How much more gravitational pull does the moon have then an ear ring on an ear?

[D H]

If I haven't made any errors, I get a factor of up to about 680 times, depending on where you are on Earth.

Looks good! I get a factor of 688 when the Moon is straight overhead, half that at moonrise/moonset.

 

[Tortise]

Gravity follows an inverse square law. Tidal effects do not. They follow an inverse cube law

The origional question is not of tidal force, but which is applying more force on the ear. If on a much bigger planet, or the sun - it wouldn't matter in the calculation of which was applying more force: a satellite the same average distance (from the ear) and mass of the moon, or the ear ring on the ear. If you take a sec. and think about it, I'm sure you can't help but to see this. But I see the point you are trying to make.

 

[D H]

The origional question is not of tidal force, but which is applying more force on the ear. If on a much bigger planet, or the sun - it wouldn't matter in the calculation of which was applying more force: a satellite the same average distance (from the ear) and mass of the moon, or the ear ring on the ear. If you take a sec. and think about it, I'm sure you can't help but to see this. But I see the point you are trying to make.

The orginal question was poorly worded. Clearly, one way to interpret the question is to look at the net forces on the ear with only with moon present versus with only the earing present.