In Science item 2, Rane Curl opined about tidal influences on
planets:
>Seems to me it doesn't matter whether the sun is rotating faster or slower
>than the planets. Tidal forces would change it rotation rate. I agree,
>though, with the angular momentum argument, but I want to know the *mechanism*
>by which the planets would be accelerated or slowed. Identify the "pushes
>and pulls". 

Hand-waving explanation follows.

First, you have to distinguish between planetary ROTATION and planetary
ORBIT.  Tides have distinct influences on both of them.  The mechanism
is the same for both, however.

If you consider planets and stars as spherically symmetrical bodies,
then you can calculate that their rotation and orbit will not be affected
by other bodies.  However, spherical symmetry is only an approximation.
Neither planets nor stars are perfectly rigid nor of zero size.  Rotating
bodies become slightly flattened spheres (Jupiter is an extreme example,
being about 12% wider across the equator than the poles).

When another body is nearby a star or planet, there is a finite difference
in the gravitational pull on the near and far sides.  This tends to pull
the star or planet into a dumbbell shape.  There's a bulge on the near
side from the greater pull than the center experiences, and another bulge
on the far side from the lessened pull due to the greater distance
("pulling" in the opposite direction).  The flow of the body is a tide.
Water flows more easily than rock, but there are tides in Earth's crust.

This does funny things to the mass distribution.  In physics-speak, the
body has a non-zero dipole moment.  We'll leave the physics-speak here.

If the two bodies aren't rotating relative to the line between them,
all of the bulges line up along the line, and nothing happens.  However,
suppose that one body *is* rotating relative to the line between them.
Then things get interesting.

Gravity tries to raise a bulge on the near and far sides of each body.
However, when the "near" and "far" sides are moving, the bulge rises,
but it rises *late*, so it doesn't line up any more.  Instead, the
bulge is carried in the direction of the relative rotation.

The near bulge, being closer to the other body than the center, pulls
on it harder.  The far bulge, being farther away than the center, pulls
on it less hard.  Thus the gravity "seen" by the other body in the
orbiting pair is skewed toward the near bulge.  The bulge will be
shifted in the direction of the relative rotation.  The body experiencing
the tide experiences a pull on the near bulge, an anti-pull on the far
bulge, the two amounting to a torque which brakes the relative rotation.
The body creating the tide falls toward the near bulge rather than
the center, which tends to pull its orbit in the direction of the other
body's rotation.  It will be pulled forward into a higher orbit if the
other body is rotating in the same direction and faster, and into a
lower orbit if the other body is rotating slower or opposite.

You can see this in the Earth-Moon system.  The Earth is now rotating
once every 24 hours (mean solar time, give or take a fraction), but
in the past it was considerably faster.  The Moon may once have rotated,
but now it is tide-locked to the Earth; it rotates once every orbit,
keeping the same face to Earth (give or take a little "libration").
The tides raised by the Moon slow the earth down and pull the Moon
into a higher, slower orbit.  Someday, in the far future, Earth will
be rotating slowly enough that there really are only 360 days per
year.  By this time, the Moon will be far enough away that total
eclipses of the Sun will be a thing of the past.

The Sun rotates every 28 days or so as well, and the Earth raises
some small tides on the Sun (and vice versa, though the solar tides
raised on Earth are much smaller than the moon's).  This tidal bulge
rotates ahead and will tend to raise Earth's orbit.  However, since
the tidal influence falls off as the *cube* of distance, the tidal
effect on Earth's orbit is very small.

Did I lose anyone with my hand-waving?

9 responses total.

I'd have to study it carefully to see if the hand waving wafts things in
the right directions. But if the effect is to pull the planet forward, in
the direction of its orbit, it will increase in speed and thereby *lower
its orbit*. This assumes that everything works out to keep the orbit
essentially circular.

No, pulling it forward makes it rise, and slow down.  If you assume
the pull is delivered at one instant, the planet will be moving
faster *at that point*.  If it was in a circular orbit before, however,
it is changed to an ellipse, and the opposite side of the orbit is
now farther from the primary.  If you consider the pull as an
infinite number of infinitesimal kicks, the increase in distance is
uniformly distributed.  It just keeps moving out.

So even though it is given forward momentum and "speeds up", when
circularized the kinetic energy is *less*, but the potential energy is
greater than the sum of the original potential energy plus the added
kinetic energy. 

I had to read that twice... I think the answer is "yes", but I
won't swear to it.

FYI, if the reference value for potential energy is object-at-infinity,
the potential energy of an object in orbit is always negative.  The
total energy is negative for an object in orbit, zero for an object
exactly at escape velocity, and positive for an object moving faster
than escape velocity.

Russ, nice description of how tides create a linkage between the rotation of 
a body such as the Earth, and the revolution of the moon. Here is a different
way to say (mostly) the same thing. 

In fact this linkage is quite complex in the Earth-moon system. The bulk 
of the tidal bulge is manifested as changing sea levels. It is also true 
that the Earth's crust is strained by the tides, but this is a much 
smaller amount.  The Earth's surface is mostly oceanic.

