War of the Worlds

2006-01-11 Thread Rob Seaman

I see Steve Allen has already supplied a thorough answer.  Interested
individuals might also scrounge through the list archives (http://
rom.usno.navy.mil/archives/leapsecs.html) since the topic has come up
before.  In fact, Demetrios Matsakis speculated on solar system wide
timescales even before this list was started.  My own skepticism over
more extreme flights of fancy is expressed under Future Directions
in http://iraf.noao.edu/~seaman/leap.

On Jan 11, 2006, at 7:01 AM, Daniel R. Tobias wrote:


If, however, this Martian second is actually defined as a
particular multiple of the SI second, then the use of leap seconds
on Mars would ultimately be necessary to account for any future
changes in the length of the Martian day.


That is indeed the issue.  Is there anybody from the geophysics side
who can comment on long term trends in Martian length of day?

I don't have an envelope large enough, but there are various issues
to consider.  The Hurtling Moons of Barsoom are much smaller than our
own and should have a negligible tidal breaking effect.  (See http://
www.freemars.org/mars/marssys.html, for instance, for their
interesting history.)  And do the Earth's oceans mediate our Moon's
breaking or is that a crustal phenomenon?  (The Earth-Moon system
should better be regarded as a double planet, than planet and
satellite.)  On the other hand, Mars passes much closer to Jupiter,
the 800 pound gorilla of the solar system, but then it is further
from King Kong - the Sun, that is - and tides are an inverse cube
effect.  But Mars is much smaller and has a smaller moment of inertia
in the first place - but then Mars is much smaller and the lever
arm to grapple with it is less pronounced.

Taken all together, one suspects that LOD(Mars) is many orders of
magnitude more constant than LOD(Earth).  One would not be
flabbergasted to be utterly wrong, however.

Hopeful news for John Carter fans:

   http://filmforce.ign.com/articles/679/679046p1.html

Rob


Re: War of the Worlds

2006-01-11 Thread John Cowan
Rob Seaman scripsit:

 I don't have an envelope large enough, but there are various issues
 to consider.  The Hurtling Moons of Barsoom are much smaller than our
 own and should have a negligible tidal breaking effect.  (See http://
 www.freemars.org/mars/marssys.html, for instance, for their
 interesting history.)  And do the Earth's oceans mediate our Moon's
 breaking or is that a crustal phenomenon?  (The Earth-Moon system
 should better be regarded as a double planet, than planet and
 satellite.)  On the other hand, Mars passes much closer to Jupiter,
 the 800 pound gorilla of the solar system, but then it is further
 from King Kong - the Sun, that is - and tides are an inverse cube
 effect.  But Mars is much smaller and has a smaller moment of inertia
 in the first place - but then Mars is much smaller and the lever
 arm to grapple with it is less pronounced.

 Taken all together, one suspects that LOD(Mars) is many orders of
 magnitude more constant than LOD(Earth).  One would not be
 flabbergasted to be utterly wrong, however.

I wouldn't be too quick to dismiss tidal braking from Phobos.  It's
awfully close to Mars, and tidal braking is as you say an inverse-cube
effect.  The formula (kai Wikipedia) is (2GMmr)/R^3, where M and m are
the masses, r is the radius of the primary, and R is the orbital radius
of the secondary.  The mass of the Earth-Moon system is eight orders of
magnitude larger than the Mars-Phobos system, and the radius of Earth
is only twice the radius of Mars, but the ratio of the cubed orbital
radii is five orders larger for Phobos than for the Moon.  So the tidal
acceleration of the Moon toward the Earth is only some three orders larger
than Phobos's toward Mars.  That puts the effect in the same ballpark.
How much difference in actual slowing can be attributed to Earth's ocean
and Mars's lack of one I don't know.

(See http://www.madsci.org/posts/archives/oct98/908453811.As.r.html for
the relevant masses and radii.)

--
Eric Raymond is the Margaret Mead   John Cowan
of the Open Source movement.[EMAIL PROTECTED]
--Bruce Perens, http://www.ccil.org/~cowan
  some years agohttp://www.reutershealth.com


Re: War of the Worlds

2006-01-11 Thread Neal McBurnett
I referenced this page, but missed the most interesting part of it:

http://www.exo.net/~pauld/physics/tides/tidalevolution.htm

 The height of a tidal bulge on a planet is proportional to the
 inverse cube of the distance between the planet and the object
 causing the tidal bulge. The torque which slows down the planet is
 proportional to the inverse sixth power of the distance.

presumably because the the same third power works both on the size of
the bulge and the differential pull on the bulge.  That suggests that
Phobos might raise a lower tide than the sun, but yet have a greater
tidal braking effect.  But I expect the speed of response of the
planet to the tidal force could still play a role in comparing the
Phobos and solar effects.

Data on the speed of change of the orbit of Phobos combined with the
conservation of angular momentum should give a good handle on the size
of the effect from Phobox.

http://en.wikipedia.org/wiki/Phobos_%28moon%29
  tidal forces are lowering its orbit, currently at the rate of about
  1.8 metres per century, and in about 50 million years it will
  either impact the surface of Mars or (more likely) break up into a
  planetary ring.

