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Something about this seems a little incorrect to me but I can't quite put my finger on it.

The moon closest to Jupiter is Io, with a diameter of $3650$ km and a mass of $8.93 \times 10^{22}$ kg, and a distance of $420,000$ km away from Jupiter. Jupiter itself has a mass of $1.898 \times 10^{27}$ kg.

This means that at any point of Io's surface, the acceleration you would experience due to Io's gravity is approximately .45 m/s^2. The gravitational acceleration from Jupiter to the center of Io is approximately .72 m/s^2. Since the radius of Io is small compared to the radius of orbit, this value is close to the value at both sides of Io (smaller than the rounding I did, probably).

Does this really mean that objects on the near side of Io to Jupiter have a net gravitational attraction to Jupiter? Would a loose rock on the surface of Io start moving to Jupiter itself?

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  • $\begingroup$ Assuming that your gravitational acceleration numbers are correct, then yes it would appear that a loose rock on the near-side surface of Io should start moving to Jupiter itself - IF one were able to somehow freeze Io's motion and stop it from orbiting around Jupiter. $\endgroup$ – user93237 Jun 13 '17 at 19:35
  • $\begingroup$ @SamuelWeir It would not be enough to cancel Io's orbital motion. You would also have to prevent the moon from accelerating toward the planet. You must think about tidal acceleration in this problem. $\endgroup$ – dmckee --- ex-moderator kitten Jun 13 '17 at 20:18
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The thing that is tripping you up in your naive analysis (beside using the diameter rather than the radius) is that the whole moon is constantly accelerating toward Jupiter, which means that loose object will only fall off the surface of the moon if their net acceleration toward the planet is larger than that of the moon.1

That means you calculate the tidal acceleration due to the planet and compare that to the acceleration due to the moon. Such a calculation is used to determine the Roche limit for an orbiting body.

The tidal acceleration toward Jupiter on the planet facing side of Io is about \begin{align} a_\text{tidal} &= \frac{f_\text{tidal}}{m_\text{loose object}} \\ &= 2 r_{Io} G \frac{m_\text{Jupiter}}{r_\text{orbit}^3} \\ &= (3.65 \times 10^6\,\mathrm{m}) \left(6.67 \times 10^{-11}\,\mathrm{\frac{N\cdot m^2}{kg^2}}\right) \frac{(1.90 \times 10^{27}\,\mathrm{kg})}{(4.20 \times 10^8\,\mathrm{m})^3} \\ &= 6.2 \times 10^{-3} \,\mathrm{m/s^2} \;, \end{align} which is much smaller than Io's surface gravity. I have assumed (incorrectly) that Io is spherical in this computation, and we have also (incorrectly) assumed that the orbit is circular. However, neither of those corrections make up a significant portion of the large difference.

1 Things can also fall of the back side if the moon accelerated toward the planet faster than the objects do, but as long as the moon is much smaller than the radius of the orbit we can use the same approximation to calculate the conditions under which that will happen.

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You accidentally plugged the diameter in for calculating $a_{Io}$ instead of the radius ($r_{Io}=1.825\times 10^6$ m). When plugging in the radius of Io for $a=G\frac{M}{r^2}$, you get $a_{Io}=1.79m/s^2$.

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