<p>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</p>
<p>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.</p>
<p>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</p>
<p>In addition, you have less centrifugal force from the orbit, to be being in an orbit that is &quot;smaller&quot; by 1821.6km but has the same period.</p>
<p>Centrufugal at Io(core) = .71813N (as expected, at Io's center of mass, its orbital centrifugal force matches Jupiter's attraction, hence an orbit)<br />
Centrifugal at IO(jupiterface) = 0.7151N
Thus the difference in perceived weight of a 1kg mass is<br />
(.71813-0.7151) + (0.7244 - 0.71813) = 0.0093N<br />
Which is a difference of 1.295% , or <strong>1 part in 77</strong></p>

CC BY-SA 4.0
2021-02-18T12:21:18.113Z
Thank you, I've already been astonished.
Greenhorn
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2021-02-18T18:45:57.387Z
This answer is incorrect. Please see my comment under the question. Perhaps you can repair this answer.
David Hammen
CC BY-SA 4.0
2021-02-18T23:12:06.390Z
@DavidHammen So he answer you DON'T provide, is better than mine? How.... quaint.
PcMan
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2021-02-18T23:23:00.007Z
@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.
David Hammen
CC BY-SA 4.0
2021-02-19T06:12:47.930Z
@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.
Greenhorn
CC BY-SA 4.0
2021-02-19T06:16:12.140Z
Exactly. I'm quite willing to accept that I made a mistake. Even several. But show them. *don't* just say neener-neener your answer stinks, I know better, without actually saying what the correct answer is or what the error(s) were. THat is just rude.
PcMan
CC BY-SA 4.0
2021-02-22T18:40:45.960Z
@Greenhorn -- I have now posted an answer. Please reject this incorrect answer.
David Hammen
CC BY-SA 4.0
2021-02-18T11:55:33.650Z
2021-02-18T12:00:42.650Z
2021-02-18T12:00:42.650Z
PcMan
<p>To answer this question, one must first define what &quot;weight&quot; is. There are two competing definitions. One, which some call &quot;actual weight&quot;, is simply the vector sum of all of the gravitational forces acting on an object. The other, which some call &quot;apparent weight&quot; 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 &quot;weight&quot; while the first concept is viewed as a fiction.</p>
<p>I'll use the latter concept, that &quot;weight&quot; 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.</p>
<p>Consider a position on the surface of Io. I'll denote</p>
<ul>
<li><span class="math-container">$\mu_J = GM_J$</span> as Jupiter's gravitational parameter. This has a value of <span class="math-container">$126686534\,\text{km}^3/\text{s}^2$</span>.</li>
<li><span class="math-container">$\mu_I = GM_I$</span> as Io's gravitational parameter. This has a value of <span class="math-container">$5959.916\, \text{km}^3/\text{s}^2$</span>.</li>
<li><span class="math-container">$r$</span> 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 <span class="math-container">$1821.6\,\text{km}$</span>.</li>
<li><span class="math-container">$R$</span> as the distance from Io's center of mass to Jupiter's center of mass. I'll use the periapsis value of <span class="math-container">$420000\,\text{km}$</span> as that will represent the extreme.</li>
