A small, polar-orbiting moon

Dave Empey wrote in message . 83...
(Bill Bogen) wrote in
om:

But an object _could_ (very small chance, I admit) be in heliocentric
orbit and yet pass over the Earth at just the right speed to enter a
circular polar orbit at 20310.8 km radius, could it not? Without
having to shed any velocity at all?

I don't think so. If the object is passing over the Earth at the
right speed to "enter" an orbit, it's already *in* orbit.

To put it another way: Imagine filming this object and then running
the film backwards:
you'd see an object in orbit around the Earth suddenly leave orbit
without gaining any velocity, which is impossible. Since Newtonian
mechanics works the same whether time runs forwards or backwards, it
would also be impossible if you run the film forwards. Therefore
entering orbit without shedding velocity is impossible.

How about entering orbit and _gaining_ velocity? Aren't some moons of
Jupiter suspected to be captured asteroids? How can a planet capture
an asteroid if the asteroid makes no velocity changes? I'm proposing
that Cynthia is an asteroid/comet that is orbiting the Sun between the
orbits of Venus and Earth when it happens to pass near Earth, is
accelerated and captured in a circular orbit.

Gary Coffman wrote in message . ..
On 24 Oct 2003 18:31:42 -0700, (Bill Bogen) wrote:
And yet Jupiter has a number of moons in pretty stable orbits, over
millenia.

So does the Sun, but that's because the parent body is *much* larger
than the satellites. So the gravitational effects of the small bodies on
each other are much smaller than the gravitational effect of the parent.

But Earth masses about 81 times Luna and Cynthia will be about 25
times closer to Earth than Luna so the Earth's influence over Cynthia
should be about 50,000 [81*25*25] times that of Luna, no? Are the
orbits of polar-orbiting artificial satellites disrupted by Luna?
Maybe Luna would simply cause Cynthia's orbit to precess?

(Jake McGuire) wrote in message . com...
(Bill Bogen) wrote in message . com...
But an object _could_ (very small chance, I admit) be in heliocentric
orbit and yet pass over the Earth at just the right speed to enter a
circular polar orbit at 20310.8 km radius, could it not? Without
having to shed any velocity at all? (I feel like a cross-examining
attorney;"You admit that my client _could_ have been carrying that
plutonium for perfectly innocent reasons?")

No.

Orbits run backwards more or less as well as they run forwards. So if
something is now in a circular orbit at 20310.8 km radius, if time is
reversed it's not going to go flying back out into interplanetary
space - it'll still be in a circular orbit.

Doesn't entropy enter into it? Cynthia would steal some energy from
Earth and speed up as it falls into circular orbit. Doesn't entropy
dictate that we can't just run the tape backwards? If we run my car
backwards, the flies aren't going to jump off the windshield and take
flight! (Admittedly lame analogy).

Joseph Hertzlinger m wrote in message link.net...
On 17 Oct 2003 04:29:24 -0700, Bill Bogen wrote:

The ancients would probably deduce that Cynthia was brighter
(relative to size) than Luna because it's closer to Earth.

I thought the brighness is proportional to the solid angle.

Luna will give off more light in total because it's larger but Cynthia
is much closer so each solid angle measure (steradian?) should be
brighter. I think Cynthia will lokk like a brighter, fast moving
Venus.

In article ,
Gary Coffman wrote:
OTOH, Earth-Moon almost has to be treated as a double planet. Any
third body in their near vicinity will be influenced more or less strongly
by both of the other bodies. That's a much more complicated situation.

Depends on the time scale. On a short time scale, you can pretty much
ignore the Moon unless your path takes you quite close to it. It's only
1/80th the mass of Earth, after all. Longer term, yes, it definitely
matters, e.g. it affects geostationary comsats' orbits despite it being
ten times farther out than they are.
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |

(Bill Bogen) writes:

(Henry Spencer) wrote in message ...
In article ,
Bill Bogen wrote:
It's the "lose some velocity" part that's hard...

But an object _could_ (very small chance, I admit) be in heliocentric
orbit and yet pass over the Earth at just the right speed to enter a
circular polar orbit at 20310.8 km radius, could it not?

