Gravitational potential energy (wtf?)
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Gravitational potential energy (wtf?)
Excuse if this has been answered elsewhere, I searched around a bit but nothing came up that answers my question.
Gravitational potential energy: What is it? After all this time I still can't comprehend potential energy (in any form, actually) being more than a placeholder or a copout designed to make conservation of energy work. Mind you I know I'm wrong, just don't understand why.
Here are some questions that hopefully when answered will give me some insight into the nature of Potential Energy, but this is not the main goal. If you have epic examples/explanations that are not these, then by all means, have at it.
Since PE = mgh (roughly), what happens to the potential energy of an object when I build a table under it? When I dig a crater out of the surface of the earth? When I change my point of reference? How is it anyway consistent at all?
Consider an object placed in a point in between several massive objects in such a way that the gravitational pull from all of them resulted in the object not moving. Does that have a potential energy?
Any clarity would be much appreciated  I've been trying to nut this sucker out for ages. No one seems to be able to explain to me what it actually is (or momentum for that matter, but that's another topic entirely and it sorta makes sense now.)
Gravitational potential energy: What is it? After all this time I still can't comprehend potential energy (in any form, actually) being more than a placeholder or a copout designed to make conservation of energy work. Mind you I know I'm wrong, just don't understand why.
Here are some questions that hopefully when answered will give me some insight into the nature of Potential Energy, but this is not the main goal. If you have epic examples/explanations that are not these, then by all means, have at it.
Since PE = mgh (roughly), what happens to the potential energy of an object when I build a table under it? When I dig a crater out of the surface of the earth? When I change my point of reference? How is it anyway consistent at all?
Consider an object placed in a point in between several massive objects in such a way that the gravitational pull from all of them resulted in the object not moving. Does that have a potential energy?
Any clarity would be much appreciated  I've been trying to nut this sucker out for ages. No one seems to be able to explain to me what it actually is (or momentum for that matter, but that's another topic entirely and it sorta makes sense now.)
Last edited by Ashbash on Wed Oct 22, 2008 12:27 pm UTC, edited 1 time in total.
 danpilon54
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Re: Gravitational potential energy (wtf?)
Gravitational potential energy is the energy you would need (kinetic energy) to escape the gravitational pull of an object.
To answer your specific questions:
When doing problems using potential energy, only the relative difference in potential energy between two points is important. If you are falling onto a table rather than the ground, using the distance to the table rather than the ground. What you are really saying is PE1 + KE1 = PE2 + KE2. If say KE1 is 0 (you fall starting from rest), then KE2 = PE1  PE2. So only the difference matters. This is approximately mgh near the earth's surface, where h is the distance between your two points.
The point you describe between two massive objects does not have 0 potential energy, but it is an equilibrium. This means the derivative of PE with respect to position is 0. Force is defined as the derivative of potential energy, so you will have 0 force. I should note that this specific point is an unstable equilibrium, as it is a maximum of the potential energy.
To answer your specific questions:
When doing problems using potential energy, only the relative difference in potential energy between two points is important. If you are falling onto a table rather than the ground, using the distance to the table rather than the ground. What you are really saying is PE1 + KE1 = PE2 + KE2. If say KE1 is 0 (you fall starting from rest), then KE2 = PE1  PE2. So only the difference matters. This is approximately mgh near the earth's surface, where h is the distance between your two points.
The point you describe between two massive objects does not have 0 potential energy, but it is an equilibrium. This means the derivative of PE with respect to position is 0. Force is defined as the derivative of potential energy, so you will have 0 force. I should note that this specific point is an unstable equilibrium, as it is a maximum of the potential energy.
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Re: Gravitational potential energy (wtf?)
You can think of gravitational potential energy as the energy a body has by virtue of its position in the local gravitational field. It's a real energy in the sense that it can be used to do work  try dropping a ball. The potential energy it had when you were holding it becomes kinetic energy as it falls.
