The sun exerts a force of 4.0 X 10^28 N on the earth, and the earth travels 9.4 X 10^11 m in its annual orbit around the sun. How much work is done by the sun on the earth in the course of a year? Explain.

none

why?

I bet Cap'nfriend will try to use a newnick in the future...

42

I came up with 42.

EDIT: So did CP

Jump off a bridge and die. We won't do your homework.

A charged particle accelerating through a magnetic field releases photons. Circular motion is constant acceleration. Why don't atomic electrons constantly emit light? The answer may surprise you!

Actually, no work is done by the sun on the Earth because there is no sun!

DUH-DUH-DUH!!!!!

Because (assuming everything is ideal) the force is always perpendicular to [direction of motion] Work requires a force to act in the same direction as [direction of motion]

Edit: I fixed it to use a more precise term

Actually the orbit could be any shape we wanted it to be; as long as the Earth returns to the same place (which it does), the work done will be zero because a gravitational field is conservative.

well if the orbit kept changing the earth wouldn't return to the same place, so like, ideal haha

42.

Work = Force * distance

The earth maintains a net distance from the sun throughout the year, and though the sun moves through the galaxy, our planet maintains it's orbit relative to the sun (which is all that matters in this case).

The sun exerts a force on the earth, but that force does not move the earth any closer to the sun over the course of a year. Therefore "distance" term in the above equation is 0.

Work = Force * 0
Work = 0

No work is done on Earth even though it's constantly accelerating. (Its total kinetic energy remains constant)

Problems like this one are why friend14 doesn't understand physics.

This is similar to how electrons in normal atomic orbits don't emit photons. When an electron inside of an atom gains energy it slips into a farther orbit (moves faster). When the electron returns to its normal orbit it emits a photon (because it loses energy). Maybe someone better at quantum physics can explain this in greater detail.

You do know that electrons don't actually orbit nuclei, right?

Don't they describe it as an electron cloud? And they have different shapes, like spheres and figure-eight type shapes and such. IIRC.

In an atom there are energy states for the electron orbits. 1s 2s 2p etc. When an electron gains enough energy, it jumps to a higher state (orbit). When the electron loses energy, the electron drops. Conservation of energy says that energy has to go somewhere. That somewhere is the photon. It's this principle in how a laser works. The atoms of the gas are simultaneously charged with energy, then the atoms all simultaneously drop energy states and thus you get a very monochromatic light and thus a laser beam.

Per the sun: If our orbit were circular, the Force exerted on the earth is at a right angle to the motion of the Earth. Thus no work is done. Work is defined by integral F dot ds. Since we're talking uniform we can just simplify to Fd*cos(theta). Since theta = pi/2 and cos(pi/2)=0, W = 0.

Yes they do.
It's a cloud of orbiting electrons. Just like a planet there is no set 'path' for it to follow. Figure 8 orbits probably not, though.

Yes, they orbit the nuclei. But the electrons has different layers of clouds, as JG pointed out.

It's like...when you say orbit, everyone things that the things being in orbit are linear to each other. Like Jon'C said.

Jon'C is right. They do orbit. They just appear as a cloud. You can't see individual electrons because of the Uncertainty Principle. Knowing both position and velocity of the electron would make Heisenburg haunt you.

oh the joys of physics, I'm bloody annoyed I missed the previous thread with CaptBevvil, do you think he would have listened to me when I mentioned that I'm doing a phd in experimental particle physics working on the ATLAS experiment at CERN....probably not (btw this is why I'm hardly on now...I have no life outside of work, grumble grumble)

anyway, what JG said is right

I wish I could setup some sort of pm that would alert me when someone comes along with stupid physics, I love de-bunking people especially when it comes to people trying to say the Newton's laws of motion are wrong.

[QUOTE=James Bond]I'm doing a phd in experimental particle physics working on the ATLAS experiment at CERN.[/QUOTE]:eek:

I think a psycho biology professor and some kid who thinks he knows everything would know a BIT more than some stupid kid getting a "PHD" in "experimental particle physics" while working on "the ATLAS experiment at CERN".

