Final answer:
The gravitational potential of a thin spherical shell is -G*M/r for points outside, -G*M/R for points inside, and at the surface. Inside, the gravitational force would be zero, allowing frictionless travel across the sphere's interior. If the sphere were spinning, centripetal forces would be present.
Step-by-step explanation:
The gravitational potential due to a thin spherical shell varies depending on the position of the point in question:
- Outside the spherical shell, the gravitational potential at a distance r from the center is given by:
V = -G*M/r
- Inside the thin spherical shell, the gravitational potential is constant and equal to the potential at the surface:
V = -G*M/R
- At the surface of the shell, the potential is also given by:
V = -G*M/R
Where G is the gravitational constant, M is the mass of the shell, and R is the radius of the shell.
If we imagine traveling inside such a shell in an elevator, we'd feel no gravitational force (g would be zero) due to the shell's mass. This would provide a unique means for traveling great distances across the inside of the sphere without the influence of gravity from the sphere itself. However, if the planet were spinning, there could still be effects due to centripetal forces.