<p>The orientation of Earth wrt the stars repeats each sidereal day (thus ~23:56h). The visible night sky only changes each (synodic 24h) day due to the different direction the sun is at when considering the star's rest frame.</p>

<p>If you have the star's position from like <em>now</em>, you simply rotate Earth for each our by 360/(23.933)° = 15.04°.
If you want to do that for an arbitrary date, take the stars' position: they come with a datum like J2000 which means their position on January 1st 2000 UTC. C calculate the sidereal days passed since then, accounting for the incomplete rotation of the current day which gives the rotation around the Earth's axis as above.</p>

<p>If you want to go into more detail and longer time, and account for centuries and longer, you will need to take into account the nutation and presession of the Earth axis.</p>

CC BY-SA 4.0
2020-02-26T22:22:23.563Z
Does this mean that the Earth rotates by 360 degrees every 23h and 56 mins?
SidS
CC BY-SA 4.0
2020-02-27T04:45:19.117Z
Yes. Sidereal day is defined as same orientation with respect to stars
planetmaker
CC BY-SA 4.0
2020-02-27T15:53:43.037Z
Perfect, thank you very much! Also, how did you calculate the angle 23.933°?
SidS
CC BY-SA 4.0
2020-02-26T21:28:06.550Z
2020-02-26T21:28:06.550Z
planetmaker
<p>If north is up, right ascension should increase left to right on the near side of the celestial sphere.
The near side of the Earth should rotate left to right with a period of <a href="https://en.wikipedia.org/wiki/Sidereal_time" rel="nofollow noreferrer">23h56m04s</a>.
The prime meridian on Earth should align with RA 0h at <a href="https://www.esrl.noaa.gov/gmd/grad/solcalc/" rel="nofollow noreferrer">12:07 UT</a> on March 20 and 3m56s earlier per day after that.</p>

<p>The constellations may be more recognizable if you differentiate the stars by apparent magnitude and show only those brighter than magnitude 6 or so.
The scale is logarithmic: magnitude 1 is 2.51 times as bright as magnitude 2, etc.
A smaller factor, e.g. 1.6 per magnitude, may result in a nicer appearance.</p>

CC BY-SA 4.0
2020-02-26T22:25:57.610Z
So the Earth rotates clockwise by 360 degrees every 23h, 56mins and 4sec? Also thanks for the magnitude tip, I will take in account the logarithmic scale
SidS
CC BY-SA 4.0
2020-02-26T22:50:39.267Z
Clockwise as viewed from the south celestial pole, counterclockwise as viewed from the north celestial pole. 360 deg in 23:56:04.
Mike G
CC BY-SA 4.0
2020-02-27T15:50:23.347Z
Thanks! Also, do I need to take into account the Earth's tilt?
SidS
CC BY-SA 4.0
2020-02-27T15:56:44.130Z
The Earth's north pole should align with the north celestial pole. The tilt is relative to the plane of its solar orbit.
Mike G
CC BY-SA 4.0
2020-02-27T15:58:27.493Z
Ok, so I don't need to take into account the tilt of the Earth?
SidS
CC BY-SA 4.0
2020-02-27T15:58:52.770Z
Not unless you're modeling its orbit around the Sun.
Mike G
CC BY-SA 4.0
2020-02-27T15:59:20.813Z
Ah right perfect, thanks!
SidS
CC BY-SA 4.0
2020-02-27T15:59:27.787Z
Declination +-90° defines the rotation axis of Earth - and the stars. So tilt is only relevant when you want to look at long-term rotation where the orientation of the axis wrt the stars changes (precession, nutation).
planetmaker
CC BY-SA 4.0
2020-02-26T21:53:30.557Z
2020-02-26T21:53:30.557Z
Mike G
<p>In my simulation, I have stars scattered around Earth, all stars are the same distance from the center of the Earth and positioned based on their right ascension and declination like so:</p>

<p><a href="https://i.sstatic.net/skJxT.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/skJxT.png" alt="Simulation of stars around Earth"></a></p>

<p>I am wondering how the movement of Earth can be simulated based on the current time when running the simulation.</p>

<p>I am using the hyg v3 database which can be found <a href="https://github.com/astronexus/HYG-Database" rel="nofollow noreferrer">here</a>. I think that this dataset has the data necessary to find out the motion of stars, but I may be wrong.</p>

<p>Also, due to the structure of this simulation, with almost 120,000 stars, it is much more efficient to rotate/move the Earth rather than move all of the stars.</p>

<p>How can I show the stars visible over a location at a certain time by simulating the motion of the stars by rotating Earth? </p>

