"total mechanical energy of Earth's orbital motion"
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"total mechanical energy of Earth's orbital motion"
The question I have on an assignment is:
"What is the total mechanical energy associated with Earth's orbital motion?" (answer in Joules)
I am stumped. It gives no other information whatsoever, and I never know if that's because the question is poorly written (which they often are), or if there's something trivial I'm missing.
Am I correct in saying that, to figure this out, you need to break out equations for Earth's kinetic and gravitational potential energy, which then requires information about its mass, orbital speed, orbital radius (assumed to be a circle), and the mass of the sun? I tried this and got a massively negative number, but I have not done gravitation energy stuff in over a year so I'm really not confident that I'm doing things right yet.
Can anyone at least clarify the question, and maybe point me in the right direction as to how to go about solving this? It seems unlikely that they'd require students to search up half a dozen values on wikipedia, but I wouldn't really be surprised considering the quality of our assignment questions. Also yes, this is a marked assignment, not that it really counts for anything significant, but I ask for no full solutions until Friday at 11:59 EST please
"What is the total mechanical energy associated with Earth's orbital motion?" (answer in Joules)
I am stumped. It gives no other information whatsoever, and I never know if that's because the question is poorly written (which they often are), or if there's something trivial I'm missing.
Am I correct in saying that, to figure this out, you need to break out equations for Earth's kinetic and gravitational potential energy, which then requires information about its mass, orbital speed, orbital radius (assumed to be a circle), and the mass of the sun? I tried this and got a massively negative number, but I have not done gravitation energy stuff in over a year so I'm really not confident that I'm doing things right yet.
Can anyone at least clarify the question, and maybe point me in the right direction as to how to go about solving this? It seems unlikely that they'd require students to search up half a dozen values on wikipedia, but I wouldn't really be surprised considering the quality of our assignment questions. Also yes, this is a marked assignment, not that it really counts for anything significant, but I ask for no full solutions until Friday at 11:59 EST please
 BlackSails
 Posts: 5315
 Joined: Thu Dec 20, 2007 5:48 am UTC
Re: "total mechanical energy of Earth's orbital motion"
Look up the velocity of the earth, look up the mass of the earth, find the kinetic energy.
You might also want the rotational kinetic energy, in which case you use the mass and the fact that the period of the earth's rotation is 24 hours.
You might also want the rotational kinetic energy, in which case you use the mass and the fact that the period of the earth's rotation is 24 hours.
Re: "total mechanical energy of Earth's orbital motion"
were is the zero point you are measuring from?
 Kobayashi_Maru
 Posts: 44
 Joined: Tue Jan 29, 2008 1:31 pm UTC
Re: "total mechanical energy of Earth's orbital motion"
The question is poorly worded, or, at the least, ambiguous. By total mechanical energy does it mean kinetic and potential. By "energy associated with the earth's orbit" does it want you to include rotational energy? If I were to interpret this in a strict, literal sense, I would say it wants the sum of the earth's orbital (not rotational) kinetic energy and the potential energy due to gravity. The potential is generally measured with a zero point at infinity, in which case the energy would indeed be massively negative. If you wanted to be a smart ass about it, you could say that you define the potential energy to be zero at the radius of the earth's orbit. In which case it wouldn't matter whether the question was asking for it or not.
Re: "total mechanical energy of Earth's orbital motion"
Mechanical energy always means kinetic and potential. I don't think this is a poorly worded question. I've seen it before a long time ago.
You need to find the kinetic energy associated with Earth rotating, then find the kinetic and potential from Earth revolving about the Sun. The potential will be due to the distance from the sun. To find the kinetic energies, use the known constants associated with orbital period, rotational period, and distance from the sun. You could probably even ignore calculus completely for this question. Just solve for the acceleration the Earth experiences from the Sun.
You need to find the kinetic energy associated with Earth rotating, then find the kinetic and potential from Earth revolving about the Sun. The potential will be due to the distance from the sun. To find the kinetic energies, use the known constants associated with orbital period, rotational period, and distance from the sun. You could probably even ignore calculus completely for this question. Just solve for the acceleration the Earth experiences from the Sun.
