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Io is the Galilean moon closest to Jupiter. Its average surface gravity is 0.183g and it has a tidal lock with Jupiter. So if you stood on the near side of Io at the equator, how much lighter would you be compared to average (say you weigh 18.3 pounds on Io, how much would you weigh due to Jupiter's gravity on Io's near side)? Jupiter has 318 Earth masses, this is why I think you'd be quite lighter on Io's near side on the equator.

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    $\begingroup$ Welcome to Astronomy SE, interesting article linked in your profile! Yes, Pluto Is A Planet Says NASA Scientist At The Site Of Its Discovery 91 Years Ago This Week $\endgroup$
    – uhoh
    Feb 18 at 11:23
  • $\begingroup$ @uhoh Thank you. Hopefully, the IAU will realize all this but I fear they aren't much keen about scientific correctness. $\endgroup$
    – Greenhorn
    Feb 18 at 11:30
  • $\begingroup$ @uhoh Io, like the Earth's moon, is a terrestrial body with an Earthlike differentiation, and this is why it has a magma mantle, and therefore, volcanism. If it was an icy body it would have water cryovolcanism like Enceladus. $\endgroup$
    – Greenhorn
    Feb 18 at 11:33
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    $\begingroup$ For giggles I ran the same calc for Earth-Sun. On the equator at noon, facing the sun, you weigh about 1/80 000th less. Much, much more due to the Earth's spin reducing your "sun orbital" velocity, than from being closer to the Sun. :) $\endgroup$
    – PcMan
    Feb 18 at 13:15
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    $\begingroup$ The selected answer is incorrect. Please deselect it. The correct answer is that the presence of Jupiter decreases weight by about 0.34% (not 1.3%) at the sub-Jupiter point, decreases weight by about the same amount at the antipode (the point on Io furthest from Jupiter), and increases weight by about 0.17% at Io's leading and trailing points. Unfortunately, my house suffered damage during the cold weather; I don't have time to write a full answer. $\endgroup$ Feb 18 at 18:45
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On the leading or trailing points of IO, you are orbiting Jupiter at the same speed as the center of mass of Io, so you experience its gravity only

On the equator, facing Jupiter, you are 1821.6 km closer to Jupiter, but still orbiting with the same period. So there is a bit of Jupiter's gravity that is not countered by Io's orbital speed.

How much? Gravity from Jupiter is more by: Assume 1kg at Io core, vs 1kg at io-jupiterface G(core) = 0.71813N G(jupiterside) = 0.7244N Differnece = 0.00627N

In addition, you have less centrifugal force from the orbit, to be being in an orbit that is "smaller" by 1821.6km but has the same period.

Centrufugal at Io(core) = .71813N (as expected, at Io's center of mass, its orbital centrifugal force matches Jupiter's attraction, hence an orbit)
Centrifugal at IO(jupiterface) = 0.7151N Thus the difference in perceived weight of a 1kg mass is
(.71813-0.7151) + (0.7244 - 0.71813) = 0.0093N
Which is a difference of 1.295% , or 1 part in 77

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  • $\begingroup$ Thank you, I've already been astonished. $\endgroup$
    – Greenhorn
    Feb 18 at 12:21
  • $\begingroup$ This answer is incorrect. Please see my comment under the question. Perhaps you can repair this answer. $\endgroup$ Feb 18 at 18:45
  • $\begingroup$ @DavidHammen So he answer you DON'T provide, is better than mine? How.... quaint. $\endgroup$
    – PcMan
    Feb 18 at 23:12
  • $\begingroup$ @PcMan My house has no water thanks to a burst pipe thanks to no heat for days. Fixing that is a higher priority than correcting a wrong answer on the internet. Give me a break. $\endgroup$ Feb 18 at 23:23
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    $\begingroup$ @DavidHammen When you'll be able to post an answer, I ask you to elaborate on why PcMan's answer is wrong and you are right, and then I can deaccept PcMan's one and accept yours. $\endgroup$
    – Greenhorn
    Feb 19 at 6:12
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To answer this question, one must first define what "weight" is. There are two competing definitions. One, which some call "actual weight", is simply the vector sum of all of the gravitational forces acting on an object. The other, which some call "apparent weight" is what can be measured by a local experiment such as an ideal spring scale or an ideal accelerometer. In general relativity, the latter concept is simply called "weight" while the first concept is viewed as a fiction.

I'll use the latter concept, that "weight" is what an ideal spring scale measures. This means one needs to address not only how much an object on the surface of Io is accelerating gravitationally toward Jupiter, but also how much Io as a whole is accelerating toward Jupiter. Since Io is close to a rigid body, it is the difference between these two accelerations that is felt.

