There's a problem with (my understanding of) gravity.
January 17, 2009 8:20 AM Subscribe
What is the contradiction in the discussion of gravitational forces in the ...
It has been a very long time indeed since Physics 101, but even with a quick refresher on various web sites this afternoon, I can't understand the contradiction in this example:
We're all familiar with the physics class demonstration of two objects of different mass falling at the same rate (in a vacuum). In any particular gravitational field, the feather and the steel ball both drop side by side. No problem.
If we repeat the same experiment on the moon, which has a weaker gravitational field due to its smaller mass, the two objects will fall more slowly, but still side by side. No problem.
But if you look at this from the other direction, and think of the planetary body as 'falling' toward the objects - which presumably it is, albeit in an infinitesimal amount - then something seems to be wrong. The body with more mass - the earth - falls towards the objects faster than the body with less mass - the moon. More mass = more acceleration.
So which is it to be: objects of different mass accelerate at the same rate in a particular gravitational field, or different rates? What's going on?
It has been a very long time indeed since Physics 101, but even with a quick refresher on various web sites this afternoon, I can't understand the contradiction in this example:
We're all familiar with the physics class demonstration of two objects of different mass falling at the same rate (in a vacuum). In any particular gravitational field, the feather and the steel ball both drop side by side. No problem.
If we repeat the same experiment on the moon, which has a weaker gravitational field due to its smaller mass, the two objects will fall more slowly, but still side by side. No problem.
But if you look at this from the other direction, and think of the planetary body as 'falling' toward the objects - which presumably it is, albeit in an infinitesimal amount - then something seems to be wrong. The body with more mass - the earth - falls towards the objects faster than the body with less mass - the moon. More mass = more acceleration.
So which is it to be: objects of different mass accelerate at the same rate in a particular gravitational field, or different rates? What's going on?
We're all familiar with the physics class demonstration of two objects of different mass falling at the same rate (in a vacuum). In any particular gravitational field, the feather and the steel ball both drop side by side. No problem.
That's not exactly true. It's approximately true. It's the same thing with the Earth moving towards the feather and steel ball.
posted by mpls2 at 8:43 AM on January 17, 2009
That's not exactly true. It's approximately true. It's the same thing with the Earth moving towards the feather and steel ball.
posted by mpls2 at 8:43 AM on January 17, 2009
Each body's mass is a linear component of the attracting force between them.
Once you're measuring acceleration using the large mass as reference, though, you're dividing the attracting force by the smaller body's mass (F==m*a) -> (a==F/m)
The large (reference) body's mass still influences the perceived acceleration, the smaller one's is not.
posted by tigrrrlily at 8:47 AM on January 17, 2009
Once you're measuring acceleration using the large mass as reference, though, you're dividing the attracting force by the smaller body's mass (F==m*a) -> (a==F/m)
The large (reference) body's mass still influences the perceived acceleration, the smaller one's is not.
posted by tigrrrlily at 8:47 AM on January 17, 2009
The gravitational force that two objects exert upon one another (at a particular distance) is proportional to the product of their masses. It is the same amount of force upon each of the two objects.
Acceleration of an object due to a force is equal to that force divided by that object's mass.
Therefore, the acceleration of one object towards a second (at a particular distance) is directly proportional to the mass of the second object.
This is why a bowling ball and a baseball fall at the same rate (if at the same distance and in vacuum) - the rate at which they fall is dependent only upon the mass of the thing that they're falling towards.
By the same token, a steel ball causes the earth to accelerate towards it with exactly the same acceleration that it would cause the moon (were it at the same distance from the moon that it is from the earth).
So, the issue here seems to be your idea that "the body with more mass - the earth - falls towards the objects faster than the body with less mass - the moon". That's simply not true.
What is true is that the earth and the steel ball get closer to each other faster than the moon and the steel ball get closer to each other. But that doesn't mean that the earth moves towards the ball faster than the moon does; it means that the sum of the earth's movement towards the ball and the ball's movement towards the earth is faster than the sum of the moon's movement towards the ball and the ball's movement towards the moon.
And the difference between those two sums is exactly the difference between the ball's movement towards the moon and the ball's movement towards the earth.
Disclaimer: This is all Newtonian. Relativity is beyond me except in a vague sense.
posted by Flunkie at 8:48 AM on January 17, 2009
Acceleration of an object due to a force is equal to that force divided by that object's mass.
Therefore, the acceleration of one object towards a second (at a particular distance) is directly proportional to the mass of the second object.
This is why a bowling ball and a baseball fall at the same rate (if at the same distance and in vacuum) - the rate at which they fall is dependent only upon the mass of the thing that they're falling towards.
By the same token, a steel ball causes the earth to accelerate towards it with exactly the same acceleration that it would cause the moon (were it at the same distance from the moon that it is from the earth).
So, the issue here seems to be your idea that "the body with more mass - the earth - falls towards the objects faster than the body with less mass - the moon". That's simply not true.
What is true is that the earth and the steel ball get closer to each other faster than the moon and the steel ball get closer to each other. But that doesn't mean that the earth moves towards the ball faster than the moon does; it means that the sum of the earth's movement towards the ball and the ball's movement towards the earth is faster than the sum of the moon's movement towards the ball and the ball's movement towards the moon.
And the difference between those two sums is exactly the difference between the ball's movement towards the moon and the ball's movement towards the earth.
Disclaimer: This is all Newtonian. Relativity is beyond me except in a vague sense.
posted by Flunkie at 8:48 AM on January 17, 2009
If I read your question correctly, you are essentially asking the following:
Suppose I place object A a distance d from the Earth. How long will it take for the the two to collide? Now suppose I do the same with an object, B, having a larger mass. Will the time be the same?