 BlackSails
 Posts: 5315
 Joined: Thu Dec 20, 2007 5:48 am UTC
Re: "total mechanical energy of Earth's orbital motion"
You can save work with the virial theorem. Calculate either the potential energy or the kinetic energy, and since the system is bound, you automatically know the other (The potential is 2 times the kinetic)
Re: "total mechanical energy of Earth's orbital motion"
The virial theorem would relate the average kinetic and average potential energies. In general, you can't just take a point and say that the potential is 2 times the kinetic. From what I can see, calculating the average energies would be much harder than needed. Just calculate each component individually here  it isn't that hard.

 Posts: 2986
 Joined: Tue May 27, 2008 10:42 pm UTC
Re: "total mechanical energy of Earth's orbital motion"
Can you take the earth as a resting point of reference? It's all relative.
It's all physics and stamp collecting.
It's not a particle or a wave. It's just an exchange.
It's not a particle or a wave. It's just an exchange.
Re: "total mechanical energy of Earth's orbital motion"
I don't think the question is ambiguous. It just requires you to correctly interpret the language physcisists use in daily life. What is probably what your teacher wants you to be able to do, and he might even have phrased the question like that on purpose.
The question reads "associated with Earth's orbital motion". So you don't need to worry about the rotation of the earth around its axis. And since the question is about orbit around the sun, the proper reference frame is that where the sun sits at the origin. So not potentional energy with respect to infinity.
The question reads "associated with Earth's orbital motion". So you don't need to worry about the rotation of the earth around its axis. And since the question is about orbit around the sun, the proper reference frame is that where the sun sits at the origin. So not potentional energy with respect to infinity.
It's one of those irregular verbs, isn't it? I have an independent mind, you are an eccentric, he is round the twist
 Bernard Woolley in Yes, Prime Minister
 Bernard Woolley in Yes, Prime Minister
Re: "total mechanical energy of Earth's orbital motion"
Technical Ben wrote:Can you take the earth as a resting point of reference? It's all relative.
No. Earth is not an inertial frame (it's undergoing accelerated motion). So that's a bad idea.
For the purpose of this question, there's only one reference frame, and one point on this frame, where both requested quantities are in fact constant.
It's one of those irregular verbs, isn't it? I have an independent mind, you are an eccentric, he is round the twist
 Bernard Woolley in Yes, Prime Minister
 Bernard Woolley in Yes, Prime Minister
 Kobayashi_Maru
 Posts: 44
 Joined: Tue Jan 29, 2008 1:31 pm UTC
Re: "total mechanical energy of Earth's orbital motion"
Sethric wrote:You need to find the kinetic energy associated with Earth rotating, then find the kinetic and potential from Earth revolving about the Sun.
Why do you include the rotational kinetic energy? I would describe the orbit of the earth by stating [imath]r(\theta)[/imath], its distance from the sun at a given angle. The energy associated with this motion does not include rotational kinetic energy. The way the question is stated it is arguable whether you should include the rotational term or not. Which is why I said it is ambiguous.
Edit

Also, I'd like to note that the two people who agree the question is unambiguous have interpreted it differently.
Last edited by Kobayashi_Maru on Thu Nov 19, 2009 4:18 pm UTC, edited 1 time in total.
 BlackSails
 Posts: 5315
 Joined: Thu Dec 20, 2007 5:48 am UTC
Re: "total mechanical energy of Earth's orbital motion"
Sethric wrote:The virial theorem would relate the average kinetic and average potential energies. In general, you can't just take a point and say that the potential is 2 times the kinetic. From what I can see, calculating the average energies would be much harder than needed. Just calculate each component individually here  it isn't that hard.
You can when the system is bound and constant (which it is, assuming the earth's orbit it circular, not elliptic).
Re: "total mechanical energy of Earth's orbital motion"
Well, I got it first try, fortunately after noticing I had forgotten to convert km to m
Just considering what we know thus far in the course, we were expected to assume a circular orbit, and we haven't done rotational energy so I doubted that would factor in (and intuitively, how does the earth's rotation have to do with the energy of its orbit?)
