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From what I've read, the look back time is about 12.9 billion light years, and the current distance to the Earendel star is approximately 28 billion light years...

How close to us was it when it emitted the light Hubble is just now detecting?

What is the formula, if there is (a relatively simple) one, for calculating this? . . . .

P.S. EDIT: Web pages like The University of Tennessee at Knoxville's (UTK.edu) and Ned Wright's clearly show lookback (light travel) times, and current distances, but not distance at the time the star or whatever was shining at us....

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  • $\begingroup$ This is probably wrong, but applying the redshift of 6.2 gives $\frac{13.7 \text{Gly}}{6.2}\approx 2.21 \text{Gly}$ $\endgroup$
    – WarpPrime
    Commented Apr 8, 2022 at 17:32
  • $\begingroup$ Thank you! Maybe that is correct.... I hope somebody else can weigh in.... Ned Wright's cosmology calculator says a red shift of 3 gives a distance at time of emission as about 2.1715 billion light years away... IF I am understanding his page correctly... $\endgroup$
    – Kurt Hikes
    Commented Apr 11, 2022 at 20:17

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The paper states that the redshift of the star is z=6.2 so the scale factor is

$$a=\frac 1{1+z}=0.13889$$

Integrating the Friedman equation (flat universe) we obtain the current distance

$\Omega_{K_0} \approx 0$

$$r=\frac c{H_0} \int_a^1 \frac{dx}{\sqrt{\Omega_{R_0}+\Omega_{M_0}x+\Omega_{\Lambda_0}x^4}}$$

$c=299792458 \ m/s$

We use the values given by the Planck Collaboration

$H_0=67.66 \ (km/s)/Mpc$

$\Omega_{R_0}=9.18\cdot 10^{-5}$

$\Omega_{M_0}=0.3110082$

$\Omega_{\Lambda_0}=0.6889$

The result of the calculation is:

$r=27.7525 \ Gly$

The initial distance is calculated by:

$$r_0=r \cdot a=3.8545 \ Gly$$

Approximately ~12.9 billion years ago, this distance of 3.85 billion light-years is what separated Earendel from the place where ~8.4 billion years later the Solar System would be born.

Best regards

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    $\begingroup$ Excellent answer! $\endgroup$
    – pela
    Commented Apr 12, 2022 at 8:04
  • $\begingroup$ Thanks! I love you! (Platonically...) $\endgroup$
    – Kurt Hikes
    Commented Apr 21, 2022 at 1:33
  • $\begingroup$ And x is proper distance, at each time period of the changing universe, right? $\endgroup$
    – Kurt Hikes
    Commented May 9, 2022 at 13:48
  • $\begingroup$ Also, why $x^4$ for lambda/dark energy? I've never seen anything to the power of four in the Friedmann equation(s) written elsewhere.... $\endgroup$
    – Kurt Hikes
    Commented May 9, 2022 at 14:31

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