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Time runs slower near a massive object. I believe this time dilation is due to the gravitational potential present near the surface when compared to a distant point. What about the center of a massive object, how does time run at the center of a massive object compared to its surface? After all at the center the acceleration is equal, lets assume that here, in all directions and cancels out.

There is a thousand general relativity questions on this site and I couldn't find anything speaking to this, but its hard to believe no one asked this or brought it up in an answer.

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Does time run slower at the center of a massive object?

Yes.

Gravitational time dilation depends on the value of the Newtonian gravitational potential, not on the gravitational acceleration. The lower the potential, the greater the time dilation. The dependence is linear when the absolute value of the Newtonian potential is much less than $c^2$, which is the case for planets and stars. When it becomes comparable to $c^2$, the concept of a Newtonian gravitational potential is no longer particularly useful.

The Newtonian gravitational potential at the center of a sphere of uniform density is more negative by a factor of 3/2 than the potential at the surface, causing greater time dilation at the center by approximately the same factor.

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    $\begingroup$ Time dilation has nothing to do with how we experience time. A second always feels like a second regardless of where I am or how fast I’m moving. Time dilation involves what another observer observes when they look at my clock, not what I observe. $\endgroup$
    – G. Smith
    Aug 19, 2020 at 23:24
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    $\begingroup$ @MarkMoralesII Well, now you have! If you don’t believe me, believe Wikipedia: “The lower the gravitational potential..., the slower time passes, speeding up as the gravitational potential increases...”. $\endgroup$
    – G. Smith
    Aug 19, 2020 at 23:48
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    $\begingroup$ If you have studied GR, you know that gravitational time dilation is determined by the $g_{00}$ component of the metric tensor. In the Newtonian limit, $g_{00}=-1-2\varphi/c^2$, where $\varphi$ is the Newtonian potential. (Misner, Thorne, and Wheeler, Gravitation, eqn (17.19)) $\endgroup$
    – G. Smith
    Aug 20, 2020 at 0:19
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    $\begingroup$ @MarkMoralesII You seem to be confusing "no gravity" with " no net gravity" inside the shell you still have gravitational attraction from all points of the shell, they just even out to "no net gravity". $\endgroup$ Aug 20, 2020 at 0:56
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    $\begingroup$ @G.Smith Huh, I didn't know that it worked that way. I guess that I have more studying to do. :) $\endgroup$ Aug 20, 2020 at 4:32

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