Except, of course, that gravitational potential is (classically) always a relative thing  you can say that object A has x more gravitational potential energy than object B, but there's no arbitrary zero point, which in practice means you define the zero wherever is convenient. This is not a problem  what matters is the difference in potential energy between two positions, not the absolute amount. If you say that a ball at height h above the floor has mgh, that's ok  if the ball ends up below the floor for some reason, it'll just have negative potential energy (once again, with reference to your zero point). It's like setting the zeroes of your coordinate axes wherever makes the problem easiest.
Now, if you dug a hole underneath the ball, and it fell, well, it'd lose more potential energy (converting it initially into kinetic energy, and then into heat, sound, etcetera). If you were holding it so that it didn't move, the gravitational potential would remain the same  after all, it hasn't changed position relative to the local field, so why should it change its potential energy? Same deal if you're holding it in the air, and then you raise the floor some how, say, with a table[1]. Gravitational potential energy is consistent, provided you keep your zero in the same spot over the course of a problem. If you move it, you've got to make allowances for that, exactly like how if you change the origin of your coordinate system in the middle of a problem you'v
Now, as to your question: If you say, have a ball suspended between two planets such that the gravitational fields from each of the planets cancelled out so the ball experienced no net force, then yes, the ball would have a gravitational potential energy. It _always_ has one. It may be convenient to definite it to be zero at that point, but it still has one. It just turns out that if you plot the gravitational potential energy of the ball in different distances between the two planets[3], that it will be a curve and the ball will be at a local maximum (peak) on the curve. This is actually an important thing to note  because the ball is at a maximum, that means the gradient of the potential is precisely zero. This isn't a coincidence  at point where a potential field has zero gradient, the net force due to that field will be zero. In fact, the force due to a field at any point is equal to minus the gradient of that field (the minus sign is a historical convention, I'm afraid).
I hope this begins to answer your questions. I apologise in advance for any mistakes I've made  it's late and I'm going easy on the stimulants at the moment.
[1] For purposes of simplicity I'm ignoring the fact that the gravitational field will be changed if you move stuff around  the effects are so small as to be insignificant in most classical situations. In this case, it's the interaction between the Earth and the ball which matters.
[2] I'm assuming we're only considering the problem in 1D  the mathematics is more complicated in more dimensions, but not fundamentally conceptually different.
Except, of course, that gravitational potential is (classically) always a relative thing  you can say that object A has x more gravitational potential energy than object B, but there's no arbitrary zero point, which in practice means you define the zero wherever is convenient. This is not a problem  what matters is the difference in potential energy between two positions, not the absolute amount. If you say that a ball at height h above the floor has mgh, that's ok  if the ball ends up below the floor for some reason, it'll just have negative potential energy (once again, with reference to your zero point). It's like setting the zeroes of your coordinate axes wherever makes the problem easiest.
Now, if you dug a hole underneath the ball, and it fell, well, it'd lose more potential energy (converting it initially into kinetic energy, and then into heat, sound, etcetera). If you were holding it so that it didn't move, the gravitational potential would remain the same  after all, it hasn't changed position relative to the local field, so why should it change its potential energy? Same deal if you're holding it in the air, and then you raise the floor some how, say, with a table[1]. Gravitational potential energy is consistent, provided you keep your zero in the same spot over the course of a problem. If you move it, you've got to make allowances for that, exactly like how if you change the origin of your coordinate system in the middle of a problem you'v
Now, as to your question: If you say, have a ball suspended between two planets such that the gravitational fields from each of the planets cancelled out so the ball experienced no net force, then yes, the ball would have a gravitational potential energy. It _always_ has one. It may be convenient to definite it to be zero at that point, but it still has one. It just turns out that if you plot the gravitational potential energy of the ball in different distances between the two planets[3], that it will be a curve and the ball will be at a local maximum (peak) on the curve. This is actually an important thing to note  because the ball is at a maximum, that means the gradient of the potential is precisely zero. This isn't a coincidence  at point where a potential field has zero gradient, the net force due to that field will be zero. In fact, the force due to a field at any point is equal to minus the gradient of that field (the minus sign is a historical convention, I'm afraid).
I hope this begins to answer your questions. I apologise in advance for any mistakes I've made  it's late and I'm going easy on the stimulants at the moment.
[1] For purposes of simplicity I'm ignoring the fact that the gravitational field will be changed if you move stuff around  the effects are so small as to be insignificant in most classical situations. In this case, it's the interaction between the Earth and the ball which matters.
