<p>The paper states that the redshift of the star is z=6.2 so the scale factor is</p>
<p><span class="math-container">$$a=\frac 1{1+z}=0.13889$$</span></p>
<p>Integrating the Friedman equation (flat universe) we obtain the current distance</p>
<p><span class="math-container">$\Omega_{K_0} \approx 0$</span></p>
<p><span class="math-container">$$r=\frac c{H_0} \int_a^1 \frac{dx}{\sqrt{\Omega_{R_0}+\Omega_{M_0}x+\Omega_{\Lambda_0}x^4}}$$</span></p>
<p><span class="math-container">$c=299792458 \ m/s$</span></p>
<p>We use the values given by the Planck Collaboration</p>
<p><span class="math-container">$H_0=67.66 \ (km/s)/Mpc$</span></p>
<p><span class="math-container">$\Omega_{R_0}=9.18\cdot 10^{-5}$</span></p>
<p><span class="math-container">$\Omega_{M_0}=0.3110082$</span></p>
<p><span class="math-container">$\Omega_{\Lambda_0}=0.6889$</span></p>
<p>The result of the calculation is:</p>
<p><span class="math-container">$r=27.7525 \ Gly$</span></p>
<p>The initial distance is calculated by:</p>
<p><span class="math-container">$$r_0=r \cdot a=3.8545 \ Gly$$</span></p>
<p>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.</p>
<p>Best regards</p>

CC BY-SA 4.0
2022-04-12T08:04:46.130Z
Excellent answer!
pela
CC BY-SA 4.0
2022-04-21T01:33:59.953Z
Thanks!  I love you!  (Platonically...)
Kurt Hikes
CC BY-SA 4.0
2022-05-09T13:48:45.353Z
And x is proper distance, at each time period of the changing universe, right?
Kurt Hikes
CC BY-SA 4.0
2022-05-09T14:31:26.277Z
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....
Kurt Hikes
CC BY-SA 4.0
2022-04-12T06:21:01.613Z
2022-04-13T07:28:19.217Z
2022-04-13T07:28:19.217Z
Albert
<p>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...</p>
<p>How close to us was it when it emitted the light Hubble is just now detecting?</p>
<p>What is the formula, if there is (a relatively simple) one, for calculating this?
.
.
.
.</p>
<p>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....</p>

CC BY-SA 4.0
2022-04-08T17:32:39.050Z
This is probably wrong, but applying the redshift of 6.2 gives $\frac{13.7 \text{Gly}}{6.2}\approx 2.21 \text{Gly}$
WarpPrime
CC BY-SA 4.0
2022-04-11T20:17:16.660Z
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...
Kurt Hikes
<star><observational-astronomy><cosmology><expansion><redshift>
How far away from us was Earendel star when it emitted the light that is just now reaching us?
Kurt Hikes
# How far away from us was Earendel star when it emitted the light that is just now reaching us?

(Asked by: username_0 on 2022-04-08)

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....

**Tags:** star, observational-astronomy, cosmology, expansion, redshift
> **Comments:**
> **username_1**: This is probably wrong, but applying the redshift of 6.2 gives $\frac{13.7 \text{Gly}}{6.2}\approx 2.21 \text{Gly}$  
> **username_0**: 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...  

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## 1 answers

### Answer by username_2 on 2022-04-12. Score: 8 | Accepted answer

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

> **Comments:**
> **username_3**: Excellent answer!  
> **username_0**: Thanks!  I love you!  (Platonically...)  
> **username_0**: And x is proper distance, at each time period of the changing universe, right?  
> **username_0**: 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....  

---

