Gravitational force when standing on an infinite disc?

Question

If a person were to stand on a flat disc of thickness 1 meter but infinite diameter, would they experience finite or infinite downward force? There is an infinite amount of mass, all of which attracts the person but most of which is also infinitely far away. Do the two "cancel out" or does the infinite mass win? For a precise definition of "infinite diameter", take the limit of a circular disc with diameter $x$ as $x$ approaches infinity, with the person standing on top of the disc's center of gravity.

Answer 1

This question can be answered by integrating the downward pull due to every annular piece of the slab. The answer is finite. Furthermore, it does not depend on how far away the person is from the slab!

Answer 2

In this case classical gravity works very similar to classical electric field.

If you remember how to compute the electric field of an infinite sheet of electric charge, you will be familiarized with this image:

https://i.ytimg.com/vi/5Z8gbUYdIdQ/hqdefault.jpg

This is the gravitational equivalence for the Gauss Law of the electric field applied to an infinite sheet.

Since for your example the disk has a thickness of 1 meter the difference would be as such:

g*2*A=-4*π*G*ρ*V

here ρ is the volumetric mass density of the disk and V the enclosed volume by the Gaussian Surface.

Since this volume is V=A*(1meter) then:

g*2*A=-4*π*G*ρ*A*1

g=-2*π*G*ρ

This would be the magnitude of the gravitational field, thus the gravitational force is:

F=-2*π*G*ρ*m

where m is the mass of the object experiencing the gravitational force, which is downwards and of finite value.

As you can see this relation does not depend on the radius of the Gaussian surface and can be applied to a infinite disk.