The answer is no, because as you're hinting at, the objects are falling towards each other, and the force on the Earth will be greater from the heavier object B.
However, you shouldn't really worry about this. The differences in the times will be proportional to the differences in the masses themselves divided by the mass of the Earth. In other words:
delta t ~= t0 * (mA-mB) / Me
Where t0 is the time it would take if the Earth were pinned down to a point. Now, the Earth's mass is 6*10^24 kg. So for objects of masses on the order of kilograms, the difference is going to be extremely tiny. You'll never ever notice it, especially since we can only measure the gravitational constant to like 5 decimal places.
Further reading: Equivalence principle, Reduced mass.
posted by dsword at 9:21 AM on January 17, 2009 [1 favorite]
Suppose I place object A a distance d from the Earth. How long will it take for the the two to collide? Now suppose I do the same with an object, B, having a larger mass. Will the time be the same?
The answer is no, because as you're hinting at, the objects are falling towards each other, and the force on the Earth will be greater from the heavier object B.
However, you shouldn't really worry about this. The differences in the times will be proportional to the differences in the masses themselves divided by the mass of the Earth. In other words:
delta t ~= t0 * (mA-mB) / Me
Where t0 is the time it would take if the Earth were pinned down to a point. Now, the Earth's mass is 6*10^24 kg. So for objects of masses on the order of kilograms, the difference is going to be extremely tiny. You'll never ever notice it, especially since we can only measure the gravitational constant to like 5 decimal places.
Further reading: Equivalence principle, Reduced mass.
posted by dsword at 9:21 AM on January 17, 2009 [1 favorite]
So in other words: Saying that two objects of different mass will fall toward the earth at the same speed is not EXACTLY true, but it is close enough to true that it can be assumed for most purposes.
posted by Catfry at 9:42 AM on January 17, 2009
posted by Catfry at 9:42 AM on January 17, 2009
First of all, ignore almost completely what damn dirty ape said. If you drop a ball bearing on the earth, the earth also falls towards the ball bearing. No ifs ands or buts. Yes, the earth is falling around the sun, but this doesn't affect that basic fact.
Let us simplify however. Given any two objects with mass (Let's be Newtonian here, since that's really the simplest domain), they are attracted to one another with the force F = GmAmB/r2 where hopefully most terms in that are relatively obvious. Both objects are attracted, both of them move. However, when you have one object (say A) much larger than the other, since F = mAa, we see that its acceleration is much smaller. If mA >> mB, then we can pretty much pretend that object A does not move, even though it does just a wee bit.
Now. As for things falling at the same speed. This assumes a different setup, usually. This assumes that you have two objects in a uniform gravitational field of some kind, like the kind that almost exists on the surface of a large, massive object. It isn't quite uniform; the gravitational field, and the corresponding force which acts on objects decreases with altitude (i.e. as the radius, r, from the center increases), but for our purposes it is close enough. In such a case we still have the following:
F = GmAmB/r2
F = GmAmC/r2
However, as the usual changes in altitude are negligible compared to the size of the radius of the Earth, most of these terms are constant, and we really have
F = G'mB
F = G'mB
where G' = GmA/r2 is the same in both equations.
However, this is essentially identical to the equation F = ma from before; note that setting these equal to one another, the masses cancel and the acceleration of the two objects is the same, which is what we should expect.
The problem is that you seem to have confused these two situations. Two objects always attract one another, if they have mass. Two objects in a uniform gravitational field fall at the same rate. The situations are not the same, and should not be confused as such.
posted by vernondalhart at 9:55 AM on January 17, 2009
Let us simplify however. Given any two objects with mass (Let's be Newtonian here, since that's really the simplest domain), they are attracted to one another with the force F = GmAmB/r2 where hopefully most terms in that are relatively obvious. Both objects are attracted, both of them move. However, when you have one object (say A) much larger than the other, since F = mAa, we see that its acceleration is much smaller. If mA >> mB, then we can pretty much pretend that object A does not move, even though it does just a wee bit.
Now. As for things falling at the same speed. This assumes a different setup, usually. This assumes that you have two objects in a uniform gravitational field of some kind, like the kind that almost exists on the surface of a large, massive object. It isn't quite uniform; the gravitational field, and the corresponding force which acts on objects decreases with altitude (i.e. as the radius, r, from the center increases), but for our purposes it is close enough. In such a case we still have the following:
F = GmAmB/r2
F = GmAmC/r2
However, as the usual changes in altitude are negligible compared to the size of the radius of the Earth, most of these terms are constant, and we really have
F = G'mB
F = G'mB
where G' = GmA/r2 is the same in both equations.
However, this is essentially identical to the equation F = ma from before; note that setting these equal to one another, the masses cancel and the acceleration of the two objects is the same, which is what we should expect.
The problem is that you seem to have confused these two situations. Two objects always attract one another, if they have mass. Two objects in a uniform gravitational field fall at the same rate. The situations are not the same, and should not be confused as such.
posted by vernondalhart at 9:55 AM on January 17, 2009
So in other words: Saying that two objects of different mass will fall toward the earth at the same speed is not EXACTLY true, but it is close enough to true that it can be assumed for most purposes.No, it is exactly true.
What is not exactly true (but close enough for most purposes) is that the distance between one object and the earth will decrease at the same rate as the distance between a different object and the earth.
There's a distinction between what I said and what you said, and it's the heart of the matter:
The distance between an object and the earth decreases due to the sum of the object falling towards the earth and the earth falling towards the object.
