<p>To really get exact times, you would have to take into account the opacity of the atmosphere, but I will be ignoring this here, and consider the light flux outside of the atmosphere of any planetary body. </p>

<p>There is no [planet] time for planets closer to the Sun than Earth, or for exoplanets orbiting brighter stars than Earth at the same distance. This  because they receive more light flux than we do. On a side note, there is an Earth time on Mercury. </p>

<p>For planets in further orbits than Earth around the Sun, [planet] time can be determined by comparing the solar flux at any place on Earth to the flux received by the planet at midday. </p>

<h1>Flux received by the planet at midday</h1>

<p>The average light flux on a planet can be determined based on the fact that the  flux varies inversely proportional to the square of the distance.
Outside Earth's atmosphere, the <a href="https://en.wikipedia.org/wiki/Solar_constant" rel="nofollow noreferrer">Solar flux is <span class="math-container">$S_0 = 1370$</span> Watts per square meter</a>, for a surface placed perpendicular to the Sun rays. </p>

<p>Since Earth is 1 AU from the Sun, the received flux for a planet at a distance of <span class="math-container">$x$</span> AU from the Sun is then given by</p>

<p><span class="math-container">$
F_{planet} = S_0 * \Big(\frac{1}{x}\Big)^2
$</span></p>

<h1>Solar intensity on Earth</h1>

<p>Based on <a href="https://dosbat.blogspot.com/2014/12/calculating-insolation-as-function-of.html" rel="nofollow noreferrer">this article</a>, the flux received from the Sun at a given location at latitude <span class="math-container">$\phi$</span> on Earth, and when the Sun is at declination <span class="math-container">$\delta$</span> in the sky can be determined to be </p>

<p><span class="math-container">$
F_{location} = \frac{S_0}{\pi} \Big(H \sin \phi \sin \delta  + sign(H) \cos \phi \cos \delta \Big)
$</span></p>

<p>where <span class="math-container">$H = \arccos \big( -\tan (\phi ) * \tan (\delta ) \big)$</span>, so <span class="math-container">$H/\pi$</span> is the length of a day. </p>

<h1>Putting it all together</h1>

<p>So given your latitude <span class="math-container">$\phi $</span> on Earth, you want to find the declination <span class="math-container">$\delta$</span> of the Sun by saying that <span class="math-container">$F_{location} = F_{planet}$</span>. </p>

<p>This gives you a trigonometric equation that you can solve numerically with a tool such as <a href="https://www.symbolab.com/solver/trigonometric-equation-calculator" rel="nofollow noreferrer">this</a>, or any other method you fancy. </p>

CC BY-SA 4.0
2020-02-21T16:36:03.473Z
Cool stuff!  Thanks for the answer!
Sam Washburn
CC BY-SA 4.0
2020-02-19T09:25:11.177Z
2020-02-20T09:09:08.223Z
2020-02-20T09:09:08.223Z
usernumber
<p>A while back NASA made a site called <a href="https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/plutotime/" rel="nofollow noreferrer">Pluto Time</a>, where you can enter your address and find that "...moment near dawn and dusk each day, where the illumination on Earth matches that of high noon on Pluto."</p>

<p>Is there a generic formula where we can calculate this for any body in the solar system?</p>

<p>How about any body around any star?</p>

CC BY-SA 4.0
2018-10-04T10:10:00.510Z
I wonder if this fits better on [Space.SE] than here?
gerrit
CC BY-SA 4.0
2018-10-04T10:17:52.313Z
Do you mean any body in place of Earth, or of Pluto in the definition of Pluto time?
Steve Linton
CC BY-SA 4.0
2018-10-04T12:01:21.690Z
@SteveLinton Any body in place of Pluto. To get an intuitive sense of the brightness.
Sam Washburn
CC BY-SA 4.0
2018-10-04T13:39:26.390Z
Obviously Mercury will be a bit of a problem, but otherwise that makes sense. Nice question.
Steve Linton
<the-sun><planet>
"Pluto Time" for any solar system body
Sam Washburn
# "Pluto Time" for any solar system body

(Asked by: username_0 on 2018-10-04)

A while back NASA made a site called [Pluto Time](https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/plutotime/), where you can enter your address and find that "...moment near dawn and dusk each day, where the illumination on Earth matches that of high noon on Pluto."

Is there a generic formula where we can calculate this for any body in the solar system?

How about any body around any star?

**Tags:** the-sun, planet
> **Comments:**
> **username_1**: I wonder if this fits better on [Space.SE] than here?  
> **username_2**: Do you mean any body in place of Earth, or of Pluto in the definition of Pluto time?  
> **username_0**: @SteveLinton Any body in place of Pluto. To get an intuitive sense of the brightness.  
> **username_2**: Obviously Mercury will be a bit of a problem, but otherwise that makes sense. Nice question.  

---

## 1 answers

### Answer by username_3 on 2020-02-19. Score: 4 | Accepted answer

To really get exact times, you would have to take into account the opacity of the atmosphere, but I will be ignoring this here, and consider the light flux outside of the atmosphere of any planetary body.

There is no [planet] time for planets closer to the Sun than Earth, or for exoplanets orbiting brighter stars than Earth at the same distance. This because they receive more light flux than we do. On a side note, there is an Earth time on Mercury.

For planets in further orbits than Earth around the Sun, [planet] time can be determined by comparing the solar flux at any place on Earth to the flux received by the planet at midday.

# Flux received by the planet at midday

The average light flux on a planet can be determined based on the fact that the flux varies inversely proportional to the square of the distance.
Outside Earth's atmosphere, the [Solar flux is $S\_0 = 1370$ Watts per square meter](https://en.wikipedia.org/wiki/Solar_constant), for a surface placed perpendicular to the Sun rays.

Since Earth is 1 AU from the Sun, the received flux for a planet at a distance of $x$ AU from the Sun is then given by

$
F\_{planet} = S\_0 \* \Big(\frac{1}{x}\Big)^2
$

# Solar intensity on Earth

Based on [this article](https://dosbat.blogspot.com/2014/12/calculating-insolation-as-function-of.html), the flux received from the Sun at a given location at latitude $\phi$ on Earth, and when the Sun is at declination $\delta$ in the sky can be determined to be

$
F\_{location} = \frac{S\_0}{\pi} \Big(H \sin \phi \sin \delta + sign(H) \cos \phi \cos \delta \Big)
$

where $H = \arccos \big( -\tan (\phi ) \* \tan (\delta ) \big)$, so $H/\pi$ is the length of a day.

# Putting it all together

So given your latitude $\phi $ on Earth, you want to find the declination $\delta$ of the Sun by saying that $F\_{location} = F\_{planet}$.

This gives you a trigonometric equation that you can solve numerically with a tool such as [this](https://www.symbolab.com/solver/trigonometric-equation-calculator), or any other method you fancy.

> **Comments:**
> **username_0**: Cool stuff!  Thanks for the answer!  

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