Science question:Orbits, acceleration and energy

[PainRack]

When the moon orbits the earth, is there work done? Its accelerating after all since its vector is changing, so, is there any energy loss?

Someone posted me this difficult question, mixed up along with a whole bunch of gravity/magnetic energy issues, I can explain the others but not this.

[Darth Wong]

When the moon orbits the earth, is there work done? Its accelerating after all since its vector is changing, so, is there any energy loss?

Someone posted me this difficult question, mixed up along with a whole bunch of gravity/magnetic energy issues, I can explain the others but not this.

There are small amounts of work done (work done via exertion of tidal forces through gravitational interaction), but not in the sense that this person is talking about.

Given a perfectly circular orbit, the velocity vector would be constantly changing direction, but not magnitude. Since energy is 1/2mv^2, the velocity vector's direction is irrelevant when computing energy.

In an elliptical orbit, it's a bit more complicated, but the end result is the same. As the moon gets closer to its host planet, its gravitational potential energy decreases, but its velocity magnitude increases, so the total energy is the same.

[Kuroneko]

The moon moves as a closed path (to a very good approximation) in the gravitational field, which is conservative (vanishing curl). Therefore the net work done is zero, although for a particle piece of the trajectory the work may be nonzero--which will then be canceled by the work done in a different part. In practice, there are effects like tidal heating, but let's ignore things like that.

[SpacedTeddyBear]

When the moon orbits the earth, is there work done? Its accelerating after all since its vector is changing, so, is there any energy loss?

Someone posted me this difficult question, mixed up along with a whole bunch of gravity/magnetic energy issues, I can explain the others but not this.

I think the net work done is zero, or very close to it since there is no change in the kinetic energy of the system. Or think of the system as an ellipse with the earth at one of the focal points. If the area swept by the moon is constant, than the net work of the system is 0.

Should've taken celestial mechanics when I had the chance......

[Kuroneko]

What level are you explaining at? Intuitively, it's enough to observe that the periodicity moon's orbit means its kinetic energy is in turn periodic, so the net power over any whole period must be zero. (As Mr. Wong notes, circular orbits have power P = F·v = 0 everywhere, as the velocity vector is always tangent to the force vector. But for elliptic orbits, this is true only for the average power.)

As an explicit work integral, a bit of calculus is required: use and Stokes' theorem to rewrite the work integrand Pdt = F·ds (v = ds/dt) as a curl of the force, (∇×F)·dA, now integrated over the area of any surface with the trajectory as a boundary. But since gravitational acceleration is the (negative) gradient of the gravitational potential, you're taking the curl of a gradient, which is always zero.

[PainRack]

thanks. Now all I have to do is hammer the fact that forces and etc doesn't equate to energy/work done.