And if anyone's curious the solution is negative on the order of 10^33. That there's a big number.
Thanks for all the replies!
Just considering what we know thus far in the course, we were expected to assume a circular orbit, and we haven't done rotational energy so I doubted that would factor in (and intuitively, how does the earth's rotation have to do with the energy of its orbit?)
And if anyone's curious the solution is negative on the order of 10^33. That there's a big number.
Thanks for all the replies!
 Kobayashi_Maru
 Posts: 44
 Joined: Tue Jan 29, 2008 1:31 pm UTC
Re: "total mechanical energy of Earth's orbital motion"
Diadem wrote:. . .the proper reference frame is that where the sun sits at the origin.
"Proper" is too strong a word to use when talking about inertial reference frames, since they are all equivalent. I'll assume you meant "natural," in which case I agree with you.
Diadem wrote:So not potentional energy with respect to infinity.
I'm not completely sure what you mean here. Do you mean the potential energy is not zero at infinity? Once again, there isn't a right or wrong here, but the standard representation of gravitational potential energy is of the form [imath]V=\frac{k}{r}[/imath], which certainly goes to zero as r goes to infinity.
Re: "total mechanical energy of Earth's orbital motion"
Diadem wrote:The question reads "associated with Earth's orbital motion". So you don't need to worry about the rotation of the earth around its axis. And since the question is about orbit around the sun, the proper reference frame is that where the sun sits at the origin. So not potentional energy with respect to infinity.
Well, actually it would be centered at the center of mass of the system (which is, of course, very much inside the sun. For Jupiter it's outside it though.) Arguably, you should measure the [notional] rotation of the sun about that point as well.
Re: "total mechanical energy of Earth's orbital motion"
Kobayashi_Maru wrote:The way the question is stated it is arguable whether you should include the rotational term or not. Which is why I said it is ambiguous.
I disagree. The question states "Orbital motion". Rotation is not orbital motion. You might argue that you have to include the motion of the earth around the centre of the galaxy, but I think it's pretty clear that the question is not meant to ask that.
Random832 wrote:Well, actually it would be centered at the center of mass of the system (which is, of course, very much inside the sun. For Jupiter it's outside it though.) Arguably, you should measure the [notional] rotation of the sun about that point as well.
Yeah it's cleary that you are allowed to ignore that. Questions like this, if you're not supposed to ignore factors like that, they will always mention that. Same thing for the eccentricity of the earth's orbit. Or the fact that the earth is not a point particle.
Kobayashi_Maru wrote:Diadem wrote:. . .the proper reference frame is that where the sun sits at the origin.
"Proper" is too strong a word to use when talking about inertial reference frames, since they are all equivalent. I'll assume you meant "natural," in which case I agree with you.
All inertial frames are equal, but some are more equal than others. In this case only one of them gives you the right answer (kinetic energy is not reference frame independent!). For me the word 'natural' implies the easiest, most obvious choice. But in this case it's stronger than that. It's not just the easiest choice, it's the only correct one.
Kobayashi_Maru wrote:Diadem wrote:So not potentional energy with respect to infinity.
I'm not completely sure what you mean here. Do you mean the potential energy is not zero at infinity? Once again, there isn't a right or wrong here, but the standard representation of gravitational potential energy is of the form [imath]V=\frac{k}{r}[/imath], which certainly goes to zero as r goes to infinity.
You're right. I messed up a bit there.
I've always hugely disliked that way of looking at potential energy though. The Earth's potential energy is clearly positive. If it were to fall inwards you'd gain kinetic energy. Where does that come from? From its potential energy of course! Implying that potential energy is positive. No, I'm told. It's negative, and it the earth falls inward it becomes more negative. Yeah right. That's mathematically convenient, I agree, but it does not make any sense at all.
Here on earth we define potential energy as mgh. That's the proper way to do it. Now write this as [imath]\int mg dh[/imath] (by the by, how do you get integrals to display nicely in latex? The dh is clearly not the same kind of thing as the mg, but I can't get that right). Then generalize to g = g(h) and you're done. Unfortunatley of course this kind of blows up at zero. Damn. That is why people define it with respect to infinity. But negative potential energy just does not make any sense.
