Tidal Torques

rafterman

I know that the moon is receding away from the Earth at about 3.8cm per year as a result of tidal torques exerted by the Earth on the moon. Will this change the length of the year,if so how do I calculate the estimated change as a result of this.

mgb_phys

No it doesn't change the length of the year because the distance of the Earth to the sun isn't changed. It does slightly change the length of the day - although tidal breaking between the Earth and the Sun has a larger effect.

joshd

If it is slowing the earth's rotation gradually, wouldn't that change the length of the day rather than the length of the year?

EDIT: ^^^ exactly. We must have posted at the same time.

rafterman

Thanks for that, That is what I thought, it must be a miss-print on my homework sheet. Confused me tho!

tim_lou

how can one conclude that the distance between the earth and the sun remains unchanged? uneven if it is changed, how can one conclude the period of the earth's orbit is completely unchanged? afterall, this is a three body problem, and the conclusion that the period of the earth depends solely on the semi-major axis doesn't hold. how can one be absolutely sure that there is not some kind of perturbation going on?

mgb_phys

The sun to Earth-moon barycentre is a 2 body problem and the separation of the Earth-moon doesn't change the barycentre.
There are more complicated higher order effects but these are negligible in the context that the question was asked.
This is true of pretty much all of physical measurements there is always some other smaller effect to take into account - the point is to know which ones matter.

D H

The first statement here is correct while the latter is not. The tidal interactions between the Moon and Earth do act to change of the day. The Sun also contributes the changes in the length of a day, but the solar contribution is orders of magnitude smaller than the lunar contribution.

Tidal forces arise from the gradient in the gravitational acceleration. Thus, while gravitational acceleration varies as the inverse square of the distance between two bodies, tidal forces vary roughly as the inverse cube of the distance. Tidal forces are a perturbation of gravity. Most of the tidal forces acting on a body result in periodic behavior. Tides go up and down. The tidal drag, a secular effect, is perturbation on top of a perturbation. The tidal drag varies as the inverse fifth power of the separation between bodies. The lunar-induced tides are more than twice as high as solar tides. The tidal drag, a fifth order effect, is almost entirely due to the Moon.

mgb_phys

Sorry - I assumed that the effect was always greatest on the smaller particle, so that in same the moon's rotation is tidally locked to Earth and so shows the same face, then the Earth will eventually become tidally locked to the Sun.

How much of the Earth-moon effect is due to the free surface water on Earth which gives a useful mechanism for tidal friction. Could you get tidal locking between two spherical objects with solid cores?

tim_lou

that is not true, the force acting on the system is not what the force would be if there is mass at the barycenter. Otherwise, the three body is trivially simply, which it isn't. (I think there was another similar discussion previously).

Regardless, there WILL be an effect on the period of the earth's orbit when the distance of the moon-earth distance increases, and I think, the point of the original question is how much of an effect is it? and if you claim that it is very negligible, please provide some details on how to show that it is negligible. I'm very curious about how one would carry out such calculation.

rcgldr

It's my understanding that it's the tidal forces on the Earth that correlate to the moon's receding. The tidal action of the Earth results in the Earths center of mass (gravity) to be drawn towards the moon. The result is the Earth rate of rotation slows down, the moon recedes, and engery is conserved.

The fact that the moon's rotation corresponds to it's rate of orbit (with some oscillation) might have required tidal action in the past to slow it down to it's current rate, but any oblong (in otherwords, not perfectly spherical) object will have a tendency to orbit so it's length is perpendicular to the direction of the orbit. Nasa twice attempted to deploy a very long cable that would hold it's orientation relative to it's orbit, but the spool jammed in both cases.