<p>This source doesn't give sufficient information to get the length of either a stellar or sidereal day.  The rotation rate of a body is not governed by the orbital parameters.  The Earth, for example, has a slowing spin rate as the Moon retreats due to tidal interactions.  This doesn't affect the Earth's orbital parameters.</p>
<p>The fields given are:</p>
<pre><code>Name        
Planet Status           
Discovered in       
Mass    
Mass*sin(i) 
Semi-Major Axis
Orbital Period
Eccentricity        
ω       
Tperi       
Radius
Inclination
Update  
Detection Method
Mass Detection Method       
Radius Detection Method     
Primary transit
Secondary transit
λ
Impact Parameter b
Time Vr=0       
Velocity Semiamplitude K
Calculated temperature
Measured temperature    
Hottest point longitude         
Geometric albedo        
Surface gravity log(g/gH)       
Alternate Names
</code></pre>

CC BY-SA 4.0
2021-01-05T16:26:03.263Z
Yeah that seems to be the conclusion. Thanks though!
OmniOwl
CC BY-SA 4.0
2021-01-05T16:16:42.020Z
2021-01-05T16:23:31.490Z
2021-01-05T16:23:31.490Z
Connor Garcia
<p>I am currently in the process of compiling a list of stars and their planets, given the information in the European Star Catalogue found <a href="http://exoplanet.eu/catalog/all_fields/" rel="nofollow noreferrer">here</a>, but I'm struggling to find out how to calculate a planets solar day. To be exact; How long, in earth hours, is the solar day of a planet found in this catalogue?</p>
<p>I know the catalogue is incomplete and that it's likely impossible to calculate the solar day for some of these planets due to this, but that's okay. I could also accept the sidereal day rather than the solar day.</p>
<p>I tried to use the formulas found <a href="http://www.celestialnorth.org/FAQtoids/dazed_about_days_(solar_and_sidereal).htm" rel="nofollow noreferrer">here</a> but I have a hard time finding out what numbers to plot into the equations, given the information available in the exoplanet catalogue. I've been doing a lot of googling and I'm not an astronomer, so it has been a bit of challenge!</p>
<p>I hope some of you can help me here.</p>

CC BY-SA 4.0
2021-01-05T14:42:14.573Z
You will need the planet's rotation speed and the diameter or circumference of the planet. Also if the planet is tidally locked, then you can just take the orbital period.
WarpPrime
CC BY-SA 4.0
2021-01-05T14:43:29.520Z
@fasterthanlight Could you elaborate what "tidally locked" means?
OmniOwl
CC BY-SA 4.0
2021-01-05T14:48:39.207Z
"Tidally locked" means an object that is always facing another object. For example, the Moon is tidally locked to the Earth as its rotational period is the same as the orbital period. Check this link: https://en.wikipedia.org/wiki/Tidal_locking
WarpPrime
CC BY-SA 4.0
2021-01-05T15:00:15.410Z
@fasterthanlight Alright, thanks! I have a planets radius in Jupiter Radius so getting a diameter is easy enough, but I'm not entirely sure how to derive the rotational speed given the information in that catalogue.
OmniOwl
CC BY-SA 4.0
2021-01-05T15:04:44.050Z
What information are you given?
WarpPrime
CC BY-SA 4.0
2021-01-05T15:08:01.880Z
There is quite a bit: Mass in MJup, Radius in RJup, Orbital Period in Days, Semi-Major Axis in AU, Orbital Inclination in degrees, angular dist. in arcsec, argument of perastron in degrees, epoch of periastron in JD, Conjonction date in JD, Primary Transit in JD, Secondary transit in JD, Sky-projected angle between planetary orbital spin and the stellar rotational spin in degrees, zero radial speed time in JD and Velocity semiamplitude in m/s
OmniOwl
CC BY-SA 4.0
2021-01-05T15:09:48.073Z
Usually you need rotational velocity to calculate it. You can check if it's tidally locked on other sites.
WarpPrime
CC BY-SA 4.0
2021-01-05T15:15:20.123Z
Yes, I understand that I'd need rotational velocity, but then since it's not provided, I'd have to derive it somehow given the information I have right? Considering that all planets listed are solid objects and not gas giants, while also not being tidally locked, how would I calculate the rotational velocity?
OmniOwl
CC BY-SA 4.0
2023-06-12T02:47:04.367Z
I don’t think we have achieved the technological prowess of being able to measure a planet’s rotation period… yet.
Pierre Paquette
<exoplanet><rotation><time>
Calculating the Solar Day of an Exoplanet?
OmniOwl
# Calculating the Solar Day of an Exoplanet?

