Jupiter’s moon Io is tidally locked. It always has the same side facing Jupiter as it orbits. Does this mean that when you are on the “light” side of Io, the gravity is less than the far side since you are being slightly pulled by Jupiter’s gravity away from Io? Or does the fact that you’re on a body orbiting the main world negate the gravitational effect compared to the orbiting body?
Tidal effects are more of a squeeze than a pull. Ehther you were below Jupiter, or on the other side of Io, you would find yourself stretched (imperceptibly) to be taller. If you are on the terminator(if that is the right word for a planet moon system), where Jupiter is right on the horizon, you will find yourself squashed (once again, imperceptibly.)
Just look at the tides on Earth. You have two bulges, one close to being under the moon, and the other on the other side of the planet.
If you are on the Jupiter facing side of Io, then Io’s gravity is pulling you towards the center of Io, and Jupiter’s gravity is pulling you away from it.
If you are on the opposite side of Io, then both Io and Jupiter’s gravity are pulling you toward the center of Io.
So more gravity on the side facing away from Jupiter.
You are not taking into account the effects of being in orbit.
If you are on the side closer to Jupiter, then you are going slower than the speed you need to go in order to stay in orbit, and so are drawn towards the center. If you are on the far side of Io, you are traveling faster than the speed you need to go in order to stay in orbit, and so are pushed away from the center.
If you are closest to Jupiter, and I am on the far side, and suddenly Io disappear, we would be in different orbits, and tend to drift further apart.
Imperceptible, though. Io is 400 million miles from Jupiter (100,000 times the radius of Earth).
Jupiter is 300 times as massive as Earth. Gravity scales to the inverse square of distance and to mass
So, gravity of Jupiter at Io is 300/(10^10) of 1 gee = 30 billionths of a gee. Add or subtract that from the gravity of Io and you don’t get much.
That seems unlikely. Earth itself is averages a little over 90 million miles from the Sun. Maybe that’s the distance from Io to the Sun? Or the Earth?
Io orbits closer to 400 thousand miles above Jupiter. Jupiter will have something like 50-100 times the gravitational impact on Io as Earth does on the Moon.
Missed the edit:
As Kbfriender notes, the gravity of Jupiter also stretches Io a tiny amount, which means that the surface of Io on the near side is slightly closer to Jupiter than it otherwise would be, and the far side is slightly farther than it otherwise would be.
The orbital effects that kbfriender mention further reduce the effect of Jupiter’s gravity because the gravity of Jupiter affects all of Io in nearly the same way. I’d have to think about it a little more to see the total effect, but what I provided above is an upper bound that is so small as to not be worth worrying about.
You’re right of course - I messed up. Io is about a quarter million miles from Jupiter (if I’ve got it right now) which gives a gravity of 0.08 gee from Jupiter at Io.
But the orbital effect that Kbfriender knocks this down considerably.
The difference between the gravity on the near side and on the far side is about 4 billionths of a gee, unless I’ve made another silly mistake.
Ok. Got my head straight on this issue.
The gravitational effect on Io from Jupiter is what is making Io orbit. This is an inverse-square effect.
The tidal effect is about 4 billionth of a gee - and is outward from Io on both the near-Jupiter and far-from-Jupiter side (as kbfriender said), and inward towards Io around the part of Io where Jupiter would be on the horizon. This is an inverse-cube effect
There will be a small deviation from the inverse cube effect that would make gravity on the near-side ever-so-slightly different from the far-side gravity - but that would be a inverse-fourth power effect, and so, really, really small.
Not much, only the most spectacular volcanos in the Solar System!
That’s a more complex tidal effect where Jupiter is losing angular velocity, and that is being transferred to the orbital velocity of its satellites.
Io gets pushed away from Jupiter by the tidal effect, essentially its gravity is raising the atmosphere of Jupiter beneath it, and Jupiter, spinning faster than Io orbits, pulls Io along into a faster (higher) orbit, and then Io transfers that energy to Europa. In the process, it gets squeezed back and forth like taffy, heating the interior. It is the change of orbital eccentricity that causes the plastic deformations that lead to its current status.
If it was just Io in orbit, even at its distance from Jupiter, it would not be having nearly the same volcanic activity, if any at all.
Is Io uniformly dense? If not, might that override any differences between the near and far sides?
Giant steps are what you take.
It’s fairly uniform in density related to depth, but there’s probably quite a bit of chugging going on under the surface, and those mass movements would probably overwhelm the difference by orders of magnitude.
In the extreme case, when a body completely fills its Roche lobes (as many binary stars do), the side near to Jupiter would come to a point, while the far side would remain rounded.
So unless we have detailed information about those movements, the OP’s question is unanswerable, right?
Yeah. When two bodies are close, all the higher order effects that don’t usually matter much, matter a lot.
I hope my legs don’t break
If it were a perfect liquid under static forces, then all points on the surface would be in equilibrium, and have the same net forces on them. Io is not a perfect liquid, and is in a constantly changing set of forces though, so Io never reaches equilibrium, and so which has more will not stay constant.
Overall, over a long enough period with sensitive enough equipment, you should, I think find that you weigh slightly* less on the side facing Jupiter. Since Io is raising the atmospheric tides on Jupiter, that mass of air is what is pulling Io faster in its orbit, so there should be a net force in that direction, so it really wouldn’t be directly under Jupiter, but rather, slightly** in the direction of Io’s orbit on the equator.
*for extreme values of slightly
** for some value of slightly
I think that it’d be the other way around: The equipotential surfaces are slightly further apart on the near side, so the field should be slightly weaker, so you should weigh less on the near side.
But, again, it’d be a tiny effect.