CC BY-SA 4.0
2020-02-26T21:13:40.933Z
The math is publicly known and as source available in programmes like stellarium
planetmaker
CC BY-SA 4.0
2020-02-26T22:23:42.363Z
@barrycarter - Hi! The source code will be available once I finish the project, I will pm you to inform once it’s public. Thanks!
SidS
CC BY-SA 4.0
2020-02-26T22:28:40.730Z
Hi @planetmaker, it would be great if you could link me to the math, I am quite new to astronomy so I am struggling a little :)
SidS
CC BY-SA 4.0
2020-02-27T04:58:22.670Z
I recommend Jean meeus -'Astronomical algorithms' https://www.amazon.com/Astronomical-Algorithms-Jean-Meeus/dp/0943396611 I used that with a few friends to write a program displaying the local sky several years ago
planetmaker
CC BY-SA 4.0
2020-02-27T15:59:05.160Z
Thanks, seems interesting, I will take a look at that.
SidS
<orbit><earth><star-systems><movement>
How Does Earth Move In Relation To The Stars
SidS
# How Does Earth Move In Relation To The Stars

(Asked by: username_0 on 2020-02-26)

In my simulation, I have stars scattered around Earth, all stars are the same distance from the center of the Earth and positioned based on their right ascension and declination like so:

[![Simulation of stars around Earth](https://i.sstatic.net/skJxT.png)](https://i.sstatic.net/skJxT.png)

I am wondering how the movement of Earth can be simulated based on the current time when running the simulation.

I am using the hyg v3 database which can be found [here](https://github.com/astronexus/HYG-Database). I think that this dataset has the data necessary to find out the motion of stars, but I may be wrong.

Also, due to the structure of this simulation, with almost 120,000 stars, it is much more efficient to rotate/move the Earth rather than move all of the stars.

How can I show the stars visible over a location at a certain time by simulating the motion of the stars by rotating Earth?

**Tags:** orbit, earth, star-systems, movement
> **Comments:**
> **username_1**: The math is publicly known and as source available in programmes like stellarium  
> **username_0**: @barrycarter - Hi! The source code will be available once I finish the project, I will pm you to inform once it’s public. Thanks!  
> **username_0**: Hi @planetmaker, it would be great if you could link me to the math, I am quite new to astronomy so I am struggling a little :)  
> **username_1**: I recommend Jean meeus -'Astronomical algorithms' https://www.amazon.com/Astronomical-Algorithms-Jean-Meeus/dp/0943396611 I used that with a few friends to write a program displaying the local sky several years ago  
> **username_0**: Thanks, seems interesting, I will take a look at that.  

---

## 2 answers

### Answer by username_1 on 2020-02-26. Score: 3

The orientation of Earth wrt the stars repeats each sidereal day (thus ~23:56h). The visible night sky only changes each (synodic 24h) day due to the different direction the sun is at when considering the star's rest frame.

If you have the star's position from like *now*, you simply rotate Earth for each our by 360/(23.933)° = 15.04°.
If you want to do that for an arbitrary date, take the stars' position: they come with a datum like J2000 which means their position on January 1st 2000 UTC. C calculate the sidereal days passed since then, accounting for the incomplete rotation of the current day which gives the rotation around the Earth's axis as above.

If you want to go into more detail and longer time, and account for centuries and longer, you will need to take into account the nutation and presession of the Earth axis.

> **Comments:**
> **username_0**: Does this mean that the Earth rotates by 360 degrees every 23h and 56 mins?  
> **username_1**: Yes. Sidereal day is defined as same orientation with respect to stars  
> **username_0**: Perfect, thank you very much! Also, how did you calculate the angle 23.933°?  

---

### Answer by username_2 on 2020-02-26. Score: 3

If north is up, right ascension should increase left to right on the near side of the celestial sphere.
The near side of the Earth should rotate left to right with a period of [23h56m04s](https://en.wikipedia.org/wiki/Sidereal_time).
The prime meridian on Earth should align with RA 0h at [12:07 UT](https://www.esrl.noaa.gov/gmd/grad/solcalc/) on March 20 and 3m56s earlier per day after that.

The constellations may be more recognizable if you differentiate the stars by apparent magnitude and show only those brighter than magnitude 6 or so.
The scale is logarithmic: magnitude 1 is 2.51 times as bright as magnitude 2, etc.
A smaller factor, e.g. 1.6 per magnitude, may result in a nicer appearance.

> **Comments:**
> **username_0**: So the Earth rotates clockwise by 360 degrees every 23h, 56mins and 4sec? Also thanks for the magnitude tip, I will take in account the logarithmic scale  
> **username_2**: Clockwise as viewed from the south celestial pole, counterclockwise as viewed from the north celestial pole. 360 deg in 23:56:04.  
> **username_0**: Thanks! Also, do I need to take into account the Earth's tilt?  
> **username_2**: The Earth's north pole should align with the north celestial pole. The tilt is relative to the plane of its solar orbit.  
> **username_0**: Ok, so I don't need to take into account the tilt of the Earth?  
> **username_2**: Not unless you're modeling its orbit around the Sun.  
> **username_0**: Ah right perfect, thanks!  
> **username_1**: Declination +-90° defines the rotation axis of Earth - and the stars. So tilt is only relevant when you want to look at long-term rotation where the orientation of the axis wrt the stars changes (precession, nutation).  

---