[2] I'm assuming we're only considering the problem in 1D  the mathematics is more complicated in more dimensions, but not fundamentally conceptually different.
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Re: Gravitational potential energy (wtf?)
An alternate term for potential energy is binding energy. It isn't just a useful device that allows the use of conservation of energy, one can weigh it if there's enough of it.
In terms of field theory, you can visualize it as a stress in the field that results from the energy that was exerted in separating the charges/masses/etc. Obtaining an intuitive feel for field theory is an exercise left for the reader.
In terms of field theory, you can visualize it as a stress in the field that results from the energy that was exerted in separating the charges/masses/etc. Obtaining an intuitive feel for field theory is an exercise left for the reader.
Re: Gravitational potential energy (wtf?)
All massive objects produce gravitational fields. A field is a function that maps points in space to force vectors. That is, you can take a "test mass", put it at some point p in space, and you the field tells you how strong the force felt by the test mass is and the direction it points.
Take two points in space, p1 and p2. How much energy is required to move the test mass from p1 to p2? We can use the work equation to answer this. Draw a line between p1 and p2 and take the path integral of the force along this line. (Assuming you know calculus, if you don't, then let me assert there is a thing called a path integral which turns children's tears into love on Earth... if you'd like an explanation, I'd be happy to provide one).
So the work equation gives you the amount of energy required to push the charge from p1 to p2. It can be positive or negative. If the number is positive, the test mass LOSES potential energy as it moves from p1 to p2. If energy is being lost in one form, it is being gained in another, and so we can use this energy to do useful things. A falling rock is losing potential energy, but it picks up speed which can allow us to slay our foes.
On the other hand, if the work is negative, it requires energy to move the object from p1 to p2. Think of a space shuttle. We *want* it in space, but it's not going to go there by itself. We need to really push hard on it. We pump more and more energy into the space shuttle until it is in orbit.
Now, orbit. That's a funny concept. Think of the moon. Gravity extends far, far from Earth. The moon definitely feels the pull of Earth. But it does not fall! From an energy standpoint, how is this possible? Well, gravity is spherically symmetrical about the Earth. That is, all points in the moon's orbit are the same distance from Earth (not quite technically true, but close enough for now). That means each point is at the same energy potential, and moving between any two points requires no expenditure of energy.
I am of course skimping on a few important points, but it should give you the gist of what potential energy is. But if you really want to understand why it works, the real answer is in understanding line integrals and fields.
Take two points in space, p1 and p2. How much energy is required to move the test mass from p1 to p2? We can use the work equation to answer this. Draw a line between p1 and p2 and take the path integral of the force along this line. (Assuming you know calculus, if you don't, then let me assert there is a thing called a path integral which turns children's tears into love on Earth... if you'd like an explanation, I'd be happy to provide one).
So the work equation gives you the amount of energy required to push the charge from p1 to p2. It can be positive or negative. If the number is positive, the test mass LOSES potential energy as it moves from p1 to p2. If energy is being lost in one form, it is being gained in another, and so we can use this energy to do useful things. A falling rock is losing potential energy, but it picks up speed which can allow us to slay our foes.
On the other hand, if the work is negative, it requires energy to move the object from p1 to p2. Think of a space shuttle. We *want* it in space, but it's not going to go there by itself. We need to really push hard on it. We pump more and more energy into the space shuttle until it is in orbit.
Now, orbit. That's a funny concept. Think of the moon. Gravity extends far, far from Earth. The moon definitely feels the pull of Earth. But it does not fall! From an energy standpoint, how is this possible? Well, gravity is spherically symmetrical about the Earth. That is, all points in the moon's orbit are the same distance from Earth (not quite technically true, but close enough for now). That means each point is at the same energy potential, and moving between any two points requires no expenditure of energy.
I am of course skimping on a few important points, but it should give you the gist of what potential energy is. But if you really want to understand why it works, the real answer is in understanding line integrals and fields.
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Re: Gravitational potential energy (wtf?)
