<p>No, this is not possible. Equatorial coordinates only specify the <em>direction</em> of a certain point in space (as seen from Earth), not the distance. During a solar eclipse, the Moon and the Sun have (approximately) the same equatorial coordinates, but the distance to the Sun is 0 or almost 1 AU.</p>
<p>It <em>is</em> possible to define <a href="https://en.wikipedia.org/wiki/Celestial_coordinate_system#Coordinate_systems" rel="nofollow noreferrer">equatorial or ecliptic coordinates with the Sun as center</a>, but even then there is no general formula. You'll need the full position (i.e. three coordinates) of the Earth (relative to the Sun) or vice versa, <strong>and</strong> the full position of the object under observation to be able to convert the coordinates.</p>

CC BY-SA 4.0
2020-07-02T21:20:04.590Z
Thanks for trying to help!  Good counter example of the sun/moon on an eclipse.
user2191367
CC BY-SA 4.0
2020-08-10T16:06:54.640Z
Hi, I have a follow-up question that I believe is answerable...
I would like to convert Ra/Dec to Heliocentric Longitude.  I realize from your reply that I can't get the Heliocentric Radius, but I would still like to convert Ra/Dec to the angle around the sun.  How would I do that?
user2191367
CC BY-SA 4.0
2020-08-10T16:26:31.620Z
That depends on where you put the origin/center of your heliocentric system.
Glorfindel
CC BY-SA 4.0
2020-08-10T16:56:50.697Z
It would be with respect to the Sun being at the center of the circle, and the angle given by the calculation would be the angle around the Sun.
Where would i be able to find the math to make that calculation of Ra/Dec -> angle around the Sun?
user2191367
CC BY-SA 4.0
2020-08-10T17:11:56.650Z
That still depends entirely on where you put the 0 angle. There is no standard definition, except perhaps the [galactic coordinate system](https://en.wikipedia.org/wiki/Galactic_coordinate_system). And it will only work for objects that are far away because of the [stellar parallax](https://en.wikipedia.org/wiki/Stellar_parallax).
Glorfindel
CC BY-SA 4.0
2020-08-10T17:25:59.510Z
True, the Zero angle needs to be defined, but I could always add whatever offset I'd need to put it where it belongs.  The overall objective is to add an 'arrow' pointing off the edge of my Solar System display to indicate what general direction a star / constellation / etc was when the user searched on it.  So it didn't have to be a perfectly exact value, just enough to get the arrow in the 'right' direction. 

i.e. the arrow would go on this display [link](https://www.hubblestars.com/resources/Solar%20System%20-%20Zoomed%20Out.png)
user2191367
CC BY-SA 4.0
2020-08-10T17:30:48.680Z
But then, why not use equatorial coordinates?
Glorfindel
CC BY-SA 4.0
2020-08-10T18:15:12.993Z
Short answer... because i barely know what i'm doing when it comes to astronomical math  ; )
But i see where you're going, Right Ascension basically gives me the 'general direction' I'm look for, right?
user2191367
CC BY-SA 4.0
2020-07-02T18:25:07.517Z
2020-07-02T18:25:07.517Z
Glorfindel
<p>Can anyone help me with the formulas to convert Equatorial coordinates (Right Ascension / Declination) into Heliocentric coordinates (Angle and Distance in relation to the Sun) ?</p>
<p>i.e. Ra/Dec of 6.7659/18.4563 ---&gt; 281.98 degrees and 0.4562 AUs</p>
<p>It feels like this is possible with no other inputs, but I can't figure it out.</p>

CC BY-SA 4.0
2020-07-02T18:24:45.813Z
Its not possible because the Earth moves relative to the sun. So you will need a date at least.  Moreover,  you need two angles (eg RA and Dec) to determine a direction.  You can't convert a direction into one angle and a distance.
James K
<solar-system><coordinate>
Calculation to Convert from Equatorial to Heliocentric coordinate systems
user2191367
# Calculation to Convert from Equatorial to Heliocentric coordinate systems

(Asked by: username_0 on 2020-07-02)

Can anyone help me with the formulas to convert Equatorial coordinates (Right Ascension / Declination) into Heliocentric coordinates (Angle and Distance in relation to the Sun) ?

i.e. Ra/Dec of 6.7659/18.4563 ---> 281.98 degrees and 0.4562 AUs

It feels like this is possible with no other inputs, but I can't figure it out.

**Tags:** solar-system, coordinate

> **Comments:**
> **username_1**: Its not possible because the Earth moves relative to the sun. So you will need a date at least.  Moreover,  you need two angles (eg RA and Dec) to determine a direction.  You can't convert a direction into one angle and a distance.  

---

## 1 answers

### Answer by username_2 on 2020-07-02. Score: 2 | Accepted answer

No, this is not possible. Equatorial coordinates only specify the *direction* of a certain point in space (as seen from Earth), not the distance. During a solar eclipse, the Moon and the Sun have (approximately) the same equatorial coordinates, but the distance to the Sun is 0 or almost 1 AU.

It *is* possible to define [equatorial or ecliptic coordinates with the Sun as center](https://en.wikipedia.org/wiki/Celestial_coordinate_system#Coordinate_systems), but even then there is no general formula. You'll need the full position (i.e. three coordinates) of the Earth (relative to the Sun) or vice versa, **and** the full position of the object under observation to be able to convert the coordinates.

> **Comments:**
> **username_0**: Thanks for trying to help!  Good counter example of the sun/moon on an eclipse.  
> **username_0**: Hi, I have a follow-up question that I believe is answerable... I would like to convert Ra/Dec to Heliocentric Longitude.  I realize from your reply that I can't get the Heliocentric Radius, but I would still like to convert Ra/Dec to the angle around the sun.  How would I do that?  
> **username_2**: That depends on where you put the origin/center of your heliocentric system.  
> **username_0**: It would be with respect to the Sun being at the center of the circle, and the angle given by the calculation would be the angle around the Sun. Where would i be able to find the math to make that calculation of Ra/Dec -> angle around the Sun?  
> **username_2**: That still depends entirely on where you put the 0 angle. There is no standard definition, except perhaps the [galactic coordinate system](https://en.wikipedia.org/wiki/Galactic_coordinate_system). And it will only work for objects that are far away because of the [stellar parallax](https://en.wikipedia.org/wiki/Stellar_parallax).  
> **username_0**: True, the Zero angle needs to be defined, but I could always add whatever offset I'd need to put it where it belongs.  The overall objective is to add an 'arrow' pointing off the edge of my Solar System display to indicate what general direction a star / constellation / etc was when the user searched on it.  So it didn't have to be a perfectly exact value, just enough to get the arrow in the 'right' direction.   i.e. the arrow would go on this display [link](https://www.hubblestars.com/resources/Solar%20System%20-%20Zoomed%20Out.png)  
> **username_2**: But then, why not use equatorial coordinates?  
> **username_0**: Short answer... because i barely know what i'm doing when it comes to astronomical math  ; ) But i see where you're going, Right Ascension basically gives me the 'general direction' I'm look for, right?  

---

