Fictious forces and calculation of energy + c.o.m. question

1. Jul 20, 2015

Electric to be

Hi, I am aware that in order for energy to be conserved the reference frame that you are using must be inertial.

However, consider the situation where you are using the reference frame of the earth and there is a decently sized planet directly above on course for collision. The two have a gravitational potential energy U between them.

My question is as the two approach, the earth is clearly not in an inertial reference frame. I know that it would be wise to use the center of mass frame in this situation, but if I were to stick to the earth, to convert to an inertial reference frame wouldn't I simply add a force equal in magnitude and opposite in direction to the earths motion to give it a zero net force? Then following this I would give the approaching planet an additional ''downward'' (in relation to the earth) force that is equal to it's mass x earth's acceleration?

Now this seems fine, but if I were to run an integral for this fictitious force plus the original gravitation force multiplied by the distance just before collision, it seems like the approaching planet receive an amount of kinetic energy gained that is much higher than the potential energy lost, violating c.o.e. . So are fictitious forces only really useful for examining relative motion and not energy transfer?

I understand using the center of mass frame would provide an inertial frame that would get rid of these problems but for the sake of the question I'm being stubborn and sticking with the earth.

Also, when calculating the binding energy needed to separate two masses in orbit, should the center of mass frame always be used to determine kinetic energy? As I understand E = K + U must equal zero or larger for the masses to seperate to 'infinity', and using a different reference frame could easily result in a kinetic energy that easily exceeds this boundary even though the two masses are certainly still bounded. If the center of mass frame is used, do I simply add the kinetic energy of each mass relative to c.o.m. and then add the (negative)gravitational potential energy?

Finally, a standard conceptual question that I have trouble visualizing. Imagine two similar masses, if they have the correct initial angular momentum, will they 'orbit' the center of mass between them or will they actually orbit each other? Should angular momentum for this system then be calculated with respect to the center of mass for each mass and then added together to determine the total angular momentum?

Thank you for any help! Excuse my typos it is late and I am very tired haha.

2. Jul 20, 2015

A.T.

If the inertial force can be described by a constant conservative force field, like in a frame of uniform linear acceleration, then you still can use CoE by accounting for potential energy of that field. But in your example the acceleration of the Earth changes, so that won’t work.

3. Jul 20, 2015

Qwertywerty

So you can't use the non-inertial frame ??

Can you not simply write pseudo force along in the line integral ?

Last edited: Jul 20, 2015
4. Jul 20, 2015

Electric to be

I don't think it would work. Imagine a more extreme case of a small mass and the earth and say you are in the small masses frame. If you wish to calculate the work done on earth and you give earth a fictious force equal to the small masses's acceleration multiplied by the earths mass and run and multiply by the distance, you will get much much more work than if you simply calculated the work from earth's (approximately) inertial frame of reference.

I guess fictitious forces just don't work for energy calculations?

5. Jul 20, 2015

jbriggs444

See the response from A.T. in post #2 above. They work fine as long as they amount to a constant, conservative force field.

For instance, you could compute potential energy from the centrifugal and Coriolis forces in a rotating frame of reference. It is straightforward to show that both centrifugal and Coriolis forces are conservative. As long as the rotation rate is fixed, potential energy can be computed.

6. Jul 20, 2015

Electric to be

Could you possibly explain further what this means exactly? In the case of the earth and a smaller mass isn't the acceleration constant? How would I apply fictitious forces from the small masses reference frame?

7. Jul 20, 2015

jbriggs444

If the distance between the earth and the smaller mass changes significantly then the acceleration from the Earth's gravity will change as well. The Earth's gravity can not then be regarded as a constant force field. Its snapshot at any instant in time will still be conservative. But it will not be constant over time.

In the small mass's reference frame there is a ficticious force equal in magnitude to the Earth's gravity and opposite in direction. The small mass is subject to both forces. The net is zero and it does not accelerate (in this frame).

The Earth is subject to the ficticious force. It accelerates at 1 gee and trades off potential energy (in the ficticious force field) for kinetic energy (in the chosen frame) at an enormous rate.

In addition, the Earth is subject to the gravitational attraction from the small mass. That does work on the Earth at a very relevant rate.

8. Jul 20, 2015

Electric to be

Alright the introduction of a fictitious force field makes more sense as there is more potential energy. Now is potential energy from the gravitational field also considered since a gravitational force still acts on the earth? How can the work be accurately calculated using potential energies from two fields?

9. Jul 20, 2015

jbriggs444

Rather than trying to decompose the problem into the potential energy of A in the gravitational field of B somehow added to the potential energy of B in the gravitational field of A, how about this instead...

Take a hypothetical nail and pound it through A into the origin of a coordinate system. Now you can compute the potential energy of B in the gravitational field of A. It is the amount of work required in the chosen coordinate system to take object B from the defined reference point (typically at infinity) to its actual location.

During this interaction, the nail did no work and no work was done on the nail. Regardless of how much stress it was under, it just sat there motionless. Zero displacement means zero work done.

Repeat the exercise but remove the nail from A and pound it through B. Now you can compute the potential energy of A in the gravitational field of B. It is the amount of work required in the chosen coordinate system to take object A from the defined reference point to its actual location.

Again, the nail does no work.
Question: Can you convince yourself that the computed potential energy will be the same in the two calculations above?

Question: Since the nail does not contribute any energy one way or the other, would it make sense to think that it is irrelevant -- that the potential energy associated with the gravitational attraction between A and B might not depend on the coordinate system but instead only depend on the separation between A and B and their respective masses?

10. Jul 20, 2015

Electric to be

Alright so essentially the total potential energy released will be:
(fictious force acting on the earth + mg) • (distance seperating) correct?

And I guess my final question is why does the field have to be constant? Can't the potential energies be calculated using an integral for work that uses force as a function of distance to account for the changing force?

11. Jul 20, 2015

jbriggs444

Yes. Though you might want to calculate the magnitude of the potential energy in question. You have a huge potential energy associated with the height of the Earth in this ficticious force field. You have a huge kinetic energy associated with the velocity of the Earth as it accelerates in this field. Suppose, that you decide to account for the inverse square law. Can you provide a rough estimate for how much error (in Joules) results from the decision to model the earth's gravity as uniform rather than inverse square over, let's say, 1 meter of change in height near sea level. How does this compare to the potential energy of a 1 kg object at 1 meter in height compared to its potential energy at sea level.
If you want potential energy to be a fixed function of position in some reference frame then the force field has to be a fixed function of position in that reference frame. If the field varies over time then the path integral for work can vary depend on when the path is traversed. Potential is then not definable as a fixed function of position.

But yes, if you want to calculate potential energy for a configuration based on the path integral of the work done on all of the objects in that configuration, that works.