And "the object falling towards the earth" is exactly the same as "some other object falling towards the earth". Not practically, but exactly. It is dependent only upon the mass of the earth (and distance, which we assume is the same).
So the difference between the rate at which one object and the earth approach each other and the rate at which another object and the earth approach each other is exactly the same as the difference between the rate at which the earth falls towards one object and the rate at which the earth falls towards the second object.
And both of those rates are virtually zero - on any human-noticeable scale, the earth doesn't even budge a smidgen, whether due to a baseball or a grand piano.
So there is no practical difference in the rates at which two objects and the earth approach each other, although there is an actual difference.
Bringing it back to your original question:
The rate at which the moon and a small ball approach each other is the sum of the rate at which the ball falls towards the moon and the rate at which the moon falls towards the ball.
The rate at which the earth and that small ball approach each other is the sum of the rate at which the ball falls towards the earth and the rate at which the earth falls towards the ball.
But the moon falls towards the ball at exactly the same rate as the earth falls towards the ball (again, if we're talking from the same distance).
So the difference between "the rate at which the moon and the ball approach each other" and "the rate at which the earth and the ball approach each other" is exactly the difference between "the rate at which the ball falls towards the moon" and "the rate at which the ball falls towards the earth".
But unlike in the case listed above, this difference is noticeable on a human scale.
posted by Flunkie at 9:57 AM on January 17, 2009
DSWord has it right. Whenever you drop something, it is being pulled towards the Earth, but the Earth doesn't stay exactly fixed - the earth "falls" a little towards the object as well. If you drop a heavier object, the Earth "falls" a little faster, so they actually will contact each other in a shorter amount of time, but the difference in times is extremely small.
It is true that all objects fall towards the earth at exactly the same rate (we call this rate "g", the "acceleration due to gravity). However, the earth falls towards the objects that we drop at a rate proportional to their mass. Part of the confusion here comes from reference frames. We tend to think of the earth as fixed, and the contact point occuring on the surface of the earth, but in reality, if you imagine some reference frame that is fixed to empty space, the earth and the dropped object both move in that reference frame. The contact point is just barely above the surface of the earth - the earth moves "up" towards this point as the object moves "down". For a heavier dropped object, the contact point is a little "higher", and since it moves towards the contact point at exactly the same rate as the lighter object, it gets there faster.
To confuse matters further, you'll get a different result if you drop the objects at the same time. In that case, it's their combined mass which is attracting the earth towards them, so they reach contact at exactly the same time.
posted by RobotNinja at 9:58 AM on January 17, 2009
It is true that all objects fall towards the earth at exactly the same rate (we call this rate "g", the "acceleration due to gravity). However, the earth falls towards the objects that we drop at a rate proportional to their mass. Part of the confusion here comes from reference frames. We tend to think of the earth as fixed, and the contact point occuring on the surface of the earth, but in reality, if you imagine some reference frame that is fixed to empty space, the earth and the dropped object both move in that reference frame. The contact point is just barely above the surface of the earth - the earth moves "up" towards this point as the object moves "down". For a heavier dropped object, the contact point is a little "higher", and since it moves towards the contact point at exactly the same rate as the lighter object, it gets there faster.
To confuse matters further, you'll get a different result if you drop the objects at the same time. In that case, it's their combined mass which is attracting the earth towards them, so they reach contact at exactly the same time.
posted by RobotNinja at 9:58 AM on January 17, 2009
To be more succinct, I think you have to think about the difference between "the ball moving towards the earth" and "the distance between the earth and the ball decreasing".
They're two fundamentally different things, and the confusion between the two seems to me to be the source of confusion behind your original question.
If I run towards you, that doesn't mean that the distance between you and me is decreasing at the same speed that I'm running. You might be running towards me, too. In fact, the distance between you and me might even be increasing, because you might be running away from me.
posted by Flunkie at 10:00 AM on January 17, 2009
They're two fundamentally different things, and the confusion between the two seems to me to be the source of confusion behind your original question.
If I run towards you, that doesn't mean that the distance between you and me is decreasing at the same speed that I'm running. You might be running towards me, too. In fact, the distance between you and me might even be increasing, because you might be running away from me.
posted by Flunkie at 10:00 AM on January 17, 2009
this doesn't need a complicated explanation for you to see where your contradiction arises. you've pretty much said it yourself in the statement of the question's premise:
"In any particular gravitational field, the feather and the steel ball both drop side by side. No problem.
If we repeat the same experiment on the moon, which has a weaker gravitational field due to its smaller mass"
(emphasis mine)
the resolution of your apparent contradiction is that the two situations are not equivalent; the gravitational field is different on the earth and on the moon.
posted by sergeant sandwich at 10:02 AM on January 17, 2009
"In any particular gravitational field, the feather and the steel ball both drop side by side. No problem.
If we repeat the same experiment on the moon, which has a weaker gravitational field due to its smaller mass"
(emphasis mine)
the resolution of your apparent contradiction is that the two situations are not equivalent; the gravitational field is different on the earth and on the moon.
posted by sergeant sandwich at 10:02 AM on January 17, 2009
If I run towards you, that doesn't mean that the distance between you and me is decreasing at the same speed that I'm running.
If it's just you and me, the only way to measure your "speed" is relative to me, which is the rate at which the distance between us is changing.
posted by mpls2 at 10:09 AM on January 17, 2009
If it's just you and me, the only way to measure your "speed" is relative to me, which is the rate at which the distance between us is changing.
posted by mpls2 at 10:09 AM on January 17, 2009
So in other words: Saying that two objects of different mass will fall toward the earth at the same speed is not EXACTLY true, but it is close enough to true that it can be assumed for most purposes.