Eh, I think the OP's real confusion was answered by Snoof, in a sorta longwinded way. The confusion the OP is experiencing seems to stem from the fact that we can define gravitational potential as simply mgh, where h is defined as "distance from an arbitrarily chosen point". Thus the bit in the OP about the potential energy changing when you hold an object out and then move a table underneath it.
The simple answer is that it *doesn't* change. As Snoof alluded to, the reason why we can define h arbitrarily is because we only care about the difference in potential energy between two objects or two locations. If an object is 2m above the floor, how much energy is gained (to be used for other purposes) if it is allowed to descend to the floor? We can measure this with reference to the center of the earth (6378.139km versus 6378.137km (roughly)), but that's dumb. Those numbers are much larger than necessary. Since gravitational potential energy varies linearly with height, we are allowed to shift the point we call zero for convenience. So we do, and say that the floor is 0. Now it's just 2m versus 0m. The actual energy doesn't change at all, but the numbers we have to deal with when calculating it are now more convenient.
We do this crap all the time. If it is at all possible to change your reference frame so that something is at 0 (or sometimes 1, depending on what you're doing), we'll do so. If we can get multiple things to be zero, hell yeah. The math is crazy simpler that way, as long as you know that this is a valid move. We'll also do things like center a rotation at (0,0) (or (0,0,0) or (0,0,0,0) or whatever, depending on what coordinates we're using) for the same reason  the math behind rotations is annoyingly complex in general, but most of it disappears if you can land it right on the zero point.
The simple answer is that it *doesn't* change. As Snoof alluded to, the reason why we can define h arbitrarily is because we only care about the difference in potential energy between two objects or two locations. If an object is 2m above the floor, how much energy is gained (to be used for other purposes) if it is allowed to descend to the floor? We can measure this with reference to the center of the earth (6378.139km versus 6378.137km (roughly)), but that's dumb. Those numbers are much larger than necessary. Since gravitational potential energy varies linearly with height, we are allowed to shift the point we call zero for convenience. So we do, and say that the floor is 0. Now it's just 2m versus 0m. The actual energy doesn't change at all, but the numbers we have to deal with when calculating it are now more convenient.
We do this crap all the time. If it is at all possible to change your reference frame so that something is at 0 (or sometimes 1, depending on what you're doing), we'll do so. If we can get multiple things to be zero, hell yeah. The math is crazy simpler that way, as long as you know that this is a valid move. We'll also do things like center a rotation at (0,0) (or (0,0,0) or (0,0,0,0) or whatever, depending on what coordinates we're using) for the same reason  the math behind rotations is annoyingly complex in general, but most of it disappears if you can land it right on the zero point.
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Re: Gravitational potential energy (wtf?)
more generally we only care about differences in anything in physics, there is no way to measure any absolute value.
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 danpilon54
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Re: Gravitational potential energy (wtf?)
All of this is true. The 0 point is arbitrary, but is generally taken (for potentials and fields) to be at infinite distance away. This makes gravitational potential always negative for simplicity. Of course if you're not worrying about anything bigger than a few meters, it helps to make the 0 point some other convenient point, like the surface of the earth, or the bottom of a ditch.
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Re: Gravitational potential energy (wtf?)
Ashbash wrote:Gravitational potential energy: What is it? After all this time I still can't comprehend potential energy (in any form, actually) being more than a placeholder or a copout designed to make conservation of energy work. Mind you I know I'm wrong, just don't understand why.
Why do you think you are wrong? Why do you think it is a copout? The fundamental rule is that you calculate the change of kinetic energy by evaluating the work done on the body, which may be a complicated integration. It may happen that for the force you are working with, the work done depends only on where the motion starts and where it ends but does not depend on the path you take. In that case you can define a function as the work done on the body between a standard point and any other, and this function is singlevalued because the work does not depend on the path, which is the only other variable it could possibly depend on. In this case it is obvious that if you define somethingyetunnamed as minus the work done getting to that point, then
kinetic energy + somethingyetunnamed is a constant. You calculate somethingyetunnamed once and you know it forever, so you can find the kinetic energy at any point in any situation whatever from then on. This procedure is way too convenient to ignore, but "somethingyetunnamed" is a darn awkward way of referring to it, and you need an easier name. Since "somethingyetunnamed" is related to kinetic energy, you will remember it better if you name if "somekindorotherenergy." "Potential " energy will do for lack of anything better. Now Potential Energy + Kinetic Energy is the same wherever the body is, or in physics shorthand, "Energy is conserved."
