The attractive force of a spherically symmetric body, such as a planet, is indistinguishable from a point source with the same mass. (Of course, no real-world object is perfectly symmetric.)
It is enough to show the potential field of the sphere is the same as the potential field from a point source of equal mass. Let the density at radius r be f(r), and integrate using spherical coordinates. Let z be a point somewhere along the north pole of the sphere. By axial symmetry, the longitude θ doesn't matter. The latitude φ does matter. Potential energy is inversely related to distance, and points in the northern hemisphere are closer to z than points in the southern hemisphere. Find the distance from z to a point at latitude φ and radius r, and take its reciprocal. Remember that latitude is 0 at the south pole, rising to 180 at the north pole. Bring in a minus sign for convention: potential is 0 at infinity and roars down to -∞ as you approach a point source. Then bring in a factor of r2×sin(φ) for spherical coordinates, and get the following integrand.
-f(r) / sqrt(z2+r2+2zr×cos(φ)) × r2×sin(φ)
Integrate with respect to φ first. The integrand is an expresssion in cos(φ), times sin(φ). That tells us what to do. Put the radical in the numerator, and wipe away -sin, it being the derivative of cos, and get this.
f(r)×r/z × sqrt(z2+r2+2zr×cos(φ))
Set φ to 0 and π, the limits of integration, and the square root reduces to z+r and z-r respectively. We must subtract the first from the second, but we have to be careful on that second square root. Is it z-r or r-z? The difference is either -2z, or -2r. This in turn is multiplied by r/z, giving either -2r2/z or -2r. Potential is suppose to be negative, so if z is below 0, the desired integrand must be -2r. By symmetry, the potential is the same when z is above 0 and still inside the sphere. Anywhere inside the sphere of radius r, the desired expression is -2r. But as you move up and away from the sphere, potential must rise to 0, thus -2r2/z. The integrand is -2rf(r) inside the sphere, and -2r2f(r)/z outside the sphere.
Integrate 2rf(r) with respect to θ if you like, and get 4πrf(r). This does not depend on z. The potential inside the shell of radius r, or any hollow sphere for that matter, is constant. There is no gradiant, and no force field. You could float around weightless in such a sphere, even if the sphere has the mass of Jupiter.
Now place z outside the shell of radius r. This time the integrand is -2r2f(r)/z. Integrate with respect to θ and get -4πr2f(r)/z. This is the surface area of the shell, times its density, divided by the distance to the center of the shell. This is the same formula you would obtain if the shell were compressed down to the origin. Since this relationship holds for every shell, it holds for the entire sphere. (You can integrate with respect to r for a formal proof, but it's pretty intuitive.) In summary, the sphere can be treated as a point source, and the gravity is the same.
If the sphere is a hollow shell, what happens when you are directly on the shell? Either formula, derived from z-r or r-z, is valid. The inside formula, independent of z, is -4πrf(r). The outside formula, evaluated at z = r, gives the same thing. Thus the potential field is continuous, as it should be. But it is not smooth. It is constant inside, then turns sharply and slopes up at z = r. Poke your head above the shell, and you suddenly feel the mighty pull of gravity. Of course, any real world shell will have some thickness, so as you climb through a hole in the shell, gravity will increase, until you are finally sitting on the surface.