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Anonymous at Sat, 23 Nov 2024 00:37:06 UTC No. 16489980
Assume you have a 3D space filled with a mass infinitely extending in all directions.
Now assume that this mass has a spherical cavity of diameter D.
What would the gravitational acceleration G be if D was the same as earth's (12700 km).
Conversely what would D have to be for G to be 9.81 m/s^2?
Which equation would describe G in this case? Considering that G(Height = D/2) = 0?
Anonymous at Sat, 23 Nov 2024 01:52:41 UTC No. 16490060
divergence theorem applied to newtonian gravity says g would be 0 inside cavity
Anonymous at Sat, 23 Nov 2024 05:44:36 UTC No. 16490226
>>16489980
>infinitely
nice try shekelstein
Anonymous at Sat, 23 Nov 2024 06:51:49 UTC No. 16490239
>>16490060
How'd this nlmake sense?
Every imaginable shell centered on the observer outside of a diameter of 2D should cancel to G=0 because it is perfectly symmetrical, right?
>>16490226
"Very large sphere centered on the observer (D = 10^10000 m)"
Happy now?
Anonymous at Sat, 23 Nov 2024 07:21:12 UTC No. 16490252
>>16490239
>How'd this nlmake sense?
newton's 2nd law is (assuming constant masses)
[math]\vec{F} = m \vec{a}[/math]
newton's law of gravitation gives the force of point source mass m' on point test mass m as
[math]\vec{F} = G \frac{m m'}{R^2}\hat{R}[/math]
which can be generalized to several point source masses m_i over
[math]\vec{F} = \sum_i G \frac{m m_i'}{R_i^2}\hat{R}_i[/math]
which can be generalized to a continuous source mass distribution rho
[math]\vec{F} = \int_V G \frac{m \rho}{R^2} dV [/math]
the test mass can be factored out to get
[math]\vec{F} = m \vec{g}[/math]
where the acceleration due to gravity g is
[math]\vec{g} = G \int_V \frac{\rho}{R^2}\hat{R}[/math]
this has exactly the same form as the electric field in electrostatics. it has zero curl, and the divergence can be worked out to be
[math]\nabla \cdot \vec{g} = 4 \pi G \rho [/math]
in integral form, this is (where M is total source mass)
[math]\int_V \nabla \cdot \vec{g} dV = 4 \pi G \int_V \rho dV[/math]
or, after applying the divergence theorem and rewriting the right integral as total mass M
[math]\boxed{\oint_S \vec{g} \cdot d \vec{A} = 4 \pi G M}[/math]
this equation says that the surface integral of the gravity field is proportional to the mass contained within the surface
consider spherical surfaces centered on the center of the cavity.
by the curlless nature of g and symmetry, g will have to be radially symmetric.
the enclosed mass is clearly zero.
therefore, g must be zero on the surface.
this applies to a sphere of any radius within the cavity.
therefore, g is zero within the cavity.
Anonymous at Sun, 24 Nov 2024 20:34:54 UTC No. 16491924
note, if the cavity wasn't spherical, >>16490252
doesn't work