Hi Brainy Astrophysic types - proabably Zeno in particular…

Been reading some Utopian myths and Agarta/hollow Earther stories in particular and got to thinking…

Say you mashed the Earth up, and reformed it with the same mass, but proportionately bigger, as a hollow ball, and decided to live on the inside. Would you still fall toward the (now empty) centre…? ie: would the epicentre of gravitational force still remain the same. Or if the ‘crust’ were thick enough and massy enough to generate a kind of evened out gravity field - acting across it’s upper and lower surfaces - could we stick to that inner surface…?

Presumeably, the Earth would still spin - would that centripetal/centrifugal force - acting to throw us outward against the inner surface - be enough to counteract the inward pull of the unmoved phantom effective gravitational epicentre - or would we have to spin it r-e-a-l-l-y fast…? ie: so fast that the crust integrity would fail…?

If on falling, due to accumulated momentum, I fall past the gravity centre, would I slow down and fall back - eventually to hang wieghtless at the exact centre of the hollow sphere…?

Yours with breathless anticipation of completely incomprehensible answers…

I’ll try that one… We’ll first forget the naive question of : how does your hollow earth hold ? It should just collapse to take the shape of a sphere… But let’s assume an unbreakable earth - it would just be another idealism…

No, the gravity field would still have the same symetry… Actually, an approximate analogy is found when you use a submarine or practice speleology : you don’t hold to the surface, you minimize your potential i.e. you go towards the center.

I think that gravitation is strong enough to fail the crust anyway… To have something that would reach the point of zero gravity (I’ll assume that g = 10 N/kg), as everybody know, centrifugal force is given by : F=mvv/r, with m your mass, v your tangential speed and r the radius. So you want a null weight, i.e. : mg = mvv/r. Forget the m. You see that your speed should be the square root of g times your radius to make your weight disappear… I’ll let you play with numbers, if you feel like it.

Eventually, yes, if there is a kind of friction/energy loss. If not, you’ll be a funny new kind of oscillator…

marc is correct. think about it tab, in your new swinging lifestyle around the earth’s now hollow core, you’d meet all the hot chicks. penetration would presumably take alot less effort too. i wouyldn’t dump that idea entirely just yet.

you don’t need to do the numbers to understand what’s the problem with centrifugal force. gravity decreases with distance from center at the 2nd power. centrifugal force increases with distance from center at the 2nd power. presuming the centers are in the same geometrical point, you’d get a combined force that just keeps on growing as you get further away. if they were skewed so that both increased or decreased with distance, you might get a rest spot. but as it is, that spot will be necessarily not in equilibrium.

further, spinning so fast would make all sorts of odd problems, like what do you do if a meteorite falls ? (because now it has it’s original energy, but also it’s relative momentum ! (since to you it appears to be spinning very fast around you))

mass on an empty shell generates a normal - looking force-field on the outside (pulls toward centre)
but on the inside all forces will cancel out exactly, net result: a zero sum
one would indeed be weightless in there

willem, bs. i could to the entire math, but it will take a while and i don;t have proper notations here.

consider the simpler problem of a circle. gravitational force depends on distance, if you take a circle and draw a point on it, as long as it’s not in the center it will be closer to a side. consequently the figure will now be symetrical with respect to a line uniting the center and the point you chose, and thus everything will cancel itself out EXCEPT the far end and the close end of that line, and consequently the point will be pulled towards the close end of that line. same thing for a sphere.

it’s out of the question that everything inside a sphere has same gravitational potential.

Yes, that’s what seems quite clear to me… But apparently there are some strange inertia explanations… Or is it Mach’s theory of stuff ? Such as, the whole universe is the source of your inertia ?

Wait, Im not sure about that. If you hit a baseball a little more than halfway to the moon, the moons gravity will pull the baseball to it. Same would go for in a sphere. You have equal mass on all sides(presumably) if an object is closer to a certain side than the other the increased proximity to a source of gravity and the decreased distance from another source of gravity would cause the object to be pulled to the closer side. I can see you point on sphereical gravity willem but the object would have to be in the exact center. (I am not including friction and air here) ( - ) <—sphere with the object in it.

willem, your math would be correct IF gravity depended liniarly with distance. it depends with the square of distance. thus, your/newton’s understanding of the matter is incorrect.

(for the record, keppler came after newton. also for the record, newton still belives sound propagates as an isothermic transformation of the air.)

edit: i don’t think she’ll sue me for one line, but i’d rather not find out ^^

as a side note,
in normal conditions, (not relativistic) newton’s laws of gravity are still a very good approximation of einstein’s relativity laws
(if i remember correctly the former is the first term of the lagrange approximation of the latter)

let’s consider a circle with the center in x0 and a point in x.

every point on the surface of the circle attracts the point x with the force

Fi = k d ^2

the long side of a triangle is L = l1 ^ 2 + l2 ^2 + 2 l1 l2 sin (a) where a is the angle.

thus, for every point on the circle d is

d = r ^ 2 + x ^ 2 + 2 r x sin (pi - a) where a is the angle of the point considered

When we integrate by angle a we get

F = integral by a (from 0 to 2 pi) of k ( r ^2 + x^2 + 2rx sin (pi-a)) which simplifies as

F = k (r ^2 + x ^2) + integral by a (from 0 to 2pi) of 2 k r x sin (pi-a)

the sin function being a periodical, and the integral being over the entire period, and the period being symetrical with respect to the Ox axis, that integral comes out 0.

no shit sherlock

seems willem is correct after all

(thanks to willem, i spent a fun half hour re-doing the above stuff, it’s been a decade since i last touched that sort of a problem. i;m not very sure why i was so convinced it must be the other way around, but anyway… to quote feynman, i had bad luck )

Look - don’t strain yourselves, knock twice on the screen for “Yes Tab, you could live on the inner surface of a sphere without falling off” and once on the screen for “No Tab, you’d be twiddling your thumbs in mid air at the centre of the sphere.”