<p>I'm trying to convert a satellite's position given in latitude/longitude decimal degrees, altitude in meters, and time in milliseconds since epoch to right ascension and the declination.</p>
<p>I have found equations for lat/lon/time to ra/dec, but none that take altitude into account.</p>

CC BY-SA 4.0
2021-12-16T15:47:32.117Z
Height does not play a role: both coordinate systems are centred in the Earth's core. So just the angles change. If you want to go to an observer's horizon system, THEN you have to know the altitude and take parallax into account.
planetmaker
CC BY-SA 4.0
2021-12-16T15:53:43.910Z
@planetmaker thank you for the response.  I thought ra/dec were independent of observer location on the earth, but I think you're saying the ra/dec output of the function would change based on observer location?
MrForum
CC BY-SA 4.0
2021-12-16T17:24:55.777Z
I disagree with planetmaker… If a satellite is above, let’s say, New York City, at an altitude of 2,000 km, it won’t be at the same angular altitude as seen from, e.g., Boston, than another satellite that is also above NYC, but at an altitude of, e.g., 20,000 km.

It is question here of parallax; the different angle to an object depending whether you’re at one place or another on the Earth. Jean Meeus explains how to correct for it in his book *Astronomical Algorithms*.
Pierre Paquette
CC BY-SA 4.0
2021-12-16T17:36:56.893Z
@Pierre in the horizon system, thus any observer: yes. But longitude, latitude system is geocentric. So altitude doesn't matter. Not does it for ra, Dec. Converting those to local hour angle and elevation, that takes its geocentric distance into account. I can recommend the same book you do 
planetmaker
CC BY-SA 4.0
2021-12-16T17:44:41.317Z
@planetmaker I agree with you on that, but presumably, OP wants to know where to look in the sky from their location in order to be able to observe said satellite. *Then*, correcting for parallax is essential.
Pierre Paquette
CC BY-SA 4.0
2021-12-16T22:12:38.727Z
Can you add two details.  1) exactly where you found the equations for lat/long/time to ra/dec  If it is linkable, please link.  If it is quotable, please quote.  2) The observer location.  It is okay to say "centre of the Earth (the default theoretical observer)" or you can edit to say "I'm trying to convert a satellite's position given in latitude/longitude decimal degrees, altitude in meters, and time in milliseconds since epoch and an observer's position given as latitude, longitude and altitude to ra/dec"
James K
<coordinate><mathematics><satellite>
Converting a Satellite's position in Lat/Lon/Alt/Time to RA/Dec
MrForum
# Converting a Satellite's position in Lat/Lon/Alt/Time to RA/Dec

(Asked by: username_0 on 2021-12-16)

I'm trying to convert a satellite's position given in latitude/longitude decimal degrees, altitude in meters, and time in milliseconds since epoch to right ascension and the declination.

I have found equations for lat/lon/time to ra/dec, but none that take altitude into account.

**Tags:** coordinate, mathematics, satellite
> **Comments:**
> **username_1**: Height does not play a role: both coordinate systems are centred in the Earth's core. So just the angles change. If you want to go to an observer's horizon system, THEN you have to know the altitude and take parallax into account.  
> **username_0**: @planetmaker thank you for the response.  I thought ra/dec were independent of observer location on the earth, but I think you're saying the ra/dec output of the function would change based on observer location?  
> **username_2**: I disagree with planetmaker… If a satellite is above, let’s say, New York City, at an altitude of 2,000 km, it won’t be at the same angular altitude as seen from, e.g., Boston, than another satellite that is also above NYC, but at an altitude of, e.g., 20,000 km.  It is question here of parallax; the different angle to an object depending whether you’re at one place or another on the Earth. Jean Meeus explains how to correct for it in his book *Astronomical Algorithms*.  
> **username_1**: @Pierre in the horizon system, thus any observer: yes. But longitude, latitude system is geocentric. So altitude doesn't matter. Not does it for ra, Dec. Converting those to local hour angle and elevation, that takes its geocentric distance into account. I can recommend the same book you do   
> **username_2**: @planetmaker I agree with you on that, but presumably, OP wants to know where to look in the sky from their location in order to be able to observe said satellite. *Then*, correcting for parallax is essential.  
> **username_3**: Can you add two details.  1) exactly where you found the equations for lat/long/time to ra/dec  If it is linkable, please link.  If it is quotable, please quote.  2) The observer location.  It is okay to say "centre of the Earth (the default theoretical observer)" or you can edit to say "I'm trying to convert a satellite's position given in latitude/longitude decimal degrees, altitude in meters, and time in milliseconds since epoch and an observer's position given as latitude, longitude and altitude to ra/dec"  

---

## 0 answers