(Asked by: username_0 on 2021-01-05)

I am currently in the process of compiling a list of stars and their planets, given the information in the European Star Catalogue found [here](http://exoplanet.eu/catalog/all_fields/), but I'm struggling to find out how to calculate a planets solar day. To be exact; How long, in earth hours, is the solar day of a planet found in this catalogue?

I know the catalogue is incomplete and that it's likely impossible to calculate the solar day for some of these planets due to this, but that's okay. I could also accept the sidereal day rather than the solar day.

I tried to use the formulas found [here](http://www.celestialnorth.org/FAQtoids/dazed_about_days_(solar_and_sidereal).htm) but I have a hard time finding out what numbers to plot into the equations, given the information available in the exoplanet catalogue. I've been doing a lot of googling and I'm not an astronomer, so it has been a bit of challenge!

I hope some of you can help me here.

**Tags:** exoplanet, rotation, time
> **Comments:**
> **username_1**: You will need the planet's rotation speed and the diameter or circumference of the planet. Also if the planet is tidally locked, then you can just take the orbital period.  
> **username_0**: @fasterthanlight Could you elaborate what "tidally locked" means?  
> **username_1**: "Tidally locked" means an object that is always facing another object. For example, the Moon is tidally locked to the Earth as its rotational period is the same as the orbital period. Check this link: https://en.wikipedia.org/wiki/Tidal_locking  
> **username_0**: @fasterthanlight Alright, thanks! I have a planets radius in Jupiter Radius so getting a diameter is easy enough, but I'm not entirely sure how to derive the rotational speed given the information in that catalogue.  
> **username_1**: What information are you given?  
> **username_0**: There is quite a bit: Mass in MJup, Radius in RJup, Orbital Period in Days, Semi-Major Axis in AU, Orbital Inclination in degrees, angular dist. in arcsec, argument of perastron in degrees, epoch of periastron in JD, Conjonction date in JD, Primary Transit in JD, Secondary transit in JD, Sky-projected angle between planetary orbital spin and the stellar rotational spin in degrees, zero radial speed time in JD and Velocity semiamplitude in m/s  
> **username_1**: Usually you need rotational velocity to calculate it. You can check if it's tidally locked on other sites.  
> **username_0**: Yes, I understand that I'd need rotational velocity, but then since it's not provided, I'd have to derive it somehow given the information I have right? Considering that all planets listed are solid objects and not gas giants, while also not being tidally locked, how would I calculate the rotational velocity?  
> **username_2**: I don’t think we have achieved the technological prowess of being able to measure a planet’s rotation period… yet.  

---

## 1 answers

### Answer by username_3 on 2021-01-05. Score: 1 | Accepted answer

This source doesn't give sufficient information to get the length of either a stellar or sidereal day.  The rotation rate of a body is not governed by the orbital parameters.  The Earth, for example, has a slowing spin rate as the Moon retreats due to tidal interactions.  This doesn't affect the Earth's orbital parameters.
The fields given are:
Name        
Planet Status           
Discovered in       
Mass    
Mass*sin(i) 
Semi-Major Axis
Orbital Period
Eccentricity        
ω       
Tperi       
Radius
Inclination
Update  
Detection Method
Mass Detection Method       
Radius Detection Method     
Primary transit
Secondary transit
λ
Impact Parameter b
Time Vr=0       
Velocity Semiamplitude K
Calculated temperature
Measured temperature    
Hottest point longitude         
Geometric albedo        
Surface gravity log(g/gH)       
Alternate Names

> **Comments:**
> **username_0**: Yeah that seems to be the conclusion. Thanks though!  

---

