How much lighter would someone be on the Jupiter-facing side of Io? 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.
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 &quot;smaller&quot; 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