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No. The fact that the Moon is in orbit around the Earth means that the pen is in perpetual free-fall. As long as the Moon is in a stable orbit, the distance t

No. The fact that the Moon is in orbit around the Earth means that the pen is in perpetual free-fall. As long as the Moon is in a stable orbit, the distance to the Earth doesn't matter. Even if, at 200,000 miles, the Earth's gravity was stronger than that of the Moon's, the pen would still fall to the Moon. Think about it.

CoreDumpling 17y ago
Despite how readily we poke fun at the answers given on the linked page, I found it interesting that I had to drill down this far to find a satisfactory answer. I liked the explanation given by my high school physics teacher: in the end, it boils down to frame of reference -- since both the moon and the pen are in orbit (aka perpetual free-fall), the earth's gravity basically becomes "background" to the situation and has no effect on the relationship between the pen and the moon. It's not the concept of gravity, the mathematical formulas, or anything else that's really hard to grasp. Looking at things from the right frame of reference is hard, and not just in physics either. I struggle with this in my daily work, even for simple things like getting widgets to align correctly relative to each other.
Qz 17y ago
On second look at the question, it doesn't even reference the moon explicitly: "If a pen is dropped on a moon..."
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.
guelo 17y ago
I don't think so, it's because of the inverse square law. If you were traveling at the same speed as the moon and at the same distance from the earth you would not be in earth's orbit because you weigh approximately 0, while the moon has enough mass to actually exert some pull on the earth. So the moon's gravity is what's keeping you orbiting around earth, you're not orbiting around the earth only the moon is and it's taking you for a ride. By my back of the envelope calculation at 384,000km the earth's pull would be .0025m/s^2 while at the the moon's surface it's pull is 1.625m/s^2. So the earth's gravitational influence would be about 1/1000th of the moon's, so the moon would be the overwhelming force causing the pen to drop.

CoreDumpling 17y ago

Despite how readily we poke fun at the answers given on the linked page, I found it interesting that I had to drill down this far to find a satisfactory answer. I liked the explanation given by my high school physics teacher: in the end, it boils down to frame of reference -- since both the moon and the pen are in orbit (aka perpetual free-fall), the earth's gravity basically becomes "background" to the situation and has no effect on the relationship between the pen and the moon. It's not the concept of gravity, the mathematical formulas, or anything else that's really hard to grasp. Looking at things from the right frame of reference is hard, and not just in physics either. I struggle with this in my daily work, even for simple things like getting widgets to align correctly relative to each other.

Qz 17y ago

On second look at the question, it doesn't even reference the moon explicitly: "If a pen is dropped on a moon..."

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.

guelo 17y ago

I don't think so, it's because of the inverse square law. If you were traveling at the same speed as the moon and at the same distance from the earth you would not be in earth's orbit because you weigh approximately 0, while the moon has enough mass to actually exert some pull on the earth. So the moon's gravity is what's keeping you orbiting around earth, you're not orbiting around the earth only the moon is and it's taking you for a ride. By my back of the envelope calculation at 384,000km the earth's pull would be .0025m/s^2 while at the the moon's surface it's pull is 1.625m/s^2. So the earth's gravitational influence would be about 1/1000th of the moon's, so the moon would be the overwhelming force causing the pen to drop.