That is all there is. Nothing about potential energy is mysterious or needs a mechanistic explanation. Conserved quantities are invaluable in physics, and you can get very famous very fast by finding one that gets used a lot. This one was found a long time ago. End of Story.
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Re: Gravitational potential energy (wtf?)
Potential energy can be thought of in a straightforward way...
It takes a great amount of your energy to pry open an old fashioned cast iron bear trap. Once you've opened it fully, you can simply spring the trap by lightly poking the release mechanism with a twig. How could such a small gesture (stick poking) cause such an energetic reaction (the trap's heavy jaws slamming shut at rapid speed)? Since the stick poking did not input enough energy to do so, it's clear that the source of the remaining required energy was the potential energy (the energy you originally put into the trap when you opened it). So in a way potential energy is a bookkeeping device like you say, but it is physically real.
It takes a great amount of your energy to pry open an old fashioned cast iron bear trap. Once you've opened it fully, you can simply spring the trap by lightly poking the release mechanism with a twig. How could such a small gesture (stick poking) cause such an energetic reaction (the trap's heavy jaws slamming shut at rapid speed)? Since the stick poking did not input enough energy to do so, it's clear that the source of the remaining required energy was the potential energy (the energy you originally put into the trap when you opened it). So in a way potential energy is a bookkeeping device like you say, but it is physically real.

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Re: Gravitational potential energy (wtf?)
Snoof wrote:Now, as to your question: If you say, have a ball suspended between two planets such that the gravitational fields from each of the planets cancelled out so the ball experienced no net force, then yes, the ball would have a gravitational potential energy. It _always_ has one. It may be convenient to definite it to be zero at that point, but it still has one. It just turns out that if you plot the gravitational potential energy of the ball in different distances between the two planets[3], that it will be a curve and the ball will be at a local maximum (peak) on the curve. This is actually an important thing to note  because the ball is at a maximum, that means the gradient of the potential is precisely zero. This isn't a coincidence  at point where a potential field has zero gradient, the net force due to that field will be zero. In fact, the force due to a field at any point is equal to minus the gradient of that field (the minus sign is a historical convention, I'm afraid).
If this is a bit mathematically confusing, or you don't know what gradients are, just think of a ball on top of a hill. Its at max potential energy but is balanced and won't fall. However, if i give it a slight nudge in one direction then it will quickly roll down one side of the hill (move towards and collide with one of the planets).
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Re: Gravitational potential energy (wtf?)
I almost always think about potential energy in terms of a hill (even now as a senior physics major). Since PE=mgh, PE is proportional to h, so the hill is the same shape as PE. You have to work with a frictionless hill though.
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Re: Gravitational potential energy (wtf?)
Thanks for the replies guys, this is starting to make sense. I'm pretty sure I get it for a localised area, say a planet, how moving reference points don't really matter because potential energy is only of concern when it changes. Ok.
However when I try and expand it out to the entire universe, my mind breaks down. I am assuming energy is conserved, and that an omniscient being can measure all forms of energy occuring at any place at a specifice time. So what should happen is at time t1, the sum total of all energies should be the same as at t2. Say we pick an arbitrary reference point, X. Now after we measure the kinetic, nuclear, heat and all other forms of energy, we calculate the PE for every object to the point X. However not everything is attracted towards X, so... what comes next? I'm thinking it's got something to do with this potential well concept?
I don't even know what I'm trying to get at really, I think I'm just having trouble expanding this to a larger system that one gravitational field, where things are nice and tidy.
I do know calculus, but I don't know what a path integral is. Is it different to a regular integral? An integral of what? Wikipedia isn't helping.
Thanks again.
However when I try and expand it out to the entire universe, my mind breaks down. I am assuming energy is conserved, and that an omniscient being can measure all forms of energy occuring at any place at a specifice time. So what should happen is at time t1, the sum total of all energies should be the same as at t2. Say we pick an arbitrary reference point, X. Now after we measure the kinetic, nuclear, heat and all other forms of energy, we calculate the PE for every object to the point X. However not everything is attracted towards X, so... what comes next? I'm thinking it's got something to do with this potential well concept?
