After seeing spectacular photographs of Jupiter, especially this one:

Shadow of IO on Jupiter by JUNO

I wondered if this would be visible with the naked eye from Jupiter orbit? Is the sunlight at Jupiter bright enough to see this in color, or would we merely see a very dim, low contrast, gray version of this once our rods had adapted to the ambient light?

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    $\begingroup$ For anyone wondering, the big black spot is a solar eclipse from Io $\endgroup$ Commented Oct 16, 2019 at 2:13
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    $\begingroup$ @BlueRaja-DannyPflughoeft Also, the shadow looks overly large relative to the planet which appears with exaggerated curvature for reasons of optics or image processing. I remember reading this is only 11% of Jupiter's surface. If you saw the shadow at that size, Jupiter's horizon would be relatively flat. $\endgroup$ Commented Oct 16, 2019 at 5:59
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    $\begingroup$ Wouldn't this question be more on topic in the astronomy stack exchange site than in this space exploration one? $\endgroup$
    – Aaron F
    Commented Oct 16, 2019 at 10:56
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    $\begingroup$ As far as I can tell, Jupiter appears orange to the naked eye from Earth. $\endgroup$ Commented Oct 16, 2019 at 13:07
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    $\begingroup$ @ point is : it looks colorful from Earth already. $\endgroup$ Commented Oct 16, 2019 at 15:22

4 Answers 4


The distance from Jupiter to the Sun is about 4.95 to 5.46 AU (Astronomical Unit, the distance of Earth to Sun). So the intensity of sunlight at Jupiter is about 1/25 of the intensity at Earth.

This light intensity is much more than neccessary to be seen in color by the naked eye enclosed in a space suit. About 1/1000 would be still bright enough.

Neptune is about 30 AU. Its intensity of 1/900 is enough for color vision.

In units of measure, the maximum illuminance on Earth's surface is about 130,000 Lux (Sun at zenith in Summer, clear sky). Above the atmosphere illuminance will be about 30 % higher.

So at Jupiter there are 5200 lx. That is comparable to a clouded sky at noon in winter with 6000 lx.

At Neptune there are 144 lx; a living room is about 100 to 300 lx. Street lighting is about 10 lx, enough to see the color of a red or yellow car.

All illuminance values are to be seen only as orders of magnitude. For human vision a factor of two for illuminance is felt as a small difference. We are able to see at a very wide range from 100,000 lx down to fractions of 1 lx.

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    $\begingroup$ Jupiter is the 4th brightest natural object in the sky (Sun, Moon, Venus), and even a powerful set of binoculars can resolve the colored bands from here on earth. If we can see that type of detail from here, then we can definitely see the banding and cloud structures from Jupiter orbit. $\endgroup$
    – Quietghost
    Commented Oct 15, 2019 at 18:45
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    $\begingroup$ You could add some comparison - the illumination available for the image above is (according to Wikipedia) about as much as on a "cloudy day in winter at noon" $\endgroup$
    – asdfex
    Commented Oct 15, 2019 at 21:15
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    $\begingroup$ @Taemyr yes, the albedo of Jupiter is 0.52, of Earth 0.367. I neglected the influence of the atmosphere of Earth, The illuminance values are just orders of magnitude. $\endgroup$
    – Uwe
    Commented Oct 16, 2019 at 10:32
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    $\begingroup$ Correct me if I’m wrong, but it sounds like the numbers you are describing would be the intensity of the Sun’s light from Jupiter, not the intensity of the Sun’s light reflected off of Jupiter as seen from orbit—which I imagine might be an order of magnitude or two lower (since Jupiter is far from a perfect reflector). You do comment that we have three orders of magnitude to play with, so the answer is probably still overall accurate (as, indeed, is suggested by @Quietghost’s comment), but the numbers provided don’t back it up perfectly it seems to me. $\endgroup$
    – KRyan
    Commented Oct 16, 2019 at 14:14
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    $\begingroup$ the maximum illuminance on Earth is about 130,000 Lux (Sun at zenith in Summer, clear sky) is that the intensity on the surface of Earth (after some has been lost to the atmosphere) - if we're trying to determine sunlight hitting Earth relative to sunlight hitting another planet, wouldn't we want to use intensity hitting the atmosphere, not the surface? $\endgroup$
    – dwizum
    Commented Oct 16, 2019 at 15:04

Jupiter is very bright and is one of the brightest things in the night sky when it is visible. Through even a small telescope (such as my own 100mm telescope) shades of dark brown, beige, cream and salmon pink are visible, even though 100x or 200x magnification dims the image. From Jupiter orbit, it would be a dazzling sight.