No, it is exactly true.
Quite right, I'd like to say I deliberately misspoke so Flunkie could come along and explain this basic feature of the misunderstanding!
posted by Catfry at 10:20 AM on January 17, 2009
No, it is exactly true.
Quite right, I'd like to say I deliberately misspoke so Flunkie could come along and explain this basic feature of the misunderstanding!
posted by Catfry at 10:20 AM on January 17, 2009
If it's just you and me, the only way to measure your "speed" is relative to me, which is the rate at which the distance between us is changing.No, this is a misapplication of a flawed English interpretation of the scientific theory of relativity to this question.
Speed can be measured relative to a whole host of things, obviously, not merely relative to the two objects involved. An observer could be on Saturn; and she could validly say that the earth is moving towards the ball at speed X and the ball is moving towards the earth at speed Y; she is not limited to merely "they're approaching each other at speed X + Y and that's all that can possibly be said".
In a more practical example, one can validly say that one car hit the other at 60 MPH while the rate at which the distance between them was decreasing by 65 MPH. And there are practical differences here, too:
The kinetic energy of a head on collision between a car going 60 and a car going 5 is different than the kinetic energy of a collision between a car going 65 and a stationary car.
This is because the kinetic energy involved in the collision is the sum of the kinetic energy of the one car with the kinetic energy of the other car, and these summands are dependent upon the square of their respective velocities, not upon their velocities linearly (it's also dependent upon their masses, of course, but let's say their masses are the same, for ease of calculation).
So the kinetic energy in the first case is proportional to 60 * 60 + 5 * 5. That's 3625.
But the kinetic energy in the second case is 65 * 65 + 0 * 0, which is 4225. Which is not 3625.
posted by Flunkie at 10:25 AM on January 17, 2009
while the rate at which the distance between them was decreasing by 65 MPHI mean, "while the distance between them was decreasing by 65 MPH".
posted by Flunkie at 10:36 AM on January 17, 2009
Response by poster: "If we repeat the same experiment on the moon, which has a weaker gravitational field due to its smaller mass"
The resolution of your apparent contradiction is that the two situations are not equivalent; the gravitational field is different on the earth and on the moon.
But this is exactly the point of (my) confusion - the fact that the gravitational field is weaker when objects of smaller mass are involved. I have read through the explanations above as carefully as I can, but can't find the resolution. Let me try to state it again, in a 'simplified' environment - deep space, one observer, and no other reference but the four objects: 'earth' 'moon' feather and ball.
1a) Observer watches from earth. Feather comes into existence and is seen to move towards the earth at some particular rate.
2a) Observer watches from moon. Feather comes into existence and is seen to move towards the moon at a rate slower than in case 1a).
3a) Observer watches from feather. Earth comes into existence above us, and is seen to move towards us. Same rate as 1a)
4a) Observer watches from feather. Moon comes into existence above us, and is seen to move towards us. Same rate as 2a)
So we have a clear demonstration of the basic premise - systems of smaller mass (the sum of the masses of the pair involved) form a weaker gravitational field. Down to this point everything is OK.
But now let's start to run it again, with the steel ball instead of the feather:
1b) Observer watches from earth. Ball comes into existence and (our textbooks say) is seen to move towards the earth at same rate as 1a)
But the sum of the masses of the pair involved is greater than 1a) and thus a stronger gravitational field is formed. The pair must move together more quickly than in 1a).
So a ball falls faster than a feather! Yet when we put them side by side and drop them off the Leaning Tower (in a vacuum) they (are said to) fall at the same rate. What gives?
(My apologies if the answer is buried in the mathematics above. I don't have so much experience reading equations like that; if the 'answer' can be expressed textually, I would perhaps be able to grasp this more readily ...)
posted by woodblock100 at 3:36 PM on January 17, 2009
The resolution of your apparent contradiction is that the two situations are not equivalent; the gravitational field is different on the earth and on the moon.
But this is exactly the point of (my) confusion - the fact that the gravitational field is weaker when objects of smaller mass are involved. I have read through the explanations above as carefully as I can, but can't find the resolution. Let me try to state it again, in a 'simplified' environment - deep space, one observer, and no other reference but the four objects: 'earth' 'moon' feather and ball.
1a) Observer watches from earth. Feather comes into existence and is seen to move towards the earth at some particular rate.
2a) Observer watches from moon. Feather comes into existence and is seen to move towards the moon at a rate slower than in case 1a).
3a) Observer watches from feather. Earth comes into existence above us, and is seen to move towards us. Same rate as 1a)
4a) Observer watches from feather. Moon comes into existence above us, and is seen to move towards us. Same rate as 2a)
So we have a clear demonstration of the basic premise - systems of smaller mass (the sum of the masses of the pair involved) form a weaker gravitational field. Down to this point everything is OK.
But now let's start to run it again, with the steel ball instead of the feather:
1b) Observer watches from earth. Ball comes into existence and (our textbooks say) is seen to move towards the earth at same rate as 1a)
But the sum of the masses of the pair involved is greater than 1a) and thus a stronger gravitational field is formed. The pair must move together more quickly than in 1a).
So a ball falls faster than a feather! Yet when we put them side by side and drop them off the Leaning Tower (in a vacuum) they (are said to) fall at the same rate. What gives?
(My apologies if the answer is buried in the mathematics above. I don't have so much experience reading equations like that; if the 'answer' can be expressed textually, I would perhaps be able to grasp this more readily ...)
posted by woodblock100 at 3:36 PM on January 17, 2009
woodblock100:
Yes, that is technically correct. The [celestial body] and the object are accelerating towards each other at a rate derived from the mass of each of them (and the distance between them at any given instant). That's the F=G(m1)(m2)/r^2 bit, which is a good equation when dealing with massive objects and large distances.