<li><span class="math-container">$\theta$</span> 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.</li>
<li><span class="math-container">$\hat r$</span> as the unit vector pointing from Io's center of mass to the point on the surface.</li>
<li><span class="math-container">$\vec r$</span> as the vector from Io's center of mass to the point on the surface. Note that the previous definitions make <span class="math-container">$\vec r = r\,\hat r$</span>.</li>
<li><span class="math-container">$\hat h$</span> as the unit vector pointing orthogonal <span class="math-container">$\hat r$</span> such that the vector <span class="math-container">$\vec R$</span> from  Io's center of mass to Jupiter's center of mass is <span class="math-container">$R\,(\cos\theta\,\hat r + \sin\theta\,\hat h)$</span>.</li>
<li><span class="math-container">$\vec l$</span> as the vector from the point in question on the surface to Jupiter's center of mass. The previous definitions make <span class="math-container">$\vec l = \vec R - \vec r = (R \cos\theta - r)\,\hat r + R\sin\theta\,\hat h$</span>.</li>
<li><span class="math-container">$l$</span> as the magnitude of <span class="math-container">$\vec l$</span> : <span class="math-container">$l^2 = R^2 - 2Rr\cos\theta + r^2$</span>.</li>
</ul>
<p>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 <span class="math-container">$\mu_J \vec l\,/\,l^3$</span> while the acceleration of Io as a whole toward Jupiter is <span class="math-container">$\mu_J \vec R\,/\,R^3$</span>. The tidal acceleration on the surface of Io due to Jupiter is thus</p>
<p><span class="math-container">$$\begin{aligned}a_\text{tidal}
&amp;= \mu_J \left(\frac{\vec l}{l^3} - \frac{\vec R}{R^3}\right) \\
&amp;= \mu_J \Biggl(\left(\frac{R \cos\theta - r}{(R^2 - 2Rr\cos\theta + r^2)^{3/2}} -\frac{\cos\theta}{R^2}\right)\hat r \,+ \\
&amp;\quad\quad\quad\left(\frac{R}{(R^2 - 2Rr\cos\theta + r^2)^{3/2}}-\frac{1}{R^2}\right)\sin\theta\,\hat h\Biggr)\end{aligned}$$</span></p>
<p>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 <span class="math-container">$2\mu_J r/R^3$</span>, 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 (<span class="math-container">$\theta\approx90^\circ$</span>), the vertical component is approximately <span class="math-container">$-\mu_J r/R^3$</span>: half the magnitude of the tidal effect at the sub-Jupiter point and its antipode, but pointing inward rather than outward.</p>
<p>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 <a href="https://en.wikipedia.org/wiki/Spaghettification" rel="nofollow noreferrer">spaghettification</a>.)</p>
<p>The question asks for a numerical value. That's easy to compute given the above. The gravitational acceleration due to Io itself is <span class="math-container">$\mu_I/r^2$</span>, which has a numerical value of <span class="math-container">$1.796\,\text{m}/\text{s}^2$</span>, directed inward. The tidal acceleration at the sub-Jupiter point is <span class="math-container">$\mu_J(1/(R-r)^2-1/R^2)$</span> which has a numerical value of <span class="math-container">$0.00627\,\text{m}/\text{s}^2$</span>, directed outward. This represents a reduction in apparent weight by 0.35%.</p>

