Gravity transition directly at the underside of a "shell planet"

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mgkii
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Does gravitational pull transition immediately from g to zero underneath the shell of a "shell planet"
I'm watching the Stanford University Lecture series: Einstein's General Theory of Relativity presented by Leonard Susskind (who incidentally has to be one of the greatest educators I've ever watched).

Whilst deriving the basic divergence equations relating acceleration, mass density, and Newton's gravitational constant, he took a small diversion to show that Gauss' Therum shows you that if you place a test-mass inside a uniformly dense planet, then "little g" is only proportional to the mass inside the radius of the sphere you place the test mass on - i.e. the effect of all of the mass "above you" cancels out. So far so good!

He then went on to extrapolate that if the mass of a planet was concentrated entirely in a shell, then something inside the shell would effectively feel no gravity. However there was some debate with the class and the example ended in rather a lot of confusion!

Sorry for the long preamble - finally to my question! Did I take the correct understanding that someone standing on the surface of a "shell earth" (where all the mass of the Earth was concentrated in an arbitrarily thin shell at the surface) would feel the full 9.8m/s/s acceleration of Earth's gravity, but if you dug through that shell then you would transition immediately to a zero m/s/s gravity? This seems to be only outcome based on what I think I've understood, but it seems so alien I need to check my logical compass!

Thanks
Matt

PS - This is only Lecture 2 of 12... please go easy on me :-)
 
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mgkii said:
Does gravitational pull transition immediately from g to zero underneath the shell of a "shell planet"

Only in the idealized case of a shell with zero thickness, which is not physically realistic.

In a realistic case of a shell with finite thickness, the "gravitational pull" continuously decreases from g to zero as you descend through the shell. Once you are inside the shell, yes, there would be zero gravity.

In the more usual technical language of relativity, the metric would continuously change as you descend through the shell, from the (curved) Schwarzschild metric for a mass ##M## (the total mass of the shell) outside to the (flat) Minkowski metric inside.
 
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Thank you. That makes complete sense now. Happy for this thread to be closed!