When you talk about a falling object on a planet-sized thing, you clump a lot of terms together, because it's a good approximation. Specifically, r in the above equation doesn't change, on any scale which we care about. r does change, but in terms of the centre of a very big thing, and a very small thing which is very close to it, we don't worry.
In which case, we can say that G, m1, and r^2 are all constants. Then the equation is F = k m, where k= G(m1)/r^2 all of which are constant as far as we're concerned. And if the force is proportional to the mass, the acceleration is constant. Feathers, bowling balls, hammers, the lot.
So, yes, the Earth/Moon/Sun does fall towards the thing, it's just not worth worrying about. To illustrate the scale, the Earth doesn't orbit the Sun, the Sun and the Earth orbit around a common centre of gravity. That centre of gravity is SO far inside the Sun that you may as well say that the Earth orbits the Sun.
Very short answer: Planetary bodies are big (in terms of both size and mass) so we can treat their position as constant relative to small things like 'planes and buildings.
posted by Wrinkled Stumpskin at 4:51 PM on January 17, 2009
Yes, that is technically correct. The [celestial body] and the object are accelerating towards each other at a rate derived from the mass of each of them (and the distance between them at any given instant). That's the F=G(m1)(m2)/r^2 bit, which is a good equation when dealing with massive objects and large distances.
When you talk about a falling object on a planet-sized thing, you clump a lot of terms together, because it's a good approximation. Specifically, r in the above equation doesn't change, on any scale which we care about. r does change, but in terms of the centre of a very big thing, and a very small thing which is very close to it, we don't worry.
In which case, we can say that G, m1, and r^2 are all constants. Then the equation is F = k m, where k= G(m1)/r^2 all of which are constant as far as we're concerned. And if the force is proportional to the mass, the acceleration is constant. Feathers, bowling balls, hammers, the lot.
So, yes, the Earth/Moon/Sun does fall towards the thing, it's just not worth worrying about. To illustrate the scale, the Earth doesn't orbit the Sun, the Sun and the Earth orbit around a common centre of gravity. That centre of gravity is SO far inside the Sun that you may as well say that the Earth orbits the Sun.
Very short answer: Planetary bodies are big (in terms of both size and mass) so we can treat their position as constant relative to small things like 'planes and buildings.
posted by Wrinkled Stumpskin at 4:51 PM on January 17, 2009
Best answer: But the sum of the masses of the pair involved is greater than 1a) and thus a stronger gravitational field is formed. The pair must move together more quickly than in 1a).
Ah, here is your misconception. The heavier ball doesn't move more quickly. The ball with the larger mass is attracted more strongly to the earth, but the ball also has more inertial mass (resistance to acceleration). That is, a heavier object takes more force to make it accelerate. It turns out, if you do the equations, that the stronger force of gravitational attraction and the stronger inertia, or resistance to acceleration, exactly balance. So, no matter the mass, all objects fall at the same speed.
This is one of the most fundamental and curious identities of our universe -- the fact that gravitation mass and inertial mass have exactly the same value. This means that the mass in the gravity equation and the mass in the F=ma equation are equal. It is possible to imagine a universe in which this were not true.
posted by JackFlash at 5:17 PM on January 17, 2009 [2 favorites]
Ah, here is your misconception. The heavier ball doesn't move more quickly. The ball with the larger mass is attracted more strongly to the earth, but the ball also has more inertial mass (resistance to acceleration). That is, a heavier object takes more force to make it accelerate. It turns out, if you do the equations, that the stronger force of gravitational attraction and the stronger inertia, or resistance to acceleration, exactly balance. So, no matter the mass, all objects fall at the same speed.
This is one of the most fundamental and curious identities of our universe -- the fact that gravitation mass and inertial mass have exactly the same value. This means that the mass in the gravity equation and the mass in the F=ma equation are equal. It is possible to imagine a universe in which this were not true.
posted by JackFlash at 5:17 PM on January 17, 2009 [2 favorites]
Response by poster: Yes, that is technically correct. The [celestial body] and the object are accelerating towards each other at a rate derived from the mass of each of them (and the distance between them at any given instant).
So if we had a tall enough Leaning Tower (in a vacuum) and dropped the two items off, the massier one would eventually move 'ahead' of the smaller one?
Then in that case, how do we resolve the 'paradox' that Galileo proposed when this whole thing first came up - when you connect the two items with a 'piece of string'?
posted by woodblock100 at 5:26 PM on January 17, 2009
So if we had a tall enough Leaning Tower (in a vacuum) and dropped the two items off, the massier one would eventually move 'ahead' of the smaller one?
Then in that case, how do we resolve the 'paradox' that Galileo proposed when this whole thing first came up - when you connect the two items with a 'piece of string'?
posted by woodblock100 at 5:26 PM on January 17, 2009
Response by poster: It turns out, if you do the equations, that the stronger force of gravitational attraction and the stronger inertia, or resistance to acceleration, exactly balance. So, no matter the mass, all objects fall at the same speed.
Oops ... I guess I should have previewed ... This one - to my layman's ear - sounds like the resolution to my question ...
posted by woodblock100 at 5:27 PM on January 17, 2009
Oops ... I guess I should have previewed ... This one - to my layman's ear - sounds like the resolution to my question ...
posted by woodblock100 at 5:27 PM on January 17, 2009
Response by poster: No; wait a minute.
Ah, here is your misconception. The heavier ball doesn't move more quickly. The ball with the larger mass is attracted more strongly to the earth, but the ball also has more inertial mass (resistance to acceleration). ... So, no matter the mass, all objects fall at the same speed.