I don't even know what I'm trying to get at really, I think I'm just having trouble expanding this to a larger system that one gravitational field, where things are nice and tidy.
I do know calculus, but I don't know what a path integral is. Is it different to a regular integral? An integral of what? Wikipedia isn't helping.
Thanks again.
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Re: Gravitational potential energy (wtf?)
well in the bigger picture you have more than one potential well, you have a dip for every mass in the universe, each of these dips extends toward infinity but they are really small unless you are really close to a mass and/or its really big
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Re: Gravitational potential energy (wtf?)
Simple: if something is not attracted to point X by gravitation, it's GPE relative to that point will be zero, or negative.
As long as you don't get confused and start thinking that negative GPE means repulsion, it will all add up eventually.
As long as you don't get confused and start thinking that negative GPE means repulsion, it will all add up eventually.
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Re: Gravitational potential energy (wtf?)
A path integral is an integral along apath, where the path is not necessarily a straight line. If I take a ball, lift it up to the clouds, move to Africa, let it drop to one meter above the gound, and then move back to where I started, how much net work have I done to overcome gravity?
A path integral would calculate force times distance for every part of that complicated path. But as we know that gravity is conservative, every path with the same start and end point takes the same amount of energy.
So instead, I can calculate along the simple path: lift the ball one meter up. And the answer is the same. Note that not every force is conservative: if I wanted to calculate the energy lost to air resistance, the different paths would give different answers. But conservative forces are really, really nice.
As for you other question about multiple gravity fields. Let's look at a 2dimensional world, so we can plot potential energy in the third dimension. If there is only one planet: the potential will look like a hole in the ground. Lowest in the middle of the planet, curving upwards inside the planet and curving back outside of it. If you go to infinity, the potential asymptotically becomes horizontal at a constant 'height' above the lowest point.The tradition is to call this level at infinity '0', and therefore all points will have negative energy, with the lowest at the planet centre.
In this picture, you can imagine potential energy like this: If you put an object (almost) infinitely far away, with zero kinetic energy, it has zero energy in total. If you drop it to any point, its total energy will still be zero, so the potential energy of that point is the negative of the kinetic energy a body gets gets by falling to it.
With this bodyatinfinity trick, it becomes easier to imagine what happens if you have more than 1 body. If you drop a body from infinity, at every point it will feel the force of both objects, but not exactly in a straight line. So its kinetic energy after falling to a certain point is the the energy it would have gotten by falling to that point with only mass 1 around, plus the energy it would have gotten from falling with only mass 2 around.
You look at the shape of the potential of both bodies, again you call the level at infinity '0', and then you add the curves of both potentials to get the total potential. So infinity is still 0, but you now have 2 holes next to each other, each a bit lower than if that object were alone. If one hole is lot deeper than the other, as with the earth and sun, you get a very deep hole with a small hole in one of its sides.
A path integral would calculate force times distance for every part of that complicated path. But as we know that gravity is conservative, every path with the same start and end point takes the same amount of energy.
So instead, I can calculate along the simple path: lift the ball one meter up. And the answer is the same. Note that not every force is conservative: if I wanted to calculate the energy lost to air resistance, the different paths would give different answers. But conservative forces are really, really nice.
As for you other question about multiple gravity fields. Let's look at a 2dimensional world, so we can plot potential energy in the third dimension. If there is only one planet: the potential will look like a hole in the ground. Lowest in the middle of the planet, curving upwards inside the planet and curving back outside of it. If you go to infinity, the potential asymptotically becomes horizontal at a constant 'height' above the lowest point.The tradition is to call this level at infinity '0', and therefore all points will have negative energy, with the lowest at the planet centre.
In this picture, you can imagine potential energy like this: If you put an object (almost) infinitely far away, with zero kinetic energy, it has zero energy in total. If you drop it to any point, its total energy will still be zero, so the potential energy of that point is the negative of the kinetic energy a body gets gets by falling to it.
