I also understand that due to the relative orbital velocity, time on Earth is dilated compared to the centers of the solar system and galaxy.
Is there a place that the Milky Way rotates around, and if so, at what approximate speed, and what would be the time dilation compared to earth?
@Thomas-Roy:
According to Einstein's general theory of relativity, significant time dilation can occur in two instances: extreme velocity (a significant fraction of the speed of light), or due to gravitational fields. (Well, any acceleration generally.) In both cases, time slows for the traveler or the person in the gravitational field relative to someone in a different reference frame.
For example, the latest atomic clocks are so accurate and sensitive that they can detect the tidal flexure of the Earth due to the presence of the Moon overhead, even though the motion is only a foot up and down. When the clock is higher, time passes more quickly for the clock since it is further from the center of the earth and experiences less gravitational field. The reverse is true for when the clock is lower.
Similar effects can be observed for speed. A satellite in a lower orbit has a higher velocity than one in a higher orbit. The lower one experiences time more slowly than the higher one. Of course, the effects of gravity are at play here too.
In terms of the cosmos, the Milky Way is a member of what Astronomers call the "local group". That is a small group of galaxies gravitationally bound together and we all orbit each other. There are ever larger scales of organization, spanning the entire universe. So, we're all moving, we're all in some kind of gravity field, no matter how big you are.
On the other hand, and more to the point of your initial question, there is a place in the cosmos called 'the great void', which is a region that spans hundreds of millions of light years with no observable matter. (There's more than one, actually, but that's the biggest we know of.) In that place, gravitational fields are likely very weak and if you're just hanging out in the middle, not moving relative to the edges, you'll probably be aging faster than anything else in the universe.
I hope this helps!
I understand that the universe is expanding. From that, I gather that the earth is moving away from some "center" point. Is there such a center point?
You are correct in understanding that the universe is expanding, but the Earth is not moving away from some center point. Imagine a balloon with some dots on it, where each dot is a star. Now blow up the balloon, increasing the space between each star. This is what is happening to the universe. Things aren't moving away from some center point, they are all simultaneously moving away from each other. So, in other words, there is no center point.
As for the question in your title, any time you are moving slower than the Earth in some frame of reference, time is passing faster for you compared to Earth. There are also time dilation effects associated with gravity, but I don't know enough relativity to comment on the subject.
Since this question appears to concern time dilations due to orbits let us look at that. If one is wanting to look at the orbit of a test mass in a central gravity field then we start with the Schwarzschild metric and its line element $$\mathrm ds^2=\left(1-\frac{2m}r\right)\,\mathrm dt^2-\left(1-\frac{2m}r\right)^{-1}\,\mathrm dr^2-r^2(\mathrm d\theta^2+\sin\theta\,\mathrm d\phi^2).$$ Here of course $m=GM/c^2$. Consider the plane of orbit fixed with $\theta=\pi/2$ which simplifies the angular part. In addition divide the entire line element by $\mathrm dt$ the coordinate time $$\left(\frac{\mathrm ds}{\mathrm dt}\right)^2=\left(1-\frac{2m}r\right)-\left(1-\frac{2m}r\right)^{-1}\left(\frac{\mathrm dr}{\mathrm dt}\right)^2-r^2\left(\frac{\mathrm d\phi}{\mathrm dt}\right)^2.$$ The last term is the velocity $v=r^2\left(\frac{\mathrm d\phi}{\mathrm dt}\right)^2$ and the left hand side is the time dilation or reciprocal of the Lorentz gamma factor. $$\Gamma=\left[\left(1-\frac{2m}r\right)-\left(1-\frac{2m}r\right)^{-1}\left(\frac{\mathrm dr}{\mathrm dt}\right)^2-v^2\right]^{-1/2}.$$ Clearly for $m=0$ this recovers the standard Lorentz gamma factor.
Now consider a weak field condition. Returning to the Schwarzschild metric the $\mathrm dt^2$ is implicitly $c^2\,\mathrm dt^2$ and so contributes far larger term than the $dr^2$ term. We may also eliminate the $\mathrm dr^2$ term if we only consider circular orbits. We then simplify our general gamma factor to $$\Gamma=\left[\left(1-\frac{2GM}{rc^2}\right)-\left(\frac vc\right)^2\right]^{-1/2}.$$ In this weak field approximation we also use the Newtonian result that the velocity is $v^2=GM/r$ and we get s simple form for the Gamma factor that is dependent only on the radius $$\Gamma=\left[\left(1-\frac{3M}{rc^2}\right)\right]^{-1/2}.$$ I include a graph of this below. This is really most valid for the time dilation factor not much less than one.
One can then input some numbers. For instance with the Earth the Schwarzschild radius is about a centimeter, and satellites orbit at around $6500\ \mathrm{km}$ or $6.5\times10^8\ \mathrm{cm}$ so an estimate for this with $3GM/rc^2\simeq2.3\times10^{-9}$ gives $1/\Gamma\simeq1-1.1\times10^{-9}$. You can similarly try the time dilation of the Earth orbiting the sun with $2GM/rc^2\simeq3\times10^3\ \mathrm m$ and you get about $1/\Gamma\simeq1-3\times 10^{-8}$
For computing time dilation due to motion in the galaxy that is a bit complicated. The reason is that stars orbit not around a central gravity field, but within a distribution of matter, most of it being dark matter. The orbits are more similar in a Newtonian sense what one sees with a two dimensional harmonic oscillator. However, for the sun and stars in our Perseid arm (I believe that is what it is called) the orbits are about $300\ \mathrm{km/s}$ and If I just us the standard Lorentz gamma equation $\gamma=(1-v^2/c^2)^{-1/2}$ I get $1/\gamma\simeq1-1\times 10^{-6}$. If I were to include gravity the factor would still be unit minus a factor within the order of magnitude of $10^{-6}$
