You are in two opposing gravitational fields, but they do not add up, but subtract, cancel, not unlike a interfering wave?. so if there is no net effect of gravity, why should there be a relativistic effect? The question was about gravity inside versus outside, if the inside effects cancel out, but the outside field does not, does that not still leave more gravitational effects in the vastly greater outside, that an expanding body might move into?

This is a common misconception for layman about Relativity; That gravitational time dilation is due to the difference "felt" locally by a clock. This is not the case. Time dilation is due to the difference in potential, or as mentioned above the specific energy per unit mass to move the clock between the points where we are considering its tick rate.

Another way to look at this is that the gravitational time dilation as viewed from a far removed point has the same magnitude as the time dilation for velocity alone for an object moving at escape velocity for the position in the gravity field of the clock.

Escape velocity is derived from the setting the sum of the specific kinetic energy and specific gravitational potential energy to 0. For points at or above the surface of mass M it looks like this.

$$0 = \frac{v^2}{2}- \frac{GM}{r}$$

Where r is the distance from the center of mass M.

Solving for v gives

$$v = \sqrt{\frac{2GM}{r}}$$

If we substitute this for v in the standard SR equation for time dilation due to relative velocity, we get

$$ T = \frac{t}{\sqrt{1-\frac{2GM}{rc^2}}}$$

which is the standard gravitational time dilation equation.

If you want to find the time dilation for the center of a uniformly dense sphere*, the term for gravitational potential at the center is

$$-\frac{3GM}{2R}$$

where R is the radius of the sphere.

Solving for escape velocity and substituting in to the SR time dilation equation like we did above gives us a gravitational time dilation at the center of

$$ T = \frac{t}{\sqrt{1-\frac{3GM}{Rc^2}}}$$

At the surface r=R, so the time dilation will be

$$ T = \frac{t}{\sqrt{1-\frac{2GM}{Rc^2}}}$$

Time dilation at the center is more even though gravity at the surface is greater.

**Large celestial object are not going to be of uniform density, as interior pressures cause a rise in density as you move towards the center. This will result in some divergence from this equation. This divergence, however, does not affect the general gist of the argument. *