How much lighter would someone be on the Jupiter-facing side of Io?

Question

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.

Answer 1

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

Answer 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.)

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%.