Unfortunately, no, because Earth's gravity will accelerate it as it
approaches. If it arrives from infinity, its speed must be at least
escape velocity (for that distance), which is about 1.4x circular-orbit
velocity.

So it doesn't arrive from infinity but rather from an orbit closer to
the Sun so Earth accelerates it to just the right velocity for a
circular polar orbit.

Orbital mechanics doesn't work that way. If it did, then as others have noted,
it would also be equally possible for polar satellites to spontaneously jump
into the corresponding outbound trajectory, which one never observes to happen
(unless the object fires a kick-motor).

Your only hope is for your "Cynthia" (or part of it) to lose energy somehow,
e.g., by making a pass through the atmosphere, or via tidal friction.
The former option requires an additional interaction with some other body
to raise "Cynthia's" perigee out of Earth's atmosphere to prevent subsequent
passages from de-orbiting it entirely. The latter option is not likely to
work in a single pass unless "Cynthia" is tidally disrupted into fragments,
some of which are moving at less than escape velocity relative to the Earth.
This is _barely_ possible if "Cynthia" is a highly elongated object whose
center of mass is moving moving almost _exactly_ at escape velocity, so that
the "inboard" fragments are moving at _just_ under escape velocity, while
the "outboard" fragments are moving at a bit over escape velocity. This
rather unlikely and rube-goldbergian situation will leave _parts_ of
"Cynthia" in a belt of highly excentric orbits --- but the fact that
"Cynthia" (and therefore its remaining fragments) must necessarily
have had a perigee inside Roche's Limit in order for tidal disruption
to have occurred makes it extremely unlikely that the fragments that remain
in orbit about the Earth will re-accrete into a single body. Most likely,
you will be left with a large number of fragments in highly eccentric orbits
that will gradually pulverize each other to dust.


-- Gordon D. Pusch

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Bill But an object _could_ (very small chance, I admit) be in heliocentric
Bill orbit and yet pass over the Earth at just the right speed to enter a
Bill circular polar orbit at 20310.8 km radius, could it not? Without
Bill having to shed any velocity at all?

What you need is some plausible delta-v for this object that occurs well
above the atmosphere, closer than the moon.

Early in the development of the solar system.

How about two objects in heliocentric orbit collide while very close to
the earth? Perhaps the two objects are a single object that broke up
during a first close encounter with earth, then passed on either side of
earth during a later encounter. Doesn't have to happen in plane, but the
angles do have to be pretty wide. It's not *likely*, of course.

The collision dumps a bunch of kinetic energy so that the resulting mass
does not have escape velocity. It explodes, some bits get kicked out of
the earth-moon system, some bits enter the atmosphere, and some bits stay
in orbit. One particularly big bit is more obvious than the rest.
Cynthia. Over time it might even accrete the rest if it were made of,
say, particularly sticky green cheese.... .

(Bill Bogen) writes:

Dave Empey wrote in message . 83...
(Bill Bogen) wrote in
om:

But an object _could_ (very small chance, I admit) be in heliocentric
orbit and yet pass over the Earth at just the right speed to enter a
circular polar orbit at 20310.8 km radius, could it not? Without
having to shed any velocity at all?

I don't think so. If the object is passing over the Earth at the
right speed to "enter" an orbit, it's already *in* orbit.

To put it another way: Imagine filming this object and then running
the film backwards:
you'd see an object in orbit around the Earth suddenly leave orbit
without gaining any velocity, which is impossible. Since Newtonian
mechanics works the same whether time runs forwards or backwards, it
would also be impossible if you run the film forwards. Therefore
entering orbit without shedding velocity is impossible.

How about entering orbit and _gaining_ velocity?

A.) That would make it even more likely to escape;
B.) What the heck do you think would it gain velocity _from_ ?!?


Aren't some moons of Jupiter suspected to be captured asteroids?

Yes, but they aren't "permanent" moons. They are in highly eccentric orbits
near the edge of Jupoter's "sphere of influence" (the rather non-spherical
region in which Jupiter's gravity dominates the Sun's, which is bounded
by a surface known as the "separatorix" that passes through L1 and L2),
and eventually, they will re-escape.


How can a planet capture an asteroid if the asteroid makes no velocity
changes?