It's one of those irregular verbs, isn't it? I have an independent mind, you are an eccentric, he is round the twist
 Bernard Woolley in Yes, Prime Minister
 Bernard Woolley in Yes, Prime Minister
Re: "total mechanical energy of Earth's orbital motion"
In school i read that potential energy = mgh. Is it different in space for some reason? The concept of negative potential energy confuses me.
Re: "total mechanical energy of Earth's orbital motion"
Well the problem with potential energy is that it's not clear how to define an absolute potential. The difference in potential energy between two objects is welldefined. And the question "How much potential energy do I gain if I move an object over a distance of x" is unambigeous. But absolute potential energy is harder. With respect to what do you measure?
Another perhaps more insightful way of looking at this is that potential energy is 'energy gained when you move an object someplace else'. That begs the question 'what place else'. The formula mgh is correct, but it is the energy gained when you move an object to the surface of the earth. If you were interested in potential energy relative to other places, that is less useful.
In studies of gravitational problems, you usually consider planets and such as pointmasses. Your neglect their physical shape, because that would needlessly complicate things. But then how do you define potential energy? Not as energy gained when traveling to the surface, clearly. The most logical choice would be energy gained when you travel the the centre of an object. To r=0. Unfortunately this is infinite. So that don't work.
So physcisists go for the 2nd most obvious choice, which is infinity. So potential energy is energy gained when you move an object infinitely far away (ie: so far away from any mass that gravitational effects become neglicible). But this means moving it against the gravitational field, so that costs energy. So the gain is negative. So potential energy is now negative.
Another perhaps more insightful way of looking at this is that potential energy is 'energy gained when you move an object someplace else'. That begs the question 'what place else'. The formula mgh is correct, but it is the energy gained when you move an object to the surface of the earth. If you were interested in potential energy relative to other places, that is less useful.
In studies of gravitational problems, you usually consider planets and such as pointmasses. Your neglect their physical shape, because that would needlessly complicate things. But then how do you define potential energy? Not as energy gained when traveling to the surface, clearly. The most logical choice would be energy gained when you travel the the centre of an object. To r=0. Unfortunately this is infinite. So that don't work.
So physcisists go for the 2nd most obvious choice, which is infinity. So potential energy is energy gained when you move an object infinitely far away (ie: so far away from any mass that gravitational effects become neglicible). But this means moving it against the gravitational field, so that costs energy. So the gain is negative. So potential energy is now negative.
It's one of those irregular verbs, isn't it? I have an independent mind, you are an eccentric, he is round the twist
 Bernard Woolley in Yes, Prime Minister
 Bernard Woolley in Yes, Prime Minister
Re: "total mechanical energy of Earth's orbital motion"
I've been reading about it, but then don't you have the same sort of problem, which is that when r=0 the potential energy equals negative infinity?
Re: "total mechanical energy of Earth's orbital motion"
Oort wrote:I've been reading about it, but then don't you have the same sort of problem, which is that when r=0 the potential energy equals negative infinity?
Only in a black hole.
Re: "total mechanical energy of Earth's orbital motion"
Tass wrote:Oort wrote:I've been reading about it, but then don't you have the same sort of problem, which is that when r=0 the potential energy equals negative infinity?
Only in a black hole.
Which is to say that if we don't treat an object (say the Earth) as a point particle (but still assuming a perfect sphere), then as you start to travel inside the object, the shells of mass on the outside no longer exert a net gravitational force. Since the gravitational potential v = Intergral{ Gm/(r^2) dr} and as you start passing inside the object the mass inside a sphere of radius r is proportional to r^3, then the integral no longer diverges as you go to r=0. Of course if all the mass is concentrated at a point (e.g. a black hole, then the integrand is always proportional to r^2 and will still diverge)
So, if the gravitational potential is not infinity at the Earth's center, why do we still want to choose v = 0 at infinity? Because it's a convienant convention that is phsically significant for all systems. If we made the potential 0 at the Earths center, that particular choice would have no significance if we were working on the moon, or Jupiter, or Alpha Centuri, but choosing v = 0 at infinity is the same choice for all of these systems.
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