If the oceans could flow around the earth frictionlessly, the bulge would
neither lead nor lag the line of centers of the two objects. The oceans
themselves are almost frictionless, but there is plenty of friction at the
shores of these oceans, and in shallow seas.

Since the Earth is rotation faster than the moons revolution (by a factor of
about 28 or 29 times), the Earth's faster rotation pushes (via friction) the
tidal bulge on the Earth out in front of the line of centers. This force has
both an action and a reaction.

The action is that since the bulge is pushed into a leading position, exerts
a force on the moon which is not central. Instead the force pushes the
moon *forward* in its orbit. It is forward because the bulge is leading rather
than lagging. This forward force in orbital mechanics always causes the object
to go into a higher orbit. A retarding force would bring it into a lower
orbit. I learned this tidbit playing spacewar on a pdp-1 in the old days, but
I digress.

The reaction is due to the fact that it is the Earth's rotational velocity
which is causing the bulge to lead. This happens at the expense of its angular
momentum. The friction force in the shallow seas on earth slows down the
Earth's rotation. 

summary of effects:
* The Earth's day continues to increase slightly.
* The number of days in a year continues to decrease slightly.
* The Earth-moon interaction has no effect on year length.
* The length of the lunar month continually increases.
* The moons orbit keeps expanding, so that total eclipses of the sun 
  become rarer and rarer, being replaced more and more
  by annular ones. Eventually there will be no more.
* Since time is now measured by atomic clocks, this secular change in the 
  length of a day can be measured directly. It was ignored for a while, but 
  now they are adding leap seconds every Jan 1, 0:00:00 (1 per year) until 
  they catch up, and then as necessary according to astronomical observations.

Glossary items relevant here:
"Line of centers" - a straight line connecting the centers of two objects.
"Central force" - a force along the line of centers (attract or repel)
"Secular" - non-periodic (and thus accumulative)
"strain"  - deformation ("stress" is force, "strain" is reaction to it)

It has been calculated recently that the earth's rotation rate has speeded
up, and the average day length decreased by 8 microseconds, over the past
half-century, because of the building of dams.  Since these are mostly
built in middle latitudes, they store water at an average reduced radius
from the axis, and this results in a more rapid rotation rate (the
"skater's spin"  effect). These articles have not discussed the slowing of
earth's rotation due to drag but I have the impression that that is much
less than the increase in rotation rate due to the redistribution of water
mass. What is the magnitude of the rotation deceleration due to earth-moon
interaction? 

Incidentally, the added leap seconds mentioned by Steve are not due
significantly to the purported slowing of the earth's rotation, but rather
to the mis-measurement of the average earth's rotation rate back when the
definition of the second was made.

This response has been erased.

"Raising the orbit" means an increase in the distance between the two
orbiting bodies. It makes more sense when one body is huge (earth) and the
other small (Mir - or even the moon). WIth your marble analogy - you have
to have a cnenral body. Put a potted delphinium there. Then the marble is
orbiting the delphinium. Increase the distance of the marble from the
delphinium to raise the orbit. 

To go "up" to a "higher" orbit you first have to increase your velocity.
Then you stop pushing. With the increased velocity you tend to move
further out, but as you move out (and your distance "up"  increases), your
velocity decreases - just as a ball thrown into the air decreases in
velocity as it rises. (You then have to do some tricky pushing to
circularize your orbit at the higher elevation.) 

The general relation for the velocity of a small object in earth orbit is
v=SQRT[GM/R], where G is the univeral gravitational constant = 6.67E-11
m^3/(s^2Kg), M is the mass of the earth (Kg), and R is the radius of the
orbit (m). Assignment: estimate the velocities of a low earth orbit
satellite (say, at 300 miles above the earth), and of the moon. M=6E24 Kg.

Rane cited all the right stuff, but I think the point got obscured...

Gravity pulls on everything with an inverse-square-law "force".
(Einstein would say that space is curved; some people say space
is flat and Einstein was bent.)  The acceleration needed to bend
the path of an object moving at speed V into a circle of radius R
is given by A = V^2/R.  When this balances the G*M/R^2 pull of
gravity, you get a perfect circular orbit.

"Up", in the orbital direction, is further away from the primary,
meaning the orbit gets bigger.  What happens if you push the
satellite "forward", meaning in its direction of orbital motion?

First thing, it speeds up a little bit.  But when that happens,
the A = G*M/R^2 of gravity is no longer sufficient to pull the
path into a circle at the greater V; the value of R increases,
and the satellite "falls away" from the primary, moving into a
higher orbit.  And if you push gently for a long time, so that
the orbit doesn't get lopsidedly elliptical, you wind up with
a new circle at a larger value of R and a smaller value of V.

If you solve algebraically for V in terms of R, and work out
the time it takes to compete a circular orbit, you'll find
that the period increases as the 3/2 power of R.  This is
the same law that was observed hundreds of years ago, but
now we have the physical explanation for it.

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