How much faster would a Clock without gravity run? Pardon the misleading title.
It is to my understanding that moving/heavy clocks run slow. The Earth itself is under gravitational influence from many sources, and is moving. Is there a way to know how much 'faster' a clock would run if those influences were removed? I don't need a precise number, an order of magnitude is fine, just to get the idea.
Any insight is appreciated.
 A: Neither effect can really be calculated in a meaningful way.
Kinematic time dilation describes the time dilation of one frame of reference relative to another. There is no preferred frame of reference, so there is no way to say what it would mean to remove the effect of kinematic time dilation.
Gravitational time dilation is a concept that makes sense in a static spacetime, and in that case the time dilation between two different points is given in terms of the difference $\Delta\Phi$ in gravitational potential as $e^{\Delta\Phi}$. Here again you have the problem of what to compare to. Do you want to compare to interplanetary space? Interstellar space? Space outside our local cluster of galaxies? As you continue this process, you reach cosmological distances, at which point you run into the problem that cosmological spacetimes aren't static, and the whole thing becomes meaningless.
This is what relativity is all about. There's no best measure of time. It's all relative.
A: The fact that a clock runs slower due to movement is an effect of special relativity where the time-dilatation of a moving frame becomes:
$\Delta t' = \gamma\Delta t$, where $\gamma = \frac{1}{\sqrt{1-(v/c)^2}}\geq1$.
So a time-step in a motionless frame ($\Delta t$) is smaller then one in a frame in motion.
To get gravitational effect we should turn to general relativity, this yields the Einstein field equation: $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R = \frac{8\pi G}{c^4}T_{\mu\nu}$ (where I left out the cosmological constant for simplicity). This equation is basically the equation for the metric of the spacetime and is in general quite hard to solve !
If the spacetime is spherical (for example a pointmass/sphere) then the solution becomes the Schwarzschild metric:
$ds^2 = \left(1-\frac{2Gm}{c^2r}\right)dt^2 - \left(1-\frac{2Gm}{c^2r}\right)^{-1}dr^2 - r^2(d\theta^2+\sin^2\theta d\phi^2)$.
If we purely consider the time-evolution in a non-moving coordinate-frame we get:
$ds^2 = \left(1-\frac{2Gm}{c^2r}\right)dt^2$.
Or to put it in the same way as above, if $\Delta t$ is the time-evolution in a flat time-space (at $r\rightarrow\infty$) and $\Delta t'$ is the time-evolution with mass then we get the relation:
$\Delta t = \left(1-\frac{Gm}{c^2r}\right)\Delta t'$.
If i were to make a first estimate I would try to simply add the deviations which would give some kind of zero'th order approximation.
