<p>If you know the length of a sideral day and the length of a year, you can determine the length of a solar day. </p>

<p>From <a href="https://en.wikipedia.org/wiki/Sidereal_time#Sidereal_days_compared_to_solar_days_on_other_planets" rel="nofollow noreferrer">Wikipedia</a>:</p>

<p><a href="https://i.sstatic.net/Tid6P.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Tid6P.png" alt="enter image description here"></a></p>

<p>After one year, there will have been one more solar day than sideral day. Therefore :</p>

<p>solar day = sideral day / (1 - (sideral day/orbital period))</p>

CC BY-SA 4.0
2020-01-29T13:18:38.720Z
From this I conclude that the farther the object is from the Sun, the smaller the difference between the length of the sidereal and of the solar day?
user30007
CC BY-SA 4.0
2020-01-29T14:45:40.420Z
@user30007 Yes, indeed.
usernumber
CC BY-SA 4.0
2020-01-29T12:57:34.250Z
2020-01-29T14:43:17.193Z
2020-01-29T14:43:17.193Z
usernumber
<p>When I only know the sidereal rotation period, how do I determine the true length of day on a body? </p>

<p>We know the Earth's sidereal day doesn't have full 24 hrs unlike the true length of day. It's similar on Mars where the sidereal day isn't as long as a Sol (about 24 hrs 40 min) either. So how do we calculate the true length of day on a planet?</p>

<p>Let's do it on the example of the planetoid Ceres: its sidereal rotation period is 9 hrs 4 min and 27 sec. How do we calculate the true length of day on Ceres?</p>

CC BY-SA 4.0
2020-01-29T12:40:31.180Z
You mean the mean Solar day? You would have to know the period around the star
Tosic
CC BY-SA 4.0
2020-01-29T17:33:03.813Z
Note that the "true" or apparent day length (the time from one high noon to the next) generally differs from that of the mean solar day. See https://en.wikipedia.org/wiki/Equation_of_time and https://physics.stackexchange.com/a/469884/123208
PM 2Ring
CC BY-SA 4.0
2020-01-30T06:29:15.297Z
@PM2Ring I was asking on the mean solar day however, what on Earth is 24 hrs and on Mars about 24 hrs 40 min.
user30007
<orbit><planet><rotation><mathematics>
How to calculate the true length of day on a celestial body?
user30007
# How to calculate the true length of day on a celestial body?

(Asked by: username_0 on 2020-01-29)

When I only know the sidereal rotation period, how do I determine the true length of day on a body?

We know the Earth's sidereal day doesn't have full 24 hrs unlike the true length of day. It's similar on Mars where the sidereal day isn't as long as a Sol (about 24 hrs 40 min) either. So how do we calculate the true length of day on a planet?

Let's do it on the example of the planetoid Ceres: its sidereal rotation period is 9 hrs 4 min and 27 sec. How do we calculate the true length of day on Ceres?

**Tags:** orbit, planet, rotation, mathematics

> **Comments:**
> **username_1**: You mean the mean Solar day? You would have to know the period around the star  
> **username_2**: Note that the "true" or apparent day length (the time from one high noon to the next) generally differs from that of the mean solar day. See https://en.wikipedia.org/wiki/Equation_of_time and https://physics.stackexchange.com/a/469884/123208  
> **username_0**: @PM2Ring I was asking on the mean solar day however, what on Earth is 24 hrs and on Mars about 24 hrs 40 min.  

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## 1 answers

### Answer by username_3 on 2020-01-29. Score: 3 | Accepted answer

If you know the length of a sideral day and the length of a year, you can determine the length of a solar day.

From [Wikipedia](https://en.wikipedia.org/wiki/Sidereal_time#Sidereal_days_compared_to_solar_days_on_other_planets):

[![enter image description here](https://i.sstatic.net/Tid6P.png)](https://i.sstatic.net/Tid6P.png)

After one year, there will have been one more solar day than sideral day. Therefore :

solar day = sideral day / (1 - (sideral day/orbital period))

> **Comments:**
> **username_0**: From this I conclude that the farther the object is from the Sun, the smaller the difference between the length of the sidereal and of the solar day?  
> **username_3**: @username_0 Yes, indeed.  

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