# Higher v but lower Energy?

1. Jul 10, 2007

### Brock

When and electron loses energy it drops in orbit of the neucleous, but does it pick up veolocity? In this lecture at 17:20 he says it does, but is he mistaken, because it would make more rounds around the neucleous because it has less distance to travel, and that would not be more velocity but probably the same. I just don't see how it would pick up more speed from it's starting point.

Last edited by a moderator: May 3, 2017
2. Jul 10, 2007

### mgb_phys

Thinking of an electron as a planet 'orbiting' a nucleus isn't a very useful picture. Even thinking of a bound electron as an object with a time varying position isn't really correct.

3. Jul 10, 2007

### belliott4488

It's true that thinking about electrons as discrete particles with well-defined positions and velocities is incorrect, but I think the professor in the video linked above was using such a classical picture just to speak loosely and give a general idea of what's going on.

Brock's original question, however, could just as easily be asked of a classical particle in orbit, however, if I'm understanding him correctly. If you think of a satellite in orbit around the Earth, it most certainly does gain velocity even as it loses energy and falls to a lower orbit. Lower orbits require greater velocity - you can think in terms of conserving angular momentum, if you know about that, or you can think about the change in potential energy that is converted to kinetic energy. As long as the object remains in orbit, lower orbits require higher velocities.

4. Jul 10, 2007

### LHarriger

When you say the electron loses energy you must be a little more specific about what energy is lost.

1) If you mean that the electron losses potential energy due to traveling closer to the nucleus then its velocity will definitely increase for the reason given by belliott: namely, conservation of angular momentum requires it. Your total energy is still conserved, its just that some potential energy has been converted to kinetic energy.

2) If you mean instead that the electron losses some of its total energy, ie: it emits a photon, then this would allow the electron to drop to a lower orbit without a velocity change. Once again, this can be shown by cons. of angular momentum. (Assuming the recoil is negligible)

Edit: All of the above is taking only classical physics are taken into consideration.

Last edited: Jul 10, 2007
5. Jul 10, 2007

### Brock

If it is pushed in, then it's veolocity is increased, and it's PE is also. If some of it's total energy is lost then it increases in RPM, is going the same speed, but lost some PE from it's charge attraction to the neucleous. So he is wrong in that video because energy = PE+KE.

6. Jul 10, 2007

### LHarriger

Pushing in implies a new force acting over a distance which increases the total energy.

No, your potential energy decreases the closer you get to the nucleus.

Above you just said that potential energy increased?

7. Jul 10, 2007

### belliott4488

Brock, you need to slow down and learn some of this stuff a little more thoroughly. If you object to something a college level instructor is saying, especially something this basic, it's much more likely that you've got a misunderstanding than that he is wrong.

For an attractive potential, PE decreases as distance decreases, at the rate 1/r for both electromagnetic and gravitational potentials.

Just figure out what velocity is required for circular orbits of radii r1 and r2, given an inverse-square law force. Set the centripetal force (k/r^2) equal to the inertial, or centrifugal, force, (mv^2/r) for each case and see what you get. You can now compare v1 and v2 to see which is bigger.

- Bruce

8. Jul 11, 2007

### Parlyne

In a stable orbit in classical gravity or E&M, the total energy, of course will be E = T + V. A result know as the Virial Theorem guarantees that in such stable orbits the average kinetic and potential energies are related by $$\langle V \rangle = -2\langle T \rangle$$. The total energy will, then, be $$E = -\langle T \rangle$$. So, if the total energy decreases (becomes more negative), the kinetic energy must actually increase. The potential simply decreases twice as much as the kinetic increases.

Last edited: Jul 12, 2007
9. Jul 11, 2007

### belliott4488

Thanks, Parlyne - that was helpful (to me, at least ). But, you did mean to say that "if the total energy decreases (becomes more negative), the kinetic energy must actually increase," correct?

10. Jul 12, 2007

### Parlyne

Yeah, that's what I was trying to say. Serves me right for trying to explain a physical idea before the coffee kicked in.

11. Jul 13, 2007

### Brock

When it's pushed in it gains PE because it's under pressure, when the pressure is released it loses energy by slowing down and moving out to a relaxed orbit. The professor is saying that it loses energy, moves in, and picks up velocity, well to pick-up v it must be forced in, in which case it would aslo pick up energy. So if it picks up v it does not lose energy like he said, unless it loses much energy falls in a closer orbit, and then gains a bit of pressure energy that pushed it in some to pick up v.

