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A small, polar-orbiting moon



 
 
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  #21  
Old October 28th 03, 05:05 AM
David M. Palmer
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Default A small, polar-orbiting moon

In article , Jake
McGuire wrote:


With close flybys in 3-body systems you can either eject or capture
something, but that's not the situation you're describing.


If Cynthia originally had a sister, a double asteroid like Hermes was
just found to be, then you have a three-body system that might allow
capture. It would take the same hand-of-god that placed Luna at just
the right size and distance to give us nice solar eclipses, but it's
possible.

Especially with an aerobrake to bleed off some energy, then the second
component circularizes Cynthia's orbit before being flung into the
utterdark. (A single asteroid aerobraking gives an orbit that passes
inside the atmosphere on subsequent passes, which quickly leads to
lithocapture and a nice iridium layer for the next intelligent species
to find.)

Of course, the protagonist realizes this after single-handedly
recapitulating the works of Galileo, Newton, Halley etc. from our time
line to develop orbital mechanics, Percival Lowell to find the cast off
sister, Shoemaker to discover that it will hit Earth in 3 years,
Goddard, Tsiolkovskii and Korolev to build a rocket, Oppenheimer and
the gang for the payload, and Bruce Willis to get the box office.


Bogen:
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.


No, brightness per steradian depends on illumination (how far from the
Sun it is--the same as the Moon is, plus a bit of Earthshine) and how
reflective it is, but not on how far away it is. (The inverse square
law, an approximation in this case, comes entirely from the solid angle
shrinking with distance.) That's why a tree nearby doesn't burn your
eyes out while a tree-covered distant mountain is other than black.

--
David M. Palmer (formerly @clark.net, @ematic.com)
  #22  
Old October 30th 03, 02:01 PM
Bill Bogen
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Default A small, polar-orbiting moon

(Gordon D. Pusch) wrote in message ...
(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.


So the sum of Earth+Cynthia energy is constant (actually
Earth+Cynthia+Luna) but that doesn't mean Cynthia's energy can't
increase at the expense of Earth's. Ex; when a probe swings around
Jupiter to take advantage of the 'slingshot effect', is it not
speeding up while Jupiter is slowing down infinitesimally?

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.


OK, I think it's finally gotten through to me: Cynthia needs a
mechanism to change her velocity as she approaches Earth orbit. I'd
rather not use aerobraking because it doesn't eventually raise her
perigee to a circular orbit. So, Cynthia is a regolith-covered comet
that somehow found itself in a Aten-esque orbit (within the Earth's
orbit around the Sun), an orbit that just reaches the Earth. Cynthia
happens to vent at just the right times and in just the right
directions to circularize it's orbit around Earth.
  #23  
Old October 30th 03, 02:11 PM
Bill Bogen
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Default A small, polar-orbiting moon

(Gordon D. Pusch) wrote in message ...
(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.)


Yes, I was partially wrong: Luna and Cynthia are equally bright per
unit of solid angle. Since Luna would appear about 12.6 times wider
than Cynthia, or 158.5 times the area, then Cynthia would only have a
magnitude (at best) of -7 as compared to Luna's -12.5. But Venus has
a magnitude of only -4.9 (at best) so Cynthia, as I said, would like
like a brighter, fast moving Venus. It should often be visible in
daylight.
 




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