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Hey, but couldn't the same argument be used to claim that tides shouldn't exist? The Earth orbits the Moon as well. That bothered me. So I checked Wikipedia:

by ced 17y ago

Hey, but couldn't the same argument be used to claim that tides shouldn't exist? The Earth orbits the Moon as well. That bothered me. So I checked Wikipedia: Thus, the tidal force depends not on the strength of the lunar gravitational field, but on its gradient (which falls off approximately as the inverse cube of the distance to the originating gravitational body).[4] [25] The solar gravitational force on the Earth is on average 179 times stronger than the lunar, but because the Sun is on average 389 times farther from the Earth, its field gradient is weaker So the distance does matter, it seems. Can anyone figure out under which conditions a pen doesn't fall to a moon? The rotational speed matters as well. If a planet were to turn fast enough that the centrifugal force is stronger than the gravitional force, it would shed matter, and a pen would be first to go. (Do such bodies exist? Anyone knows? Presumably, it would have to be solid to stand a chance) [This is all quite off-topic, of course. Yay for random thoughts on physics]

pmjordan 17y ago
So the distance does matter, it seems. Can anyone figure out under which conditions a pen doesn't fall to a moon? When there is a force holding the moon in place that doesn't act on the pen as well, or the pen is really far from the moon. Orbit is a special case, so assume earth, moon and pen start at rest. They will all start accelerating towards their combined centre of mass. The pen essentially doesn't contribute at all to the location of the centre of mass; If moon and earth had quasi-zero radius, the pen would hit the object first on whose side of the centre of mass (CM) it started on. (the objects' acceleration will be such that their individual CMs would hit the CM at exactly the same time, so the objects nearest the CM will accelerate slowest) earth - pen - CM - moon -> pen hits earth first. earth - CM - pen - moon -> pen hits moon first. Earth is bigger than the moon, which is much bigger than the pen, so this won't quite be true, but the pen will still have to be quite far from the moon to begin with for it to make a difference. (distance pen-moon vs. radius of moon vs. distance moon-earth) The original question is clearly the latter case. In fact the pen and the moon are so close compared to any third objects, and the pen's mass so irrelevant, that you can treat them as being in the moon's frame of reference. So unless there's a force acting on either the moon or the pen which isn't acting on the other (this can never be true for gravity), the pen will always drop to the moon. Because this is true in general, it is also true in orbit. Aside: Note that there cannot be any tides on the moon because the moon itself rotates around its own axis at the same rate as it rotates around earth. Also, the water involved in tides doesn't start floating off - it is still very much attracted to the earth, and the deformation of such a gigantic body of water is extremely slight - metres of deformation of a shell with a radius of ~6300km. You won't notice the pen's reaction to that sort of force.

by ced 17y ago

pmjordan 17y ago

So the distance does matter, it seems. Can anyone figure out under which conditions a pen doesn't fall to a moon? When there is a force holding the moon in place that doesn't act on the pen as well, or the pen is really far from the moon. Orbit is a special case, so assume earth, moon and pen start at rest. They will all start accelerating towards their combined centre of mass. The pen essentially doesn't contribute at all to the location of the centre of mass; If moon and earth had quasi-zero radius, the pen would hit the object first on whose side of the centre of mass (CM) it started on. (the objects' acceleration will be such that their individual CMs would hit the CM at exactly the same time, so the objects nearest the CM will accelerate slowest) earth - pen - CM - moon -> pen hits earth first. earth - CM - pen - moon -> pen hits moon first. Earth is bigger than the moon, which is much bigger than the pen, so this won't quite be true, but the pen will still have to be quite far from the moon to begin with for it to make a difference. (distance pen-moon vs. radius of moon vs. distance moon-earth) The original question is clearly the latter case. In fact the pen and the moon are so close compared to any third objects, and the pen's mass so irrelevant, that you can treat them as being in the moon's frame of reference. So unless there's a force acting on either the moon or the pen which isn't acting on the other (this can never be true for gravity), the pen will always drop to the moon. Because this is true in general, it is also true in orbit. Aside: Note that there cannot be any tides on the moon because the moon itself rotates around its own axis at the same rate as it rotates around earth. Also, the water involved in tides doesn't start floating off - it is still very much attracted to the earth, and the deformation of such a gigantic body of water is extremely slight - metres of deformation of a shell with a radius of ~6300km. You won't notice the pen's reaction to that sort of force.