Obviously.

Okay, I've got a physics question.

If a light wave is the combination of an electric field and a magnetic field, why doesn't the light wave decrease in amplitude (and thus energy, as E[sub]wave[/sub] α f[sup]2[/sup]A[sup]2[/sup]) with the distance from the point of origin like the fields do?

The light wave is not a field. The light field does decrease in intensity.

But not in energy, right? Is there any math simple enough to explain this to me?

If you put a lightbulb at the center of a perfect sphere, as the radius of the sphere increases the amount of electromagnetic radiation per square centimetre decreases. This is because the total output of light from a light bulb is constant. Intensity of the light reaching a particular point in space is, similar to the function of a radius for a sphere - an inverse exponential relationship between the distance and the initial power of the light source.

The light doesn't lose energy. There's simply the same amount of energy being spread over a much larger area.

"light" is just an EMF. The further you go away from the source of the EMF, the weaker it becomes. The math you want are Maxwell equations, Ampere's Law, and Gauss' Law. Those deal with the relationship between E and B fields.

I think I get the concept of the same amount of energy being spread over a larger radius, but I'll go look at those equations. Thanks.

Or maybe I'll assume that the equations are wrong and disprove them.

Make sure you use concepts such as velocital physics and free energy!

I was trying to explain to Iz about the fact that two light beams passing each other have a relative velocity of c. I think I broke her brain.

[QUOTE=James Bond]I'm doing a phd in experimental particle physics working on the ATLAS experiment at CERN.[/QUOTE]

[QUOTE=James Bond]I'm doing a phd in experimental particle physics working on the ATLAS experiment at CERN....[/QUOTE]

you know the thing is, the best for me at the moment isn't that I'm working on ATLAS, but that I have my own office...well, kinda, I share it with 3 other post-grads , but still, my own office in the physics department 24/7

I even have my name on the departmental website too linky (look for James Poll), I'll have my ugly mug up there soon as they are moaning at me to get my photo done. I also have a little sign-in board where I have to move it from In to Out depending on where I am...amazing how the little things seem more important.

Sorry to jump back to the middle of the thread, but electrons don't orbit the nucleus.

A given electron in an atom exists as a probability cloud around the nucleus; its position is not defined except at the moment you measure it. It's not just that we don't know the electron's position: the position actually isn't defined to a single point. Therefore you can't talk about motion or orbits in a classical sense; 'motion' is meaningless to something without a location.

If an electron were to orbit an atom in a classical sense, it would constantly lose energy because accelerating charged particles constantly emit electromagnetic waves. Thus the electron would spiral into the nucleus; stable, planet-like orbits aren't possible. The first response to this problem was the Bohr model of the atom, but that was already viewed as incorrect by 1930.

This really isn't controversial stuff; they teach you that the Bohr model isn't right in high school chemistry. Take a look at the Wikipedia articles on atomic orbitals, the Bohr model, quantum mechanics, or anything else that's related if you don't believe me.

Your problem here is that you're assuming that a light wave will emanate from a stationary charged particle. You're thinking of electric fields that surround a stationary charged particle, whose magnitudes do drop off proportional to the inverse of r squared.

This is a completely different case from electromagnetic waves. Say you get a light wave started (by jiggling a particle up and down). Then ignore the initial particle completely, and just look at the electric and magnetic fields in space that's completely free of all particles. Maxwell's equations allow a wave to keep traveling on its own in free space because, in effect, the light wave keep recreating itself: the changing electric field creates a changing magnetic field, which creates a changing electric field, and so on. The amplitude of the wave doesn't diminish because there's no reason why it should. The energy carried by the wave doesn't have anywhere to go till the wave hit something; it can't just disappear because that would violate conservation of energy.

Ugh. I have AP Physics next trimester. That's Monday. Thanks for reminding me.