Re: War of the Worlds
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 --Eve Maler http://www.ccil.org/~cowan
Re: War of the Worlds
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 right now Cheers, Neal McBurnett http://bcn.boulder.co.us/~neal/
Re: War of the Worlds
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 > > actuall
Re: War of the Worlds
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
War of the Worlds
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
