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Imagine a supermassive hollow shell in space, and also imagine there is an object at the center of this shell. How does the force of gravity affect the body inside the shell?

My reasoning is that the shell will pull outward on the body equally in all directions. Now do the forces cancel out essentially doing nothing to the body or will the forces act to pull the body apart? The essence of my question is whether or not two opposing gravitational forces cancel eachother or if they act together to "rip apart" something.

As an analogy, if I pull equally hard on two sides of a rope the rope would experience two equal and opposite forces but would still get ripped apart if I pull hard enough. Does gravity act in the same manner?

Here's a crudely drawn diagram of the setup: enter image description here

The object at the center is not necessarily a human, consider a small non-massless particle.

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  • $\begingroup$ Tidal forces might play a role, but other than that, I think that the forces should cancel. $\endgroup$
    – Jonas
    Jun 13 at 10:26
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    $\begingroup$ the premiss of the question is wrong , it is a theorem that "If the object is a spherically symmetric shell (i.e., a hollow ball) then the net gravitational force on a body inside of it is zero." see phys.libretexts.org/Bookshelves/University_Physics/… page "the shell theorem" $\endgroup$
    – anna v
    Jun 13 at 10:29
  • $\begingroup$ @Jonas see my comment and link $\endgroup$
    – anna v
    Jun 13 at 10:29
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    $\begingroup$ Possible duplicate: Is spacetime flat inside a spherical shell? $\endgroup$
    – Qmechanic
    Jun 13 at 10:40
  • $\begingroup$ Is your shell spherically symmetric, with uniform density and thickness? $\endgroup$
    – PM 2Ring
    Jun 13 at 11:54
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The net gravitational force on any mass inside a spherical shell is zero. This is a consequence of Newtons' shell theorem. As per this link it states

"In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to astronomy.

Isaac Newton proved the shell theorem and stated that:

A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

This means there will be no net force on the body, and it will not be torn apart.

As an analogy, if I pull equally hard on two sides of a rope the rope would experience two equal and opposite forces but would still get ripped apart if I pull hard enough. Does gravity act in the same manner?

No. As explained above.

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    $\begingroup$ The rope analogy also fails for the following reason: force is exerted on the ends of the rope. Gravity is exerted on the entire rope, so there is no tension or force exerted on one end that is not also exerted on the other. $\endgroup$ Jun 13 at 18:59

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