Cylinder

The figure below shows how, for cylinders longer than they are thick, most of the forces from symmetricly spaced pairs of masses mostly cancel out on the "equator".

The on-axis gravitational field g of a thin disk of mass M and radius R and a distance x from the center of one end is

g=(2GM/R²)[1-(x/√(x²+R²))].

The cylinder may be viewed as a stack of infinitesmal disks as shown in the figure below:

The mass density of the cylinder (assuming it is uniform) is

ρ=M/(πR²L),

so the mass of the disk of thickness dx is

dM=ρπR²dx=Mdx/L.

So, the field dg due to the thin disk is

dg=(2GdM/R²)[1-(x/√(x²+R²))]=(2GM/(LR²))[1-(x/√(x²+R²))]dx

To get the net field, just integrate from x=0 to x=L. The result is

g_end=[GM/(RL)][(2L/R)+½-½√(R²+L²)/R].

Gauss's law for gravitational fields is

∫g∙dA=4πGM_enc

where M_enc is the mass enclosed by the Gaussian surface. Refer to the figure below.

If the length of the cylinder is much longer than its radius, the field near the center will be very nearly constant and radial. So an appropriate Gauss's surface would be a concentric cylinder of radius r and length L' where L>>L'. As shown above, the mass density is ρ=M/(πR²L), and so M_enc=ML'/L. So,

∫g∙dA=4πGM_enc≈g(2πrL')=4πGML'/L.

To get the field at the surface, let r=R:

g_equator≈2MG/(RL)

Now approximate g_end and compare with g_equator for L>>R

g_end=[GM/(RL)][(2L/R)+½-½√(R²+L²)/R]≈[GM/(RL)][3L/(2R)]=[3GM/(2R²)]>>2MG/(RL)

and therefore the field at the ends will be much larger than the field at the equator.

If R>>L, the field at the center of the end approaches the field in the center of a charged disk which is zero by symmetry. Therefore the field at the edge will be larger than the field at the ends.

Here is a rough sketch of what the field at the surface would look like for L>R. I have just drawn a fraction, but the whole field would be axially symmetric. The vectors are representative of the force you would feel if you were walking on the surface. So, walking from the equator to a pole would be like walking uphill but you would also get heavier. If you placed a small mass on the surface, it would only be in equilibrium at the equator (stable equilibrium) or at a pole (unstable equilibrium).