<p>No it is saying that they will calculate the deflection of star light at the limb of the sun.</p>
<blockquote>
<p>If we plug in the numbers  (We will use the published values of G, M and c)</p>
</blockquote>
<blockquote>
<p>and use the fact that one radian is 206265 arcseconds (we will multiply the answer by this number to change the units)</p>
</blockquote>
<blockquote>
<p>we get for a radius of the sun ... (we will calculate the deflection at the limb of the sun)</p>
</blockquote>
<p>The radius used is not calculated, it is taken from the literature.</p>

CC BY-SA 4.0
2024-08-04T16:20:49.890Z
Thanks James. Yeah I was guess that.  "it is saying that they will calculate the deflection of star light at the limb of the sun.", but why should it be at the limb of the sun? It could be anywhere depending on the observer and target positions ,shouldn't it?
ABC
CC BY-SA 4.0
2024-08-04T21:37:08.690Z
You could calculate the deflection of light passing at any distance from the centre of the sun. Of course not for light that hits the sun! But it would be possible to calculate the deflection at 1 or 2 or 10 radii.  But light that just passes the limb of the sun will be deflected the most, so the value 1.75 arcsec is the maximum deflection of light by the gravity of the sun.
James K
CC BY-SA 4.0
2024-08-04T06:29:17.643Z
2024-08-04T06:29:17.643Z
James K
<p>On this website they describe how the &quot;deflection&quot; is measured during the solar eclipse.
<a href="https://eclipse2017.nasa.gov/testing-general-relativity" rel="nofollow noreferrer">https://eclipse2017.nasa.gov/testing-general-relativity</a></p>
<p>I'm confused about the following:<a href="https://i.sstatic.net/A25nOcS8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/A25nOcS8.png" alt="enter image description here" /></a></p>
<p>The last sentence, are they saying they measured the solar radius using the deflection? How?</p>

CC BY-SA 4.0
2024-08-04T06:26:03.740Z
Can you retype the text from the image.  Images don't get searched or indexed, and cause difficulty for screen readers.
James K
<observational-astronomy><general-relativity><gravitational-lensing>
Light deflection during solar eclipse
ABC
# Light deflection during solar eclipse

(Asked by: username_0 on 2024-08-04)

On this website they describe how the "deflection" is measured during the solar eclipse.
<https://eclipse2017.nasa.gov/testing-general-relativity>

I'm confused about the following:[![enter image description here](https://i.sstatic.net/A25nOcS8.png)](https://i.sstatic.net/A25nOcS8.png)

The last sentence, are they saying they measured the solar radius using the deflection? How?

**Tags:** observational-astronomy, general-relativity, gravitational-lensing
> **Comments:**
> **username_1**: Can you retype the text from the image.  Images don't get searched or indexed, and cause difficulty for screen readers.  

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## 1 answers

### Answer by username_1 on 2024-08-04. Score: 3 | Accepted answer

No it is saying that they will calculate the deflection of star light at the limb of the sun.

> If we plug in the numbers (We will use the published values of G, M and c)

> and use the fact that one radian is 206265 arcseconds (we will multiply the answer by this number to change the units)

> we get for a radius of the sun ... (we will calculate the deflection at the limb of the sun)

The radius used is not calculated, it is taken from the literature.

> **Comments:**
> **username_0**: Thanks James. Yeah I was guess that.  "it is saying that they will calculate the deflection of star light at the limb of the sun.", but why should it be at the limb of the sun? It could be anywhere depending on the observer and target positions ,shouldn't it?  
> **username_1**: You could calculate the deflection of light passing at any distance from the centre of the sun. Of course not for light that hits the sun! But it would be possible to calculate the deflection at 1 or 2 or 10 radii.  But light that just passes the limb of the sun will be deflected the most, so the value 1.75 arcsec is the maximum deflection of light by the gravity of the sun.  

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