Re: [Phys-L] Moon's orbit

[John Denker <jsd@av8n.com>]

On 10/17/2016 06:58 PM, Anthony Lapinski wrote:
> The Sun pulls on the Moon with about twice the force that the Earth pulls
> on it. So why doesn't the Moon get pulled away from the Earth? I realize
> this is complicated. Is there a "simple" explanation I can tell high school
> students?

The simple answer goes like this:
 -- Yes, the moon gets pulled toward the sun.
 -- No, it does not get pulled /away/ because the earth gets pulled, too.

The compound system (earth + moon) is pulled toward the sun just
enough to keep it in its annual orbit around the sun.

If you want the simple answer, that's all there is to it.

===========================
At the next level of detail:

Executive summary:  Gravitational acceleration is one thing;  tidal
stress is something else.

In accordance with Einstein's principle of equivalence, any uniform
gravitational field is equivalent to an accelerated reference frame.
As another way of saying almost the same thing:  Locally, to first
order, any gravitational field is equivalent to an accelerated
reference frame.

So to an excellent approximation, the acceleration caused by the
sun vanishes if you work in a frame that is accelerated along with
the earth (or, better, accelerated along with the center-of-mass
of the earth/moon system).

The equivalence principle tells us that only thing that could possibly
matter is the nonuniformity, i.e. the /difference/ between the solar
gravitational acceleration at the earth versus at the moon.  That
difference is verrrrry much smaller than the aforementioned factor
of two.  To leading order it can be considered a tidal stress, of
a kind that slightly perturbs the shape of the moon's orbit around
the earth.  Even so, when averaged over a month, the leading-order
term vanishes, i.e. it does not cause a cumulative increase in the
earth-moon distance.

> Several sites mentioned the Hill sphere (never heard of this before).

It's not a particularly deep or important concept.  It just
has to do with which way something falls if you drop it.

>  Is this related to the Roche limit?

The Hill sphere (aka Roche sphere) is *not* at all closely related
to the Roche limit.  Not in concept and not in typical size.
They both have to do with gravity, but that's about as close as
the relationship gets.

The Roche limit *is* important.  It is directly related to the
original simple question.  Note the contrast:
  -- The gravitational acceleration due to the sun falls off
   like 1/r^2, but does not matter.
  -- The tidal stress due to the sun falls of like 1/r^3, and
   does matter.

If you got very close to the sun (much closer than the annual
orbit of Mercury) then the tidal stress from the sun would be
enough to shred the monthly orbit of the earth/moon system.

In contrast, as things really stand, the earth/moon system is
trapped in a gravitational well, i.e. a gravitational potential,
and the tidal stress from the sun is too small (by many orders
of magnitude) to pop them out of the well.