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I suppose the answer would be a range as the Jupiter-facing side of Ganymede passed from Jupiter's light side to its shadow side. I'd also like to know how much light the sun projects onto the surface of Ganymede as well. I'm also interested in day night cycle characteristics.

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    $\begingroup$ The free program Stellarium does such simulations, but I'm not sure how accurate it is. $\endgroup$ – barrycarter Mar 4 '15 at 1:18
  • $\begingroup$ Apparently it is at least as accurate as my calculation. Beware though the difference between visual magnitude and bolometric flux - not a great problem here, though a Stellarium visual mag. will not include the intrinsic IR from Jupiter. $\endgroup$ – Rob Jeffries Mar 4 '15 at 19:59
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How much? Well how accurately do you need it? How do you want it quantifying? And in what wavelength range?

Jupiter scatters a fraction of its incident sunlight. It also has its own luminosity (predominantly in the infrared).

A quick calculation:

The solar constant (flux at 1 au) is about 1370 W/m$^{2}$. Jupiter is situated about 5.2 au from the Sun (it varies by about +/- 5%) and has a radius of 70,000 km. The albedo is about 0.34.

Thus it receives about $7.8\times 10^{17}$ W from the Sun and radiates about $2.6\times 10^{17}$ W back into space. Assuming this is done more-or-less isotropically into a hemmisphere, then Ganymede, at a distance of 1,070,000 km from Jupiter, receives a flux of only 0.1 W/m$^2$ multiplied by the fraction of the sunlit hemisphere that can be seen.

This compares with the $\sim 50$ W/m$^2$ it receives from the Sun!

This surprising result (to me) puts into perspective all the simulations you see of things in orbit around Jupiter. The planet is still pretty faint compared to the Sun.

I'd be grateful if someone could double-check the sums!

[The intrinsic infrared luminosity of Jupiter is less than a fifth of what it receives from the Sun, so this would increase, but not double the received power.]

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    $\begingroup$ Stellarium says "full Jupiter" on Ganymede would have magnitude -16.02 brightness, significantly brighter than our own full moon appears to us, but nowhere near as bright as the Sun (about -23.11 magnitude on Ganymede). I'm not sure how they compute this, but stellarium.org/doc/head/index.html may have more. Skimming that site, it appears they make an effort to use published papers to determine what the sky looks like. $\endgroup$ – barrycarter Mar 4 '15 at 15:23
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    $\begingroup$ @barrycarter They just scale the absolute magnitudes for the relative distances. I presume these are visual magnitudes, rather than bolometric, but since I assumed a unchanged reflection spectrum, they should be comparable. My factor of 500 is almost exactly 7 mag, precisely agreeing with Stellarium. Case closed. $\endgroup$ – Rob Jeffries Mar 4 '15 at 19:57
  • $\begingroup$ Looks good to me! $\endgroup$ – uhoh Apr 11 '17 at 14:33
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    $\begingroup$ @uhoh - Nice way to link to your answer that's one inch below your comment. Haha. $\endgroup$ – iMerchant Apr 12 '17 at 0:00
  • $\begingroup$ @iMerchant ya I know. If more answers are added over time, they can move up and down, so I try to use epoch-agnostic referencing. $\endgroup$ – uhoh Apr 12 '17 at 0:40
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This is a back-of-the-envelope analysis to double-check @RobJeffries' self-described case-closed answer posted (above).

Light received by Jupiter and Ganymede:

$$I_{Jup} = r_{Jup}^2$$ $$I_{Gan} = r_{Gan}^2$$

Fraction of Jupiter's received light received by Ganymede:

$$f = a_{Jup}\frac{1}{2}\frac{r_{Gan}^2}{4 R_{Gan}^2}$$

The $\frac{1}{2}$ because Ganymede is lit by Jupiter from the side, half is invisible from Earth. I'm likewise surprised that it's so small, until realizing that Jupiter is only 7.5 degrees wide in Ganymede's sky. It's those darn NASA images with the satellites dwarfed by Jupiter that's throwing us off perhaps

$$\frac{f \times I_{Jup}}{I_{Gan}} = \frac{a_{Jup}}{8}\left( \frac{r_{Jup}}{R_{Gan}} \right)^2$$

$$\frac{a_{Jup}}{8}\left( \frac{r_{Jup}}{R_{Gan}} \right)^2 = \frac{0.34}{8}\left( \frac{70,000}{1,070,400} \right)^2 \approx \text{1.8E-03}$$

Frighteningly close to @RobJeffries' answer. Yay for envelope backs!

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