With this bodyatinfinity trick, it becomes easier to imagine what happens if you have more than 1 body. If you drop a body from infinity, at every point it will feel the force of both objects, but not exactly in a straight line. So its kinetic energy after falling to a certain point is the the energy it would have gotten by falling to that point with only mass 1 around, plus the energy it would have gotten from falling with only mass 2 around.
You look at the shape of the potential of both bodies, again you call the level at infinity '0', and then you add the curves of both potentials to get the total potential. So infinity is still 0, but you now have 2 holes next to each other, each a bit lower than if that object were alone. If one hole is lot deeper than the other, as with the earth and sun, you get a very deep hole with a small hole in one of its sides.
Re: Gravitational potential energy (wtf?)
Something in space that doesn't seem gravitationally attracted to some other object still has a potential energy relative to it. You can break the problem up into chunks to find it. First, how much energy would it take to move that object in space to a point where it WOULD be attracted to the other? Second, how much potential energy does the moved object have at that point? Subtract the first answer from the second answer, and you'll have its gravitational potential for the original positions.
It's kind of like finding the potential energy of a bike rider on a hill relative to a point on the OTHER side of the hill. He's not rolling towards the point you're interested in at first, but with a certain amount of energy you can get him to a point where he is.
Aero/astro engineers do something related to this when calculating how much energy it takes to get a satellite from the surface of the Earth to a stable orbit around Mars. It's not as if something in Martian orbit can ever fall back to earth, but the equations still have meaningful results.
It's kind of like finding the potential energy of a bike rider on a hill relative to a point on the OTHER side of the hill. He's not rolling towards the point you're interested in at first, but with a certain amount of energy you can get him to a point where he is.
Aero/astro engineers do something related to this when calculating how much energy it takes to get a satellite from the surface of the Earth to a stable orbit around Mars. It's not as if something in Martian orbit can ever fall back to earth, but the equations still have meaningful results.
Occam's Quandary: any idea can be made to sound like the simpler one.
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Re: Gravitational potential energy (wtf?)
Any conservative force can be described as the gradient of some scalar (it's potential). In local gravity, F=mg
[math]F=mg\hat{y}=\nabla V\Rightarrow V=mg\bar{y}+C[/math]
Where C is an arbitrary constant. Now, the work needed to move an object between two points in space is
[math]W=\int_{a}^{b}\bar{F}\bullet d\bar{l}[/math]
Where of course
[math]W=\int_{a}^{b}\nabla V\bullet d\bar{l}[/math]
And then the gradient theorem tells us that
[math]\int_{a}^{b}\nabla V\bullet d\bar{l}=V(b)V(a) = mg(y_{b}y_{a}) = W[/math]
In other words, to lift an object from height a to b, all you have to do is look at the change in gravitational potential (that is, potential energy.) Also, most importantly, it is only meaningful in comparison to something. The arbitrary constant C is inescapable, but it doesn't matter since it goes away when you subtract the PE of one point in space from the PE of another.
[math]F=mg\hat{y}=\nabla V\Rightarrow V=mg\bar{y}+C[/math]
Where C is an arbitrary constant. Now, the work needed to move an object between two points in space is
[math]W=\int_{a}^{b}\bar{F}\bullet d\bar{l}[/math]
Where of course
[math]W=\int_{a}^{b}\nabla V\bullet d\bar{l}[/math]
And then the gradient theorem tells us that
[math]\int_{a}^{b}\nabla V\bullet d\bar{l}=V(b)V(a) = mg(y_{b}y_{a}) = W[/math]
In other words, to lift an object from height a to b, all you have to do is look at the change in gravitational potential (that is, potential energy.) Also, most importantly, it is only meaningful in comparison to something. The arbitrary constant C is inescapable, but it doesn't matter since it goes away when you subtract the PE of one point in space from the PE of another.
I edit my posts a lot and sometimes the words wrong order words appear in sentences get messed up.
 You, sir, name?
 Posts: 6983
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Re: Gravitational potential energy (wtf?)
Thanks. The fora do not seem to like my TeXcraft.
I edit my posts a lot and sometimes the words wrong order words appear in sentences get messed up.
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