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 Pope Hilarius II Nov 9, 2008 This may or may not be a dumb question. We all know bigger celestial bodies have a stronger gravity on account of their mass, but doesn't mass density play a role, too? For instance, if you were to compress Jupiter's mass into an Earth-sized sphere, would it still have the same gravitational pull? In addition, since black holes are theoretically inescapable but have no mass in the conventional sense, why do they exert gravitational pull? # ? Apr 17, 2021 21:11
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 roomforthetuna Mar 22, 2005 I don't need to know anything about virii! My CUSTOM PROGRAM keeps me protected! It's not like they'll try to come in through the Internet or something! The first question the answer is it's mass and inverse distance squared. If you compressed Earth into a point and then checked the gravity force at one now-Earth-radius away from that point, it would be *almost* the same (but very slightly more because the parts of the force that were previously at a 30 degree angle are now all in the same direction). If you checked the gravity of point-earth near its *new* radius, the force would be huge, because the distance part is squared - every time you halve the distance, the force is multiplied by 4. (Conversely, if you went inside regular Earth to head for that smaller radius, the distance factor doesn't work the same way because all the earth that's above you is cancelling out the gravity from the earth that's below you - so a less-dense body can't exert the same force as a more-dense body, but it's not because the density changes the rule, it's just because you can't put as much mass at as short a distance with a less dense material.) Edit: or more specific to the Jupiter question, if you squashed Jupiter into earth size, the gravity *at a Jupiter radius distance* would be very similar, but at the surface would be much larger, because it's closer. roomforthetuna fucked around with this message at 12:41 on Apr 18, 2021 # ? Apr 18, 2021 12:38

#### roomforthetuna posted:

If you compressed Earth into a point and then checked the gravity force at one now-Earth-radius away from that point, it would be *almost* the same (but very slightly more because the parts of the force that were previously at a 30 degree angle are now all in the same direction).

Actually it is exactly the same; when you consider the compressed earth some parts get closer and some get further away, but when you do the math it all cancels out and you can just treat gravity as coming from the centre of mass. (for a spherical body, which earth isn't exactly, but you get what I mean)

 DamnCanadian Jan 3, 2005 Perpetuating the stereotype since 1978. According to Newtonian mechanics, the “force” of gravity you feel is based on two factors: the mass of the object exerting the force, and your distance from the object’s center of mass. If you had two objects of equal mass (but one had more volume, i.e. less mass density), you would feel the same gravitational effect from both if you are the same distance from each object’s center of mass. I say “force” at the beginning, because gravity is just how we experience the warping of spacetime caused by any object with mass, but Newtonian mechanics allows us to treat it as a force. There are some other factors if you want to consider relativistic effects, but I’d have to look that up. e: partially beaten # ? Apr 18, 2021 16:58

#### Hallucinogenic Toreador posted:

Actually it is exactly the same; when you consider the compressed earth some parts get closer and some get further away, but when you do the math it all cancels out and you can just treat gravity as coming from the centre of mass. (for a spherical body, which earth isn't exactly, but you get what I mean)
Oh yeah, I forgot that thing. It seems very unintuitive to me - the closer/further away aspect intuitively makes sense (incorrectly!), but the angled fragments seems like it wouldn't be right.

Like if you have two 1 ton tennis balls and a weightless stick holding them apart as your 'planet', aligned horizontally, and you're one radius 'up' from the center of mass, your distance to those masses is greater than a radius *and* you're getting a sqrt(2)-related fraction of the forces canceled out by that fraction being in opposite directions. (Because the forces are southwest for one ball and southeast for the other.) So your resulting 'southward' force for that object is way less than it would be for a 2 ton ball in the middle of that diagram.

But I guess it's just a whole bunch of weird math where the part of the sphere that's really close to you is exerting so much more force (because of the distanced squared thing) that it's enough to precisely make up for the reduction of the more distant parts *and* the angled parts. So that shortcut only works for spheres and is completely misleading for any other shape.

 Hallucinogenic Toreador Nov 21, 2000 Whoooooahh I'd beNothin' without youBaaaaaa-by Another fun counter-intuitive example is that if you're inside a hollow spherical shell all the mass cancels out and you experience zero g, even if you're not at the centre. https://en.wikipedia.org/wiki/Shell_theorem # ? Apr 18, 2021 22:04

#### Hallucinogenic Toreador posted:

Another fun counter-intuitive example is that if you're inside a hollow spherical shell all the mass cancels out and you experience zero g, even if you're not at the centre. https://en.wikipedia.org/wiki/Shell_theorem
Nice. That is indeed the exact same kind of thing, where it uncomfortably feels like an impossibly unlikely coincidence for the math to work out that way.