In the case of planets is mass the only thing that determines how fast an object travels as it orbits another?

Last edited: Jul 13, 2007
12. Jul 13, 2007

### belliott4488

Okay, this is wrong for several reasons. First of all, pressure, which is force per unit area, has no relevance here; I'll assume you mean a force. Next, if you apply a force to reduce the radius of the orbit, you don't gain potential energy. You're thinking of pushing against some kind of restoring force as if there's a potential field associated with it, but that applies only to conservative forces. The "force" you "push" against in this case is the inertial centrifugal force, which is not a conservative force at all. In other words, even if you could simply "push" on an orbiting object to get it "move in", you could not get back the work you did in the way you can with a conservative force field, so you don't gain potential energy.
Last, the way you reduce the radius of a circular orbit is to apply a force in the negative velocity direction, which slows down the orbiting object, reducing its kinetic energy. (Actually, you do it twice - once to convert to an elliptical orbit with perigee at the lower radius and apogee at the original radius, and then again at the perigee location to lower the apogee to the new radius, converting to a circular orbit again - it's called a "Hohmann Transfer" if you want to look it up.) Although you've slowed down the object with the application of the first force (which is why it now follows a lower trajectory than before), it still speeds up as it falls, so that its velocity at perigee is higher than the original velocity. After the second application of force, the velocity at perigee is reduced so that the object now follows a lower trajectory and doesn't reach the prior apogee height. Its energy is now lower than what it started with, but its velocity is higher because that's true for any circular orbit with a smaller radius - Kepler's Third Law.
Nope - go back to your basic lessons in orbital dynamics (I'm assuming you're studying this stuff). The mass of an object cancels out when solving for its motion in a gravitational field - that's why two objects of different mass fall at the same rate, disregarding air resistance. The same holds true for orbits (which are sometimes described as "freefall that keeps missing the Earth"). Orbits in a given gravitational field are determined by the initial position and velocity vectors, and that's all. Two objects of different mass will follow identical orbits provided they both start at the same place with the same velocity.
Now, you might be thinking of the mass of the central body, in which case, yes, the field depends on that, and therefore so do the velocities of any orbiting bodies. But their velocities can't be determined from the mass of the central body alone - you need to know the details of their orbits.
If you're just limiting yourself to circular orbits, however, then all you need to know is the radius, and you can get the velocity (look at my earlier post on balancing the central force with the inertial, or centrifugal force). For example, a handy approximation for a circular orbit 7000 km from the center of the Earth is to give the object in question a velocity of 7 km/sec perpendicular to the radial vector. It's not exactly circular, but it's decent for back-of-the envelope calculations. The point is that once you have the central field, the radius alone is enough to determine the velocity in a circular orbit.
Hope that helps some,
Bruce

Last edited: Jul 13, 2007
13. Jul 13, 2007

### olgranpappy

Good answers, all. My two cents would be that perhaps this issue would be easier for Brock to understand if it is explained in terms of concrete numbers. For example

Consider an electron in the ground state of hydrogen:
1. Its kinetic energy is 13.6 eV
2. Its potential energy is exactly 0.0 eV. (I have redefined my energy scale such that the ground state potential is exactly zero, which I am perfectly free to do and which allows us to only consider positive numbers)
3. Its total energy is 13.6 eV + 0.0 eV = 13.6 eV

Consider an electron in the first excited state of hydrogen:
1. Its kinetic energy is: 3.4 eV
2. Its potential energy is: 20.4 eV
3. Its total energy is: 3.4 eV + 20.4 eV = 23.8 eV

So, we see that the total energy of an electron in the first excited state of hydrogen is indeed greater than the energy of an electron in the ground state of hydrogen. I.e. 23.8 eV is greater than 13.6 eV.

In order to make a transition from the first excited state to the ground state the electron must loose (23.8 eV - 13.6 eV) = 10.2 eV of energy.

How to loose this energy? Emit a photon of energy 10.2 eV. Which, if I remember correctly, corresponds to a wavelength of about 121 nanometers... and indeed, there is a very nice line right at 121 nanometers in the absorption spectrum of hydrogen (I.e., the far "right" of the Lyman series).

Cheers.