# “Pluto Time” for any solar system body

A while back NASA made a site called Pluto Time, 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?

• I wonder if this fits better on Space Exploration than here? – gerrit Oct 4 '18 at 10:10
• Do you mean any body in place of Earth, or of Pluto in the definition of Pluto time? – Steve Linton Oct 4 '18 at 10:17
• @SteveLinton Any body in place of Pluto. To get an intuitive sense of the brightness. – Sam Washburn Oct 4 '18 at 12:01
• Obviously Mercury will be a bit of a problem, but otherwise that makes sense. Nice question. – Steve Linton Oct 4 '18 at 13:39

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, 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, 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, or any other method you fancy.

• Cool stuff! Thanks for the answer! – Sam Washburn Feb 21 '20 at 16:36