CC BY-SA 4.0
2021-02-23T07:08:56.190Z
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?.
Roger Wood
CC BY-SA 4.0
2021-02-23T09:06:32.837Z
@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](https://physics.stackexchange.com/a/141981/52112) 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.)
David Hammen
CC BY-SA 4.0
2021-02-23T09:15:26.957Z
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.
David Hammen
CC BY-SA 4.0
2021-02-23T18:41:27.230Z
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.
Roger Wood
CC BY-SA 4.0
2021-02-23T18:46:16.300Z
This popped up when I googled "Is Io in hydrostatic equilibrium?" https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2000JE001367
Roger Wood
CC BY-SA 4.0
2021-02-22T18:22:03.190Z
2021-02-22T18:22:03.190Z
David Hammen
<p>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.</p>

CC BY-SA 4.0
2021-02-18T11:23:27.993Z
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](https://www.forbes.com/sites/jamiecartereurope/2021/02/15/yes-pluto-is-a-planet-says-nasa-scientist-at-the-site-of-its-discovery-91-years-ago-this-week/)
uhoh
CC BY-SA 4.0
2021-02-18T11:30:50.977Z
@uhoh Thank you. Hopefully, the IAU will realize all this but I fear they aren't much keen about scientific correctness.
Greenhorn
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2021-02-18T11:33:28.123Z
@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.
Greenhorn
CC BY-SA 4.0
2021-02-18T11:49:22.397Z
Can't you simply take the gravity formula and the relevant mass, radi and distances?
Alchimista
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2021-02-18T11:50:12.900Z
I have an answer, about 0.25% less on the near side, but I want to check my maths before posting because I'm not completely confident in the amount of rounding that I've done
James K
CC BY-SA 4.0
2021-02-18T11:51:19.823Z
@Alchimista Yes, but you need to consider that Io's orbit is a accelearting frame of reference, and the centrifugal force almost exactly balances Jupiter's gravity
James K
CC BY-SA 4.0
2021-02-18T11:57:25.040Z
@JamesK well OK but I don't see anything different than standard. Like it would be in a terrestrial case. I mean, I am not sure that in spite of the exotic location this belongs to Astronomy. It seems like an exercise of physics. Anyway is fine for me. I don't even know who OP is, so an answer can certainly be wrote. Check the rounding and we'll know the value :)
Alchimista
CC BY-SA 4.0
2021-02-18T12:22:41.193Z
@Alchimista No, I was just curious since Io is so close to a so massive planet. And why would you want to know who I am?
Greenhorn
CC BY-SA 4.0
2021-02-18T12:29:36.740Z
@Greenhorn because you won't ask for somebody doing a calculation for you, normally. I am not interested on who you are, I meant OP could be someone just starting. Normally one would ask your question in a different way, like I am doing so and so, are you folks spotting something wrong? All the calculations here are standard. This way you are obliged to ask again if instead of Io you choose the moon. This is reflected by the answer you got. You can't use it for anything else than Jupiter - Io.
Alchimista
CC BY-SA 4.0
2021-02-18T13:01:03.500Z
@Alchimista I've been curious about Jupiter-Io. Since this is the most extreme known example I guess (concerning spherical bodies), all other examples would have weaker tidal forces, such as Earth-Moon.
Greenhorn
CC BY-SA 4.0
2021-02-18T13:15:09.407Z
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. :)
PcMan
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2021-02-18T14:04:54.037Z
@Greenhorn yes I understand that
Alchimista
CC BY-SA 4.0
2021-02-18T18:45:23.023Z
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.
David Hammen
CC BY-SA 4.0
2021-02-18T20:28:36.033Z
I think David has done a more accuate analysis than me, so I'll wait till he's fixed his house and checked his numbers.  0.34% looks like the correct value rather than my 0.25% since I was playing fast and loose with rounding.
James K
CC BY-SA 4.0
2021-02-18T21:40:12.747Z
Regarding the linked article in the OP's profile, I have lost **all** respect for Dr. Alan Stern. *Stern’s argument against the decision to relate Pluto to “dwarf planet” status boils down to this: it doesn’t make scientific sense.* Stern knows full well that the distinction makes eminent sense; Stern was the first author of one the key papers that noted the marked distinction between the eight planets and the not-quite planets (aka dwarf planets). The only difference is that he wanted to call the eight planets "uberplanets" and the dwarf planets "unterplanets".
David Hammen
CC BY-SA 4.0
2021-02-19T06:10:43.067Z
@DavidHammen Just because there are other quite massive bodies in the planets' "neighbourhood" doesn't change what _they_ are _themselves_. Dr. Stern always made clear the unterplanets are planets too. Until 2006 it was clear a planet is a substellar, spheroidal and geologically differentiated body either in direct orbit around a star or rogue. However, they should have (re-)classified Ceres a planet long before 2006, as soon as stronger telescopes proved its sphericity.
Greenhorn
CC BY-SA 4.0
2021-02-22T18:36:34.030Z
@Greenhorn The problem with "sphericalness" as a metric is that there is no clear boundary at the lower end of the mass scale. It's such a lousy metric that no one has proposed a specific definition. A problem also exists at the upper end of the mass scale; super-Jupiters, brown dwarfs, and red dwarfs form a continuum. The IAU "solved" this latter issue by defining planets as objects that orbit the Sun. BTW, this topic (what is / is not a planet) is off-topic for this site. There's not one thing we can do, and you will find people on both sides of the debate here. We try to keep things calm.
David Hammen
CC BY-SA 4.0
2021-02-22T21:54:13.583Z
Related?: [Are we slightly lighter during the day and slightly heavier at night, owing to the force of the Sun's gravity?](https://physics.stackexchange.com/questions/121775/are-we-slightly-lighter-during-the-day-and-slightly-heavier-at-night-owing-to-t)
Keith McClary
<gravity><tidal-forces><io>
How much lighter would someone be on the Jupiter-facing side of Io?
Greenhorn
# How much lighter would someone be on the Jupiter-facing side of Io?

(Asked by: username_0 on 2021-02-18)

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.