Jupiter is on a slightly eccentric orbit, which means that the size of its
"sphere of influence" varies depending on how far it instantaneously is
from the sun. The "captured" moons have orbital periods on the order
of a terrestrial _year_; they were "captured" because they passed just
inside L1 or L2 on a trajectory almost alone the separatrix; by the time
they got back to the separatrix, it had expaned slightly, "trapping" them
on highly eccentric, precessing orbits. Eventually, they will again happen
to have an apojove close to L1 or L2 at a time when the separatrix is small,
and they will escape.


I'm proposing that Cynthia is an asteroid/comet that is orbiting the Sun
between the orbits of Venus and Earth when it happens to pass near Earth,
is accelerated and captured in a circular orbit.

orbital mechanics doesn't work that way. It will have to interact with
something, such as the Earth's atmosphere of via tidal friction to become
"trapped;" this will leaving it in a highly eccentric orbit, that will
require further interactions to circularize it so that it does not simply
continue to lose energy via atmospheric or tidal friction and come crashing
down to leave a large, hot, ecosphere-destroying hole in the ground.


-- Gordon D. Pusch

perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;'

(Bill Bogen) writes:

(Jake McGuire) wrote in message . com...
(Bill Bogen) wrote in message . com...
But an object _could_ (very small chance, I admit) be in heliocentric
orbit and yet pass over the Earth at just the right speed to enter a
circular polar orbit at 20310.8 km radius, could it not? Without
having to shed any velocity at all? (I feel like a cross-examining
attorney;"You admit that my client _could_ have been carrying that
plutonium for perfectly innocent reasons?")

No.

Orbits run backwards more or less as well as they run forwards. So if
something is now in a circular orbit at 20310.8 km radius, if time is
reversed it's not going to go flying back out into interplanetary
space - it'll still be in a circular orbit.

Doesn't entropy enter into it? Cynthia would steal some energy from
Earth and speed up as it falls into circular orbit.

Neither Entropy nor Conservation of Energy work that way. "Cynthia" doesn't
"steal" energy. The Earth / "Cynthia" system _HAS_ a certain amount of
total energy, which is conserved; because gravity is a pairwise interaction,
that total energy can't be attributed solely to either Earth or "Cynthia,"
but only to the system as a whole.


Doesn't entropy dictate that we can't just run the tape backwards?

Only if friction or other dissipative forces are important. Why do you
expect your "Cynthia" to run into "friction" ???


If we run my car backwards, the flies aren't going to jump off the
windshield and take flight! (Admittedly lame analogy).

Indeed it is VERY lame.

A much better analogy is: Someone throws a superduperball (tm) at your windshield,
and it bounces off. Because a superduperball (tm) is extremely elastic, energy
is almost perfectly conserved during this collision. Because energy is almost
perfectly conserved in this collision, it is virtually impossible to determine
whether you are watching the film "forwards" or "backwards," solely from observing
the collision.

Now: Gravitational collisons are even =MORE= "elastic" than any "superduperball"
could ever possibly be. Absent some other dissipative force, such as atmospheric
or tidal friction, any purely gravitational trajectory is perfectly reversible.


-- Gordon D. Pusch

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(Bill Bogen) writes:

Joseph Hertzlinger m
wrote in message
link.net...
On 17 Oct 2003 04:29:24 -0700, Bill Bogen wrote:

The ancients would probably deduce that Cynthia was brighter
(relative to size) than Luna because it's closer to Earth.

I thought the brighness is proportional to the solid angle.

Luna will give off more light in total because it's larger but Cynthia
is much closer so each solid angle measure (steradian?) should be
brighter. I think Cynthia will lokk like a brighter, fast moving
Venus.

You have that exactly backwards. Both Luna an "Cynthia" are receiving the
same intensity of sunlight, so unless "Cynthia's" surface has a much higher
albedo than Luna's, they will reflect light with roughly equal intensity.
Hence, their surfaces will look about equally bright per unit solid angle.
How much _total_ light the reflect will depend on how large they are,
but not the amount of light per unit solid angle. "Cynthia" will only
appear "brighter" if it is highly reflective, or if it is so close that it
appears to subtend a larger solid angle than the moon. (Note that the latter
would be a Very Bad Thing, as that would imply it would be raising Huge Tides.)


-- Gordon D. Pusch

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