How Thick Would an Infinite Plate Need to Be to Mimic Earth's Gravity?

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To mimic Earth's gravity with an infinite plate, the required thickness H must be 2/3 of Earth's radius. The discussion revolves around applying Gauss's Law for gravitational fields to derive this relationship. Initially, there was confusion about incorporating thickness into the model, as previous lessons only covered infinite plates without thickness. The user successfully equated the gravitational integrals of both models to find the solution. This approach demonstrates the application of gravitational principles to theoretical scenarios.
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Homework Statement



Imagine the Earth is indeed not a sphere of radius r, but an infinite plate of thickness H. What value of H is needed to allow the same gravitational acceleration to be experiences as on the surface of the actual Earth? (Assume the Earth's density is uniform and equal in the two models.)

r Earth = 6370km
g = 9.81 N/kg
m Earth = 5.91 * 10^24 kg

Homework Equations



Gauss's Law?

The Attempt at a Solution



So far I have tried to set up gauss's law(for gravitation instead) as Integral(g dot dA) for the spherical earth, and set it equal to the Integral(g dot dA) of a infinite plate version of earth. I get h = r * sqrt(2pi), which is not correct.

Can anyone tell me if I am in the general correct direction with my idea to use gauss's law for gravitational fields? My main trouble is the thickness of an infinite plate. How is this possible, in class we only learned about infinite plates w/o thickness, and parallel plates. Is this a special version of parallel plates, and if so, how does one bring H into the equation?
 
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Can you just use Newton's Law of gravitation? Basically, you can write the Earth's mass in terms of the thickness, then equate the two equations, solving for t?
 
Okay, so I figured it out and no longer need help.

What I did was find the integral(g dot dA) of the spherical earth, which was 4piGm, and set that equal to the integral(g dot dA) of the infinite plate earth. Evaluating the integral, and doing some tricky substitutions with p = m/v, I found that the H(thickness) = 2/3 R(earth)
 
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