But what is the earth - compared to the moon in these examples - but a heavier ball? If increasing the mass of one of the objects from 'feather mass' to 'ball mass' doesn't make it fall any faster (because of the increased inertia), why should increasing the mass of the object at the other side of the pair - from 'moon mass' to 'earth mass' - make a difference?
Going by this description, every pair of objects in the universe, with no matter what masses, would 'fall' towards each other at exactly the same rate. But that isn't what actually happens - things falling on the moon do fall more slowly than they fall on earth.
Maybe I'll wake up tomorrow morning and realize that these are 'dumb' questions, but at the moment, I still don't grok this ... (and looking over the answers upthread, I would bet I'm not the only one ...)
posted by woodblock100 at 11:39 PM on January 17, 2009
Ah, here is your misconception. The heavier ball doesn't move more quickly. The ball with the larger mass is attracted more strongly to the earth, but the ball also has more inertial mass (resistance to acceleration). ... So, no matter the mass, all objects fall at the same speed.
But what is the earth - compared to the moon in these examples - but a heavier ball? If increasing the mass of one of the objects from 'feather mass' to 'ball mass' doesn't make it fall any faster (because of the increased inertia), why should increasing the mass of the object at the other side of the pair - from 'moon mass' to 'earth mass' - make a difference?
Going by this description, every pair of objects in the universe, with no matter what masses, would 'fall' towards each other at exactly the same rate. But that isn't what actually happens - things falling on the moon do fall more slowly than they fall on earth.
Maybe I'll wake up tomorrow morning and realize that these are 'dumb' questions, but at the moment, I still don't grok this ... (and looking over the answers upthread, I would bet I'm not the only one ...)
posted by woodblock100 at 11:39 PM on January 17, 2009
Best answer: why should increasing the mass of the object at the other side of the pair - from 'moon mass' to 'earth mass' - make a difference
because, the object at the other side of the pair has its own inertial mass, which dictates how much it accelerates. i imagine the confusion comes in because somewhere deep in your brain, you're thinking of the earth or moon or whatever as stationary, and having trouble getting over the conceptual hurdle that both things move towards each other, no matter what two things they are.
the fundamental problem here probably has to do with frame of reference. it complicates things somewhat to put yourself on the feather or on the moon or whatever. try instead to put yourself floating out in space somewhere, far away from either object, so that you're not moving along with either one. then go through this little derivation with me, i promise it won't hurt and isn't as tedious as it looks:
it is probably easiest, in the context of this question, to look at the basic force equation in a less complicated way. you can group the G constant and the r2 distance factor altogether and consider it all as just another constant, but only right at the moment you let go of the feather. (after you let go, r will start changing, and it stops being strictly constant.) nevertheless i think that's a reasonable constraint to place on things given the question you're asking.
written in the above way, the force in the feather-earth system is simply a multiplication:
Ffeather = (some stuff) * mfeather * mearth
Fearth = (some stuff) * mfeather * mearth
this force is the same mutual attractive force that's applied both to the earth, and to the feather, but in opposite directions. the thing to take home here is that the force is the same on both objects.
ok, so what about the acceleration?
Ffeather = mfeather * afeather
Fearth = mearth * aearth
the accelerations are not the same, because there's a mass term in there that's specific to whichever object you're talking about. if you take the force equations up top there, and plug it into these two equations that are simply F = ma, you get the following two expressions for the acceleration of the earth and feather:
mfeather * afeather = (some stuff) * mfeather * mearth
mearth * aearth = (some stuff) * mfeather * mearth
now, in these new equations, there's a mass term on each side that you can cancel. but be careful, because it's a different mass in the two equations. pay attention to the subscripts.
if you do the canceling, you get:
afeather = (some stuff) * mearth
aearth = (some stuff) * mfeather
so there, you see the feather accelerates by some thing times a really huge number, mearth. and the earth accelerates by the same thing times a really tiny number, mfeather in fact if you divide the top equation by the bottom equation, then you'll get a nice pretty form for the ratio of the accelerations:
(afeather / aearth) = (mearth / mfeather)
in other words, the feather accelerates more than the earth, by the same factor that the earth is heavier than the feather. again, pay attention to which subscript is where. the enormous difference between the mass of the earth and the feather makes the earth's acceleration so small compared to that of the feather that it's basically undetectable. i think from how you phrased the question that you have some understanding of this already.
you can repeat this analysis for any two pair of objects - feather and moon, earth and moon, what have you, and it will be correct. do this, and write down what afeather and so on are in the feather-moon system. compare the accelerations to the feather-earth system, or the earth-moon system, or whatever. i think if you work carefully through what happens in each case, and plug in some actual numbers for the masses so that you can compare, it might help straighten out some of the confusion.
posted by sergeant sandwich at 1:26 AM on January 18, 2009
because, the object at the other side of the pair has its own inertial mass, which dictates how much it accelerates. i imagine the confusion comes in because somewhere deep in your brain, you're thinking of the earth or moon or whatever as stationary, and having trouble getting over the conceptual hurdle that both things move towards each other, no matter what two things they are.
the fundamental problem here probably has to do with frame of reference. it complicates things somewhat to put yourself on the feather or on the moon or whatever. try instead to put yourself floating out in space somewhere, far away from either object, so that you're not moving along with either one. then go through this little derivation with me, i promise it won't hurt and isn't as tedious as it looks:
it is probably easiest, in the context of this question, to look at the basic force equation in a less complicated way. you can group the G constant and the r2 distance factor altogether and consider it all as just another constant, but only right at the moment you let go of the feather. (after you let go, r will start changing, and it stops being strictly constant.) nevertheless i think that's a reasonable constraint to place on things given the question you're asking.
written in the above way, the force in the feather-earth system is simply a multiplication:
Ffeather = (some stuff) * mfeather * mearth
Fearth = (some stuff) * mfeather * mearth
this force is the same mutual attractive force that's applied both to the earth, and to the feather, but in opposite directions. the thing to take home here is that the force is the same on both objects.
ok, so what about the acceleration?