#### roomforthetuna posted:

Nice. That is indeed the exact same kind of thing, where it uncomfortably feels like an impossibly unlikely coincidence for the math to work out that way.

It's not a coincidence. The shell theorem is a mathematical consequence of the inverse square law, which is a consequence of energy conservation & the gravitational field being a conservative vector field. Energy conservation is itself a consequence of time translation symmetry and you can go on and on. These aren't coincidences so much as they are different ways of mathematically expressing the same relationship.

#### Morbus posted:

It's not a coincidence. The shell theorem is a mathematical consequence of the inverse square law, which is a consequence of energy conservation & the gravitational field being a conservative vector field. Energy conservation is itself a consequence of time translation symmetry and you can go on and on. These aren't coincidences so much as they are different ways of mathematically expressing the same relationship.
It's specifically the "perfect sphere" aspect of it that seems like an unlikely coincidence, that doesn't really sound mentioned in what you said. Like the same wouldn't be true of being inside a cube shell or a pyramid shell. And I'm *fairly* sure the one where you're on the outside where the force is equivalent to the force from a point of equal mass at the center of mass only works with a perfect solid sphere, and does not work with the perfect shell sphere, which only makes it feel even weirder. (And the one where if you're inside it's zero g doesn't work with a filled sphere, only with a zero-thickness sphere shell... or maybe any thickness but you can't intersect the shell?)

I mean sure it's derivable from the math, but it still seems bizarre and coincidental that it works out to perfectly match up to interactions with a primitive shape - and not even the *same* primitive shape in both things, just one that looks the same from the outside.

#### Pope Hilarius II posted:

This may or may not be a dumb question.

We all know bigger celestial bodies have a stronger gravity on account of their mass, but doesn't mass density play a role, too? For instance, if you were to compress Jupiter's mass into an Earth-sized sphere, would it still have the same gravitational pull?

In addition, since black holes are theoretically inescapable but have no mass in the conventional sense, why do they exert gravitational pull?

I don't think anyone has addressed the last part yet, but black holes do have mass; that's why they're described as e.g. stellar-mass or supermassive. Mass is one of the few measurable qualities black holes do have.

#### roomforthetuna posted:

And I'm *fairly* sure the one where you're on the outside where the force is equivalent to the force from a point of equal mass at the center of mass only works with a perfect solid sphere, and does not work with the perfect shell sphere, which only makes it feel even weirder.

It's been way too long for me to reproduce the math, but when I learned this stuff I remember proving that a shell was equivalent to a point mass first and then using that to show that a solid sphere is as well, as you can think of a solid sphere as a collection of infinitesimally thin concentric shells.

#### quote:

(And the one where if you're inside it's zero g doesn't work with a filled sphere, only with a zero-thickness sphere shell... or maybe any thickness but you can't intersect the shell?)

Likewise you can show that an infinitesimally thin shell has zero net gravity on the inside, so any spherical (hollow!) shell with real physical thickness does as well.

#### Mr. Peepers posted:

Likewise you can show that an infinitesimally thin shell has zero net gravity on the inside, so any spherical (hollow!) shell with real physical thickness does as well.
But that's definitely not true if you intersect the shell, because that would approximately include "a guy 20 feet down in a mineshaft", and that guy definitely doesn't experience anything resembling zero g.

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#### roomforthetuna posted:

But that's definitely not true if you intersect the shell, because that would approximately include "a guy 20 feet down in a mineshaft", and that guy definitely doesn't experience anything resembling zero g.

True! Now (pretending here that the planet is a perfect uniform sphere) you can split the whole of the planet into 2 regions: all the matter below 20 feet down from the surface would be a solid sphere, and would exert a gravitational force on you like you'd expect. All the matter on the whole planet above the 20-feet-below-the-surface level now acts like a spherical shell surrounding you, and would therefore exert a net zero gravitational force! As you move towards the center of the planet, more and more of the planet becomes part of the zero-g spherical shell that surrounds you and less makes up the solid sphere that actually exerts a net gravitational force. It's actually a linear decrease as you move towards the center of the planet, i.e. if you tunnel down 25% of the way towards the core you'll feel 25% less gravity than you would at the surface.