Consider a position on the surface of Io. I'll denote

  • $\mu_J = GM_J$ as Jupiter's gravitational parameter. This has a value of $126686534\,\text{km}^3/\text{s}^2$.
  • $\mu_I = GM_I$ as Io's gravitational parameter. This has a value of $5959.916\, \text{km}^3/\text{s}^2$.
  • $r$ as the distance from Io's center of mass to the point on the surface. Ignoring Io's non-spherical shape, this has a mean value of $1821.6\,\text{km}$.
  • $R$ as the distance from Io's center of mass to Jupiter's center of mass. I'll use the periapsis value of $420000\,\text{km}$ as that will represent the extreme.
  • $\theta$ as the angle between the line from Io's center of mass to the point on the surface and the line from Io's center of mass to Jupiter's center of mass.
  • $\hat r$ as the unit vector pointing from Io's center of mass to the point on the surface.
  • $\vec r$ as the vector from Io's center of mass to the point on the surface. Note that the previous definitions make $\vec r = r\,\hat r$.
  • $\hat h$ as the unit vector pointing orthogonal $\hat r$ such that the vector $\vec R$ from Io's center of mass to Jupiter's center of mass is $R\,(\cos\theta\,\hat r + \sin\theta\,\hat h)$.
  • $\vec l$ as the vector from the point in question on the surface to Jupiter's center of mass. The previous definitions make $\vec l = \vec R - \vec r = (R \cos\theta - r)\,\hat r + R\sin\theta\,\hat h$.
  • $l$ as the magnitude of $\vec l$ : $l^2 = R^2 - 2Rr\cos\theta + r^2$.

Using Newton's law of gravitation, the acceleration of a test mass toward Jupiter at the point of interest on the surface of Io is $\mu_J \vec l\,/\,l^3$ while the acceleration of Io as a whole toward Jupiter is $\mu_J \vec R\,/\,R^3$. The tidal acceleration on the surface of Io due to Jupiter is thus

$$\begin{aligned}a_\text{tidal} &= \mu_J \left(\frac{\vec l}{l^3} - \frac{\vec R}{R^3}\right) \\ &= \mu_J \Biggl(\left(\frac{R \cos\theta - r}{(R^2 - 2Rr\cos\theta + r^2)^{3/2}} -\frac{\cos\theta}{R^2}\right)\hat r \,+ \\ &\quad\quad\quad\left(\frac{R}{(R^2 - 2Rr\cos\theta + r^2)^{3/2}}-\frac{1}{R^2}\right)\sin\theta\,\hat h\Biggr)\end{aligned}$$

Note that the horizontal component vanishes at 0°, 89.8°, and 180°. (The horizontal component is not quite zero at 90°, but it is very close to zero.) The vertical component at 0° and 180° are very close to the same value, approximately $2\mu_J r/R^3$, pointing outward. Tidal effects reduce the apparent weight of an object at the points where Jupiter is directly overhead and directly underfoot. At locations where Jupiter is on the horizon ($\theta\approx90^\circ$), the vertical component is approximately $-\mu_J r/R^3$: half the magnitude of the tidal effect at the sub-Jupiter point and its antipode, but pointing inward rather than outward.

The net effect is to stretch Io along the line connecting the centers of mass of Jupiter and Io, and to squeeze Io along its waist. This stretching and squeezing can tear bodies apart in more extreme gravitational circumstances. The greatest extremes occur near black holes, where closely items can be squeezed into strings of spaghetti and then torn apart. (The name for these extreme tidal effects is spaghettification.)

The question asks for a numerical value. That's easy to compute given the above. The gravitational acceleration due to Io itself is $\mu_I/r^2$, which has a numerical value of $1.796\,\text{m}/\text{s}^2$, directed inward. The tidal acceleration at the sub-Jupiter point is $\mu_J(1/(R-r)^2-1/R^2)$ which has a numerical value of $0.00627\,\text{m}/\text{s}^2$, directed outward. This represents a reduction in apparent weight by 0.35%.

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  • $\begingroup$ Io is tidally locked and presumably close to being in hydrostatic equilibrium (it looks pretty squishy). The ellipsoidal shape is quoted on Wikipedia with radii 1829.7, 1819.2, 1815.8 km. Presumably the first number is the elongation along the Io-Jupiter axis. 1829.7 km is 0.45% longer than the mean radius of 1821.6 km. So Io's field will be roughly 0.9% less (inverse square) at the sub-Jupiter point. This is bigger than the 0.35% from Jupiter's field and would add to it giving 1.25% lighter. Does this seem correct?. $\endgroup$
    – Roger Wood
    Feb 23 at 7:08
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    $\begingroup$ @RogerWood I didn't want to get into the details Io's non-spherical shape. This means that gravity on Io due to Io is not quite inverse square, and it means that your simple calculation is not quite correct. See, for example, this answer to the question "Why is Earth's gravity stronger at the poles?" The gravitational acceleration due to Io at Io's sub-Jupiter point will be somewhat greater than a simple inverse square relationship would indicate. (How much, I don't know.) $\endgroup$ Feb 23 at 9:06
  • $\begingroup$ Regarding Io being in hydrostatic equilibrium, that too is not quite correct. Any planet-like object that exhibits geological activity is not in hydrostatic equilibrium. (That includes the Earth.) Hydrostatic equilibrium is a nice approximation; I myself use it a lot. But is an approximation, just as is assuming spherical gravity. Io is a particularly interesting case. The tidal stresses induced by Jupiter act to circularize Io's orbit, while the Laplace resonances with Europa and Ganymede tend to make Io's orbit more elliptical. This creates some very interesting hysteresis loops. $\endgroup$ Feb 23 at 9:15
  • $\begingroup$ That's a really nice answer regarding the Earths gravity! I assume J2 takes the opposite sign for a prolate spheroid and the net effect is to similarly partially cancel the increased radius. The argument about geologic activity cuts both ways. It can both cause the equilibrium to be upset but can also provide the mechanism to ensure equilbrium. $\endgroup$
    – Roger Wood
    Feb 23 at 18:41
  • $\begingroup$ This popped up when I googled "Is Io in hydrostatic equilibrium?" agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2000JE001367 $\endgroup$
    – Roger Wood
    Feb 23 at 18:46

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