Ffeather = mfeather * afeather
Fearth = mearth * aearth
the accelerations are not the same, because there's a mass term in there that's specific to whichever object you're talking about. if you take the force equations up top there, and plug it into these two equations that are simply F = ma, you get the following two expressions for the acceleration of the earth and feather:
mfeather * afeather = (some stuff) * mfeather * mearth
mearth * aearth = (some stuff) * mfeather * mearth
now, in these new equations, there's a mass term on each side that you can cancel. but be careful, because it's a different mass in the two equations. pay attention to the subscripts.
if you do the canceling, you get:
afeather = (some stuff) * mearth
aearth = (some stuff) * mfeather
so there, you see the feather accelerates by some thing times a really huge number, mearth. and the earth accelerates by the same thing times a really tiny number, mfeather in fact if you divide the top equation by the bottom equation, then you'll get a nice pretty form for the ratio of the accelerations:
(afeather / aearth) = (mearth / mfeather)
in other words, the feather accelerates more than the earth, by the same factor that the earth is heavier than the feather. again, pay attention to which subscript is where. the enormous difference between the mass of the earth and the feather makes the earth's acceleration so small compared to that of the feather that it's basically undetectable. i think from how you phrased the question that you have some understanding of this already.
you can repeat this analysis for any two pair of objects - feather and moon, earth and moon, what have you, and it will be correct. do this, and write down what afeather and so on are in the feather-moon system. compare the accelerations to the feather-earth system, or the earth-moon system, or whatever. i think if you work carefully through what happens in each case, and plug in some actual numbers for the masses so that you can compare, it might help straighten out some of the confusion.
posted by sergeant sandwich at 1:26 AM on January 18, 2009
Response by poster: I think if you work carefully through what happens in each case ...
Thank you (and the other respondents) for all the time spent on this; I think I'm getting a handle on it. Let me try to describe actions that correspond to the equations you have given:
Free-floating large planet and small feather; there is a gravitational attraction between them (as a system), with the force equal both ways. On release, because of their unequal masses, the resulting accelerations are unequal; the feather gets 'up to speed' first, and does most of the moving (but not all of it). Run a videotape of the event, carefully noting time to impact.
Run the clock back, and do it all again, this time with a steel ball instead of a feather. The gravitational attraction is now larger, due to the increased mass of the system, but the force still acts equally both ways. On release, the same sort of thing happens as in the first example, but while the acceleration of the ball - being worked out by the same type of equation:
aball = (some stuff) * mearth
... comes out to be the same value as that of the feather, the acceleration of the earth, being dependent on the mass of the ball, is now greater:
aearth = (some stuff) * mball
Run a videotape of the event, carefully noting time to impact.
Time to impact should be shorter in the second example: the small object is accelerating at the same rate in both cases, but the value of the earth's acceleration is greater in the second case. The actual difference is obviously going to be staggeringly small, but surely it is 'real'. The heavier item should hit the ground first.
(You know, I'm honestly not trying to be obstinate here; when I started to read that obviously carefully prepared explanation, I fully anticipated that it would 'solve' the puzzle for me, and that I would nod my head in agreement, say 'thank you', and leave this alone. I'm sorry that I'm not able to do that ...)
posted by woodblock100 at 3:12 AM on January 18, 2009
Thank you (and the other respondents) for all the time spent on this; I think I'm getting a handle on it. Let me try to describe actions that correspond to the equations you have given:
Free-floating large planet and small feather; there is a gravitational attraction between them (as a system), with the force equal both ways. On release, because of their unequal masses, the resulting accelerations are unequal; the feather gets 'up to speed' first, and does most of the moving (but not all of it). Run a videotape of the event, carefully noting time to impact.
Run the clock back, and do it all again, this time with a steel ball instead of a feather. The gravitational attraction is now larger, due to the increased mass of the system, but the force still acts equally both ways. On release, the same sort of thing happens as in the first example, but while the acceleration of the ball - being worked out by the same type of equation:
aball = (some stuff) * mearth
... comes out to be the same value as that of the feather, the acceleration of the earth, being dependent on the mass of the ball, is now greater:
aearth = (some stuff) * mball
Run a videotape of the event, carefully noting time to impact.
Time to impact should be shorter in the second example: the small object is accelerating at the same rate in both cases, but the value of the earth's acceleration is greater in the second case. The actual difference is obviously going to be staggeringly small, but surely it is 'real'. The heavier item should hit the ground first.
(You know, I'm honestly not trying to be obstinate here; when I started to read that obviously carefully prepared explanation, I fully anticipated that it would 'solve' the puzzle for me, and that I would nod my head in agreement, say 'thank you', and leave this alone. I'm sorry that I'm not able to do that ...)
posted by woodblock100 at 3:12 AM on January 18, 2009
I think you have it now. Remember the gravity equation:
F = GmAmB/r2
Notice there are two masses, A and B in the equation. As long as you leave mass A the same, the earth example, then all experiments dropping various objects B will have the same result because the gravitational mass of B and the inertial mass of B always balance out.
