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Time according to the gravity of Sagittarius A?

They found the factor of time on the varying distances from the SMBH, I am curious on how to turn those factors into relatable terms such as: At 1 km from the event horizon ____ time passes on earth in an hour.

I'm just not sure how to figure this out.

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marked as duplicate by Carl Witthoft, Chappo, James K, peterh, Glorfindel Nov 21 '18 at 13:12

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    $\begingroup$ Hi, Blake, welcome to Astronomy Stack Exchange! I'm not sure what you mean by "reliable"; did you mean "relatable" instead - referring to more common units understandable on a human scale? Please add a bit more detail to your question to address that. Thanks! $\endgroup$ – HDE 226868 Nov 20 '18 at 4:36
  • $\begingroup$ Thank you and yes, common units. I just want to be able to take the factor of time and explain how much time would pass on earth at a certain distance from the event horizon of a black hole, in hours, days and/or years. I'm sure this is a simple question but I can't remember what to do with the factors. I'm not a mathematician. Haha $\endgroup$ – Blake Frazier Nov 20 '18 at 5:53
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    $\begingroup$ So what are you asking? Your link gives a very simple formula for time dilation as a function of radius (relative to the Schwartzchild radius). So look up the $r_s$ for the black hole of interest and you're done. $\endgroup$ – Carl Witthoft Nov 20 '18 at 15:43
  • $\begingroup$ @JamesK So 3500 hours pass on Earth as 1 hour passes 1km from the horizon. $\endgroup$ – PM 2Ring Nov 21 '18 at 3:17
  • $\begingroup$ I think @JamesK has it backwards and PM 2Ring has it right. $\endgroup$ – Phil N DeBlanc Nov 21 '18 at 7:42
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I am curious on how to turn those factors into relatable terms such as: At 1 km from the event horizon ____ time passes on earth in an hour.

The key here is to understand what the formula relates to. The (all important) observer is in this case a notional observer so distant that curvature (and hence gravitational effects) are negligible.

That observer is the "base clock" we reference. Everything is dilated relative to that clock.

That means that the dilation on Earth (not strictly speaking an ideal distant observer) has to be taken into account to get an answer in the form you mean. So starting with the formula you had :

$$t = t_\infty \sqrt{1 - \frac{r_s}{r}}$$

We want this relative to Earth so we need :

$$t_e = t_\infty \sqrt{1 - \frac{r_s}{r_e}}$$

And combining youy get :

$$t = t_e \sqrt{ \frac {1 - \frac {r_s} r} {1 - \frac {r_s} {r_e}} } = t_e \sqrt { \frac { r_e(r-r_s) } { r(r_e-r_s) } }$$

Now in the specific case of Earth and Sagitarius A*, $r_e$ is huge so the dilation factor for Earth is as close to one as makes no difference in practice, so you don't really need to go to all that trouble and you can use $t_\infty \approx t_e$.

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