Now suppose that the Earth had the same mass and radius as it currently does, but all of the mass was concentrated into a thin (1 ft. thick) hollow spherical shell below your feet. Also suppose there was a small hole in the shell, just big enough four you to fit through. Compare the force of gravity on you outside the shell to the force of gravity if you stepped inside the shell.

The force of gravity at the shell will be extremely great on me due to the huge mass collapsed into the small radius.

At the center of the shell, the gravitational forces all around should cancel out, giving me a feeling of weightlessness; which will be a lesser force compared to that felt while standing on the shell.

Explanation:

For the collapsed earth:

mass = 5.972 × 10^24 kg

radius = 1 ft

according to Newton's gravitation law, the force of gravity due to two body with mass is given as

Fg = GMm/

Where Fg is the gravitational force between the two bodies.

G is the gravitational constant

M is the mass of the earth

m is my own mass

R is the distance between me and the center of the earths in each case

For the case where I stand on the shell:

radius R will be 1 ft

Fg = GMm/

Fg = GMm

For the case where I stand stand inside the shell, lets say I'm positioned at the center of the shell. The force of gravity due to my mass will be balanced out by all other masses around due to the shell of the hollow earth. This cancelling will produce a weightless feeling on me.

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.