert zero net pull when I tear open a package of snacks. Yet the package is destroyed in the process.