Uwe has provided a very good analysis of why Jupiter should be visible in color from Jupiter orbit, and Ags1 has pointed out that you can see the different colors through even a small telescope.

I wish to offer an even simpler demonstration that you would see the colors. Go out at night and look at Jupiter in the sky. You can use software such as Stellarium to figure out where it will be. You will probably perceive Jupiter as clearly being yellow, maybe very slightly orange.

Since it is bright enough to perceive the colors when seen from Earth, you should certainly be able to perceive them much closer, in Jupiter orbit.


Let's double check that the color of Jupiter will be visible over a large range of distances. (tl;dr it will!)

I calculated the relative brightness of Jupiter normalized to 5.2 AU the average distance from Earth assuming 100% illumination, and then the relative surface brightness.

This assumes the eye remains dark-adapted.

Anywhere beyond roughly 1.5 AU, Jupiter is unresolved by our eye (assuming a nominal 1 arcminute resolution) so the apparent surface brightness increases as we move closer to Jupiter, until we can start resolving the disk. After that the apparent surface brightness (intensity per unit solid area) doesn't increase, for the same reason that a wall does not get brighter as we walk closer to it.

So as we move from 6.2 AU to about 1.5 AU Jupiter's apparent surface brightness will only increase, and after that it remains roughly constant. So I propose to agree with several other answers that the colors of Jupiter will be visible anywhere from Earth to Jupiter orbit.

The four dots represent (from left to right):

location                                   distance (1E+06 km)
---------------------------------          -------------------
Io's orbit                                       0.42
Jupiter's Hill sphere (max orbit)               53.2
5.2 - 1 AU (inferior conjunction)              628.5 
5.2 + 1 AU (superior conjunction)              928.5

enter image description here

Python script for the plot

import numpy as np
import matplotlib.pyplot as plt

R_jupiter          = 69900.                # km
a_jupiter          = 778.5E+06             # km
a_io               = 421.8E+03             # km
AU                 = 150E+06               # km
G_Msun, GM_jupiter = 1.327E+20, 1.267E+17  # m^3/s^2

minres = (1/60.) * np.pi/180  # minimum resolution of human eye ~ 1 arcminute

R_Hill = a_jupiter * (GM_jupiter / (3.*G_Msun))**(1./3.)  

print('R_Hill for Jupiter (million km): ', R_Hill/1E+06)

print('Jupiter at 5.2 AU, minres (radians): ', 2*R_jupiter/a_jupiter, minres)

distance = np.logspace(np.log10(a_io), np.log10(a_jupiter+AU), 2001)
size     = 2*R_jupiter / distance

apparent_size               = np.sqrt(minres**2 + size**2)
relative_brightness         = (a_jupiter/distance)**2
apparent_surface_brightness = relative_brightness * (minres / apparent_size)**2

points                             = [a_jupiter+AU, a_jupiter-AU, R_Hill, a_io]
ipoints                            = [np.argmax(distance >= p) for p in points] # I'm lazy
relative_brightness_points         = relative_brightness[ipoints]
apparent_surface_brightness_points = apparent_surface_brightness[ipoints]

plt.plot(distance, relative_brightness)
plt.plot(points, relative_brightness_points, 'ok')
plt.plot(distance, apparent_surface_brightness)
plt.plot(points, apparent_surface_brightness_points, 'ok')
plt.text(3E+06, 0.8, 'apparent surface brightness', fontsize=16)
plt.text(3E+06, 1E+05, 'relative brightness', fontsize=16)
plt.xlabel('distance (km)', fontsize=16)

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