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=*(2*GM*/*R*^{2})[1-(*x*/√(*x*^{2}+*R*^{2}))].

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*/(*πR*^{2}*L*),

so the mass of the disk of
thickness d*x *is

d*M*=*ρπR*^{2}d*x=M*d*x*/*L.*

So, the field d*g*
due to the thin disk is

d*g=*(2*G*d*M*/*R*^{2})[1-(*x*/√(*x*^{2}+*R*^{2}))]=(2*GM*/(*LR*^{2}))[1-(*x*/√(*x*^{2}+*R*^{2}))]d*x*

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

*g*_{end}=[*GM*/(*RL*)][(2*L*/*R*)+½-½√(*R*^{2}+*L*^{2})/*R*].

Gauss's law for
gravitational fields is

∫*g∙*d*A*=4*πGM*_{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*/(*πR*^{2}*L*),
and so *M*_{enc}=*ML*'/*L*. So,

∫*g∙*d*A*=4*πGM*_{enc}≈*g*(2*πr**L*')=4*πGML*'/*L.*

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

*g*_{equator}≈2*MG*/(*RL*)

Now approximate *g*_{end}
and compare with* g*_{equator} for* L>>R*

*g*_{end}=[*GM*/(*RL*)][(2*L*/*R*)+½-½√(*R*^{2}+*L*^{2})/*R*]≈[*GM*/(*RL*)][3*L*/(2*R*)]=[3*GM*/(2*R*^{2})]>>2*MG*/(*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).