But its too late to do that math tonight

-Neal

On Wed, Jan 11, 2006 at 11:41:27PM -0700, Neal McBurnett wrote:
 On Wed, Jan 11, 2006 at 11:44:13PM -0500, John Cowan wrote:
  I wouldn't be too quick to dismiss tidal braking from Phobos.  It's
  awfully close to Mars, and tidal braking is as you say an inverse-cube
  effect.  The formula (kai Wikipedia) is (2GMmr)/R^3, where M and m are
  the masses, r is the radius of the primary, and R is the orbital radius
  of the secondary.  The mass of the Earth-Moon system is eight orders of
  magnitude larger than the Mars-Phobos system, and the radius of Earth

 I assume you mean the mass of phobos vs the mass of the moon, not the
 systems, since that is what fits in the raw numbers and equations you
 provide.  But that is less than 7 orders of magnitude different, as I
 read your reference.

  is only twice the radius of Mars, but the ratio of the cubed orbital
  radii is five orders larger for Phobos than for the Moon.  So the tidal
  acceleration of the Moon toward the Earth is only some three orders larger
  than Phobos's toward Mars.  That puts the effect in the same ballpark.

 But the tides from the sun are very significant on earth, and much
 more pronounced on mars.

  (See http://www.madsci.org/posts/archives/oct98/908453811.As.r.html for
  the relevant masses and radii.)

 A quick google search for   mars  tides  yields much more useful
 and interesting answers.

  http://www.findarticles.com/p/articles/mi_m1134/is_5_112/ai_102275148

  the solid-body tides on Mars--caused by the Sun, not by a Martian
  satellite--are large enough to indicate that at least part of that
  planet's core is liquid.
  ...
  Early in the study, the investigators realized only a liquid core
  could give rise to a tidal bulge capable of having the observed
  gravitational effect on the spacecraft. And how much bulge is that?
  About a third of an inch.
  Fluid core size of Mars from detection of the solar tide, Science
  300:299-303, April 11,2003)

 But of course we need to treat the web with some skepticism.  I doubt
 this tidbit got it right about what causes the tides (Phobos vs the
 sun):

 http://ganymede.ipgp.jussieu.fr/GB/projets/netlander/
  Another way to proceed will be to measure tides produced by Phobos,
  one of Mars' moons. Those tides are 10 times lower than the tides
  produced by the Earth' Moon.

 As for changes in the length of the day, we have to look at the
 mechanism by which tides relate to the slowing of the day:

 http://www.exo.net/~pauld/physics/tides/tidalevolution.htm

  There are also tides in the solid earth. The tidal bulge is about 1
  meter high. The moon pulls up this tidal bulge on the earth, there
  is a time delay between the pull of the moon and the time when the
  tidal bulge reaches its maximum height. During this time the
  rotation of the earth carries this tidal bulge around the planet in
  the direction of rotation.

  The moon then pulls on the mass of the tidal bulge and slows the
  rotation of the earth.

 So the degree of slowing is affected by both the size of the bulge,
 how delayed the bulge is, and the angular velocity of the body giving
 rise to the tides, making it harder to compare the effects of the sun
 with rapidly-moving phobos.

 That is what would relate to this aspect of your question:

  How much difference in actual slowing can be attributed to Earth's ocean
  and Mars's lack of one I don't know.

 I also note that the axial orientation of Mars changes widely back and
 forth, which would clearly affect the long term effects due to the sun:

   While Earth's tilt varies from 23 to 25 degrees, the Red Planet's
   actually shifts from 15 to 40 degrees over a 100 million year
   period

 I don't see a handy reference to pull all that together 

Re: War of the Worlds

2006-01-11 Thread John Cowan
Neal McBurnett scripsit:

  I wouldn't be too quick to dismiss tidal braking from Phobos.  It's
  awfully close to Mars, and tidal braking is as you say an inverse-cube
  effect.  The formula (kai Wikipedia) is (2GMmr)/R^3, where M and m are
  the masses, r is the radius of the primary, and R is the orbital radius
  of the secondary.  The mass of the Earth-Moon system is eight orders of
  magnitude larger than the Mars-Phobos system, and the radius of Earth

 I assume you mean the mass of phobos vs the mass of the moon, not the
 systems, since that is what fits in the raw numbers and equations you
 provide.  But that is less than 7 orders of magnitude different, as I
 read your reference.

Actually I didn't mean either one: I meant the mass of the primary times
(not plus) the mass of the secondary, the Mm in the formula.  So
the mass of Mars times the mass of the Phobos is ~ 10^39 kg, whereas
the mass of Earth times the mass of the Moon is ~ 10^47 kg:  eight
orders of magnitude, as I said.  Sorry for the misstatement.

--
Even a refrigerator can conform to the XML  John Cowan
Infoset, as long as it has a door sticker   [EMAIL PROTECTED]
saying No information items inside.   http://www.reutershealth.com
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