My contribution to the thread is below. And, if you’ll forgive a little self-promotion, I've also written a couple of educational physics books, which are available at the links given:
Mind Openers 1.0: A Conceptual Review of Modern Physics https://www.amazon.com/gp/product/B08BHYYVZ1?notRedirectToSDP=1&ref_=dbs_mng_calw_0&storeType=ebooks
and
Mind Openers 2.0: A Conceptual Reinterpretation of Modern Physics https://www.amazon.com/gp/product/B08BKRWL38?notRedirectToSDP=1&ref_=dbs_mng_calw_1&storeType=ebooks
Cheers!
<PERSON>
Time Dilation Effects for Stars on the Outer Edge of The Milky Way First, we tackle gravitational time dilation: whereby stars closer to the center will age more slowly because they are in stronger gravity. Ignoring galactic rotation to simplify, general relativity tells us that t’ = t (1-2GM/rc^2)^1/2 where t’ is the dilated (slowed) rate of time passage and t would be the rate of time passage if the mass M (of the galaxy in this case) was not present – or at an “infinite distance from it” – and r is the distance from the center of the mass of the galaxy. We will assume the point in question lies at the outer edge – so that the entire mass of the galaxy lies inside it. We will ignore any hypothetical dark matter - which in any case is thought to lie farther out in "a halo". 562
Using binomial expansion, where (1+x)^n is approximately equal to 1+nx, to simplify we get t’ = t (1 - GM/rc^2) = t – tGM/rc^2 Letting the change in time between the two situations equal ∆t, then ∆t = t’ – t = t – t - tGM/rc^2 = tGM/rc^2 Therefore ∆t/t = GM/rc^2 Going with orders of magnitude to get a rough sense of it, let M = the mass of the Milky way = approximately 10^12 solar masses, which equals about 10^42 kilograms. And, let r = 60,000 light years (a broadly accepted radius for the Milky Way - but subject to debate depending on where we define the "edge") which equals 5.6 x 10^20 meters. Therefore ∆t/t = (6.673 x 10^-11 m3/kg-s2 x 10^42kg)/5.6 x 10^20 meters x 9 x 10^16 m2/s2 = 1.32 x 10^-6
And, if we accept the age of the Milky way at about 13.20 billion years, then 1.32 x 10^-6 x 13.2 x 10^9 years = 17.42 x 10^3 years or approximately 17,000 years In other words, the stars near the center would be about 17,000 years younger than those at the outer edge due to gravitational time dilation.
Now to address velocity time dilation: whereby stars moving faster with respect to the center will age more slowly. The speed of the outer stars of the Milky Way is about 210 km per second wrt to the center of the galaxy. From special relativity we get t’ = t (1 – v^2/c^2)^-1/2 Using binomial expansion to simplify we get: t’ = t (1 + ½ v^2/c^2) and rearranging for ∆t/t, as we did before, ∆t/t = ½ v^2/c^2 = ½ (2.1 x 10^5/3 x 10^8)^2 = 2.45 x 10^-7
And 2.45 x 10^-7 x 13.2 x 10^9 years = 3234 years less time passed because of the star's motion. So, the net time dilation for a star at the edge of the galaxy would be roughly equal to the years gained by gravitational time dilation minus years lost due to velocity time dilation which, rounding off to the nearest thousand, is about 17,000 – 3000 = 14,000 years 921