**Tags:** gravity, tidal-forces, io

> **Comments:**
> **username_1**: 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](https://www.forbes.com/sites/jamiecartereurope/2021/02/15/yes-pluto-is-a-planet-says-nasa-scientist-at-the-site-of-its-discovery-91-years-ago-this-week/)  
> **username_0**: @username_1 Thank you. Hopefully, the IAU will realize all this but I fear they aren't much keen about scientific correctness.  
> **username_0**: @username_1 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.  
> **username_2**: Can't you simply take the gravity formula and the relevant mass, radi and distances?  
> **username_3**: I have an answer, about 0.25% less on the near side, but I want to check my maths before posting because I'm not completely confident in the amount of rounding that I've done  
> **username_3**: @username_2 Yes, but you need to consider that Io's orbit is a accelearting frame of reference, and the centrifugal force almost exactly balances Jupiter's gravity  
> **username_2**: @username_3 well OK but I don't see anything different than standard. Like it would be in a terrestrial case. I mean, I am not sure that in spite of the exotic location this belongs to Astronomy. It seems like an exercise of physics. Anyway is fine for me. I don't even know who OP is, so an answer can certainly be wrote. Check the rounding and we'll know the value :)  
> **username_0**: @username_2 No, I was just curious since Io is so close to a so massive planet. And why would you want to know who I am?  
> **username_2**: @username_0 because you won't ask for somebody doing a calculation for you, normally. I am not interested on who you are, I meant OP could be someone just starting. Normally one would ask your question in a different way, like I am doing so and so, are you folks spotting something wrong? All the calculations here are standard. This way you are obliged to ask again if instead of Io you choose the moon. This is reflected by the answer you got. You can't use it for anything else than Jupiter - Io.  
> **username_0**: @username_2 I've been curious about Jupiter-Io. Since this is the most extreme known example I guess (concerning spherical bodies), all other examples would have weaker tidal forces, such as Earth-Moon.  
> **username_4**: 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. :)  
> **username_2**: @username_0 yes I understand that  
> **username_5**: 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.  
> **username_3**: I think username_5 has done a more accuate analysis than me, so I'll wait till he's fixed his house and checked his numbers.  0.34% looks like the correct value rather than my 0.25% since I was playing fast and loose with rounding.  
> **username_5**: Regarding the linked article in the OP's profile, I have lost **all** respect for Dr. Alan Stern. *Stern’s argument against the decision to relate Pluto to “dwarf planet” status boils down to this: it doesn’t make scientific sense.* Stern knows full well that the distinction makes eminent sense; Stern was the first author of one the key papers that noted the marked distinction between the eight planets and the not-quite planets (aka dwarf planets). The only difference is that he wanted to call the eight planets "uberplanets" and the dwarf planets "unterplanets".  
> **username_0**: @username_5 Just because there are other quite massive bodies in the planets' "neighbourhood" doesn't change what _they_ are _themselves_. Dr. Stern always made clear the unterplanets are planets too. Until 2006 it was clear a planet is a substellar, spheroidal and geologically differentiated body either in direct orbit around a star or rogue. However, they should have (re-)classified Ceres a planet long before 2006, as soon as stronger telescopes proved its sphericity.  
> **username_5**: @username_0 The problem with "sphericalness" as a metric is that there is no clear boundary at the lower end of the mass scale. It's such a lousy metric that no one has proposed a specific definition. A problem also exists at the upper end of the mass scale; super-Jupiters, brown dwarfs, and red dwarfs form a continuum. The IAU "solved" this latter issue by defining planets as objects that orbit the Sun. BTW, this topic (what is / is not a planet) is off-topic for this site. There's not one thing we can do, and you will find people on both sides of the debate here. We try to keep things calm.  
> **username_6**: Related?: [Are we slightly lighter during the day and slightly heavier at night, owing to the force of the Sun's gravity?](https://physics.stackexchange.com/questions/121775/are-we-slightly-lighter-during-the-day-and-slightly-heavier-at-night-owing-to-t)  

---

## 2 answers

### Answer by username_4 on 2021-02-18. Score: 2 | Accepted answer

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**

> **Comments:**
> **username_0**: Thank you, I've already been astonished.  
> **username_5**: This answer is incorrect. Please see my comment under the question. Perhaps you can repair this answer.  
> **username_4**: @username_5 So he answer you DON'T provide, is better than mine? How.... quaint.  
> **username_5**: @username_4 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.  
> **username_0**: @username_5 When you'll be able to post an answer, I ask you to elaborate on why username_4's answer is wrong and you are right, and then I can deaccept username_4's one and accept yours.  
> **username_4**: Exactly. I'm quite willing to accept that I made a mistake. Even several. But show them. *don't* just say neener-neener your answer stinks, I know better, without actually saying what the correct answer is or what the error(s) were. THat is just rude.  
> **username_5**: @username_0 -- I have now posted an answer. Please reject this incorrect answer.  

---

### Answer by username_5 on 2021-02-22. Score: 2

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](https://en.wikipedia.org/wiki/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%.

> **Comments:**
> **username_7**: 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?.  
> **username_5**: @username_7 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](https://physics.stackexchange.com/a/141981/52112) 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.)  
> **username_5**: 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.  
> **username_7**: 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.  
> **username_7**: This popped up when I googled "Is Io in hydrostatic equilibrium?" https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2000JE001367  

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