<p>If the RA and Dec in question are geocentric, then they are relative to the <a href="https://en.wikipedia.org/wiki/Earth-centered_inertial" rel="nofollow noreferrer">ECI</a> origin.
The difference between heliocentric and geocentric positions is significant for any solar system object; on a baseline of 1.0 au, a trans-Neptunian object at 40 au has a parallax of 1.4°.</p>
<p>However, for near-Earth objects, the observer's location on Earth is also significant.
An asteroid passing at 0.01 au has a parallax of 0.24° on a baseline of 6370 km.
The Minor Planet Center lists <a href="https://minorplanetcenter.net/db_search" rel="nofollow noreferrer">observations</a> and generates <a href="https://minorplanetcenter.net/iau/MPEph/MPEph.html" rel="nofollow noreferrer">ephemerides</a> in topocentric coordinates.</p>

CC BY-SA 4.0
2021-04-25T02:34:53.650Z
2021-04-25T02:34:53.650Z
Mike G
<p>I'm new to the forum.
This might be a silly question. I had a confusion about how RA and Dec are calculated. So, from what I understand RA is the angle made by the position vector of an object(position vector as measured with respect to ECI's origin) with vernal equinox direction. Declination is the angle made by the same position vector with the celestial equator(coincides with Earth's equator).
I wanted to make sure that the position vector of the object is considered with respect to ECI's origin. If that is in fact the case, then wouldn't the RA and Dec of a Near-Earth Object depend on the position of the Earth in its orbit ? (If it's a far away star, then the effect of orbital motion of Earth is negligible). Please correct me if I'm fundamentally mistaken in my understanding. Thanks!</p>

CC BY-SA 4.0
2021-04-25T13:26:30.857Z
Astronomy SE and other Stack Exchange sites aren't forums. They are places where you can ask questions regarding a topic related to the site, as well as contribute to the community's repository of helpful information.
WarpPrime
<declination><right-ascension>
RA and Dec of Near-Earth-Objects
Prajwal
# RA and Dec of Near-Earth-Objects

(Asked by: username_0 on 2021-04-24)

I'm new to the forum.
This might be a silly question. I had a confusion about how RA and Dec are calculated. So, from what I understand RA is the angle made by the position vector of an object(position vector as measured with respect to ECI's origin) with vernal equinox direction. Declination is the angle made by the same position vector with the celestial equator(coincides with Earth's equator).
I wanted to make sure that the position vector of the object is considered with respect to ECI's origin. If that is in fact the case, then wouldn't the RA and Dec of a Near-Earth Object depend on the position of the Earth in its orbit ? (If it's a far away star, then the effect of orbital motion of Earth is negligible). Please correct me if I'm fundamentally mistaken in my understanding. Thanks!

**Tags:** declination, right-ascension
> **Comments:**
> **username_1**: Astronomy SE and other Stack Exchange sites aren't forums. They are places where you can ask questions regarding a topic related to the site, as well as contribute to the community's repository of helpful information.  

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## 1 answers

### Answer by username_2 on 2021-04-25. Score: 1

If the RA and Dec in question are geocentric, then they are relative to the [ECI](https://en.wikipedia.org/wiki/Earth-centered_inertial) origin.
The difference between heliocentric and geocentric positions is significant for any solar system object; on a baseline of 1.0 au, a trans-Neptunian object at 40 au has a parallax of 1.4°.

However, for near-Earth objects, the observer's location on Earth is also significant.
An asteroid passing at 0.01 au has a parallax of 0.24° on a baseline of 6370 km.
The Minor Planet Center lists [observations](https://minorplanetcenter.net/db_search) and generates [ephemerides](https://minorplanetcenter.net/iau/MPEph/MPEph.html) in topocentric coordinates.


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