If you then change the experiment by changing mass A, the moon example, then all of the masses B fall at the new rate.
posted by JackFlash at 9:46 AM on January 18, 2009
F = GmAmB/r2
Notice there are two masses, A and B in the equation. As long as you leave mass A the same, the earth example, then all experiments dropping various objects B will have the same result because the gravitational mass of B and the inertial mass of B always balance out.
If you then change the experiment by changing mass A, the moon example, then all of the masses B fall at the new rate.
posted by JackFlash at 9:46 AM on January 18, 2009
Time to impact should be shorter in the second example: the small object is accelerating at the same rate in both cases, but the value of the earth's acceleration is greater in the second case. The actual difference is obviously going to be staggeringly small, but surely it is 'real'. The heavier item should hit the ground first.
yes, speaking absolutely strictly, this is correct - in a simple two-body system with an attractive force, everything else being the same, the heavier test object will hit the ground ever so slightly sooner than the lighter one.
it IS also strictly correct to say that in an ideal, uniform gravitational field, all objects will fall at precisely the same speed. but as you've pointed out, the earth moves too, which means the gravitational field changes slightly during the fall, and it does so by a different amount for the two objects.
this is one of those examples of the distinction between a theoretical construct such as "a perfectly uniform gravitational field" (which doesn't really exist anywhere except on paper), and situations in reality for which that construct is a good approximation. the notion that two objects of dissimilar masses will fall at equal rates is an approximation due to the large mass of the planet. this is the approximation that's being assumed by e.g. JackFlash above me.
it's a damn good one though, considering that what you're throwing out in the approximation will only show up down in the 26th decimal point or so. there are lots of other things going on around any experiment (other people dropping things elsewhere on earth, say.. or continental drift, or whatever) that are much more significant than one part in 1026.
this is a significant mental hurdle to overcome for a lot of people - the notion that the "laws of physics" like newtonian gravitation are not immutable, perfect descriptors of a mathematical universe. they are models - models that describe the behavior of things, subject to certain constraints, constraints which are usually true. the models stop making sense under certain conditions (edge cases, limits of very large or very small, and so on).
so the "puzzle" is i think not an inherent contradiction, nor is it a failure of you to understand the "falling objects in a uniform field" model, but it's an example of how the model can break down at the edges. (of course, this is when the discoverin' starts to happen! (: )
posted by sergeant sandwich at 4:44 PM on January 18, 2009
yes, speaking absolutely strictly, this is correct - in a simple two-body system with an attractive force, everything else being the same, the heavier test object will hit the ground ever so slightly sooner than the lighter one.
it IS also strictly correct to say that in an ideal, uniform gravitational field, all objects will fall at precisely the same speed. but as you've pointed out, the earth moves too, which means the gravitational field changes slightly during the fall, and it does so by a different amount for the two objects.
this is one of those examples of the distinction between a theoretical construct such as "a perfectly uniform gravitational field" (which doesn't really exist anywhere except on paper), and situations in reality for which that construct is a good approximation. the notion that two objects of dissimilar masses will fall at equal rates is an approximation due to the large mass of the planet. this is the approximation that's being assumed by e.g. JackFlash above me.
it's a damn good one though, considering that what you're throwing out in the approximation will only show up down in the 26th decimal point or so. there are lots of other things going on around any experiment (other people dropping things elsewhere on earth, say.. or continental drift, or whatever) that are much more significant than one part in 1026.
this is a significant mental hurdle to overcome for a lot of people - the notion that the "laws of physics" like newtonian gravitation are not immutable, perfect descriptors of a mathematical universe. they are models - models that describe the behavior of things, subject to certain constraints, constraints which are usually true. the models stop making sense under certain conditions (edge cases, limits of very large or very small, and so on).
so the "puzzle" is i think not an inherent contradiction, nor is it a failure of you to understand the "falling objects in a uniform field" model, but it's an example of how the model can break down at the edges. (of course, this is when the discoverin' starts to happen! (: )
posted by sergeant sandwich at 4:44 PM on January 18, 2009
Response by poster: yes, speaking absolutely strictly, this is correct - in a simple two-body system with an attractive force, everything else being the same, the heavier test object will hit the ground ever so slightly sooner than the lighter one.
Great! Now I can sleep again!
I'm certainly not about to push for more clarification/explication/etc. but I'm going to be nursing this concept in the back of my mind for quite some time yet, I think. Although in the examples I proposed, the value for the planetary acceleration is vanishingly small, and 'makes no difference', I wonder what happens when the thing plays out at cosmic scales. But there too, I suppose it doesn't matter; the universe is 'analog', and the numbers don't have to come out 'perfectly' ...
Thanks!
posted by woodblock100 at 5:46 PM on January 18, 2009
Great! Now I can sleep again!
I'm certainly not about to push for more clarification/explication/etc. but I'm going to be nursing this concept in the back of my mind for quite some time yet, I think. Although in the examples I proposed, the value for the planetary acceleration is vanishingly small, and 'makes no difference', I wonder what happens when the thing plays out at cosmic scales. But there too, I suppose it doesn't matter; the universe is 'analog', and the numbers don't have to come out 'perfectly' ...
Thanks!
posted by woodblock100 at 5:46 PM on January 18, 2009
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The earth is "falling" around the sun, not to towards the dropped objects. The dropped objects are being attracted to the earth through gravity and falling into it. The earth is not moving up towards it. The earth is too busy being pulled toward the sun. The sun's gravitational field is so powerful that those objects you are dropping are also being pulling to the sun. They are not pulling the earth up to it. Their gravitational fields have been canceled out by the larger two objects: the sun and the earth.
posted by damn dirty ape at 8:39 AM on January 17, 2009