Tidal locking
Charles R Martin
elsewhere.
Let us assume for purposes of discussion that through the application of
appropriate magic (by Clarke's 3rd Law) the earth is rather suddenly strippped
of the Moon and transported to an orbit around a handy super-Jovian planet in
the livable volume around another star.
Clearly the superJovian will be big enough that we'd expect to see the Earth's
rotation eventually phase-lock. (If this isn't clear, then assume that too.)
How long would this take?
What would be the transient effect as it happened?
What would happen climatically ... climactically ... what would happen to the
weather? ;-)
Brett Evill
> Clearly the superJovian will be big enough that we'd expect to see the Earth's
> rotation eventually phase-lock. (If this isn't clear, then assume that too.)
>
> How long would this take?
That would depend on how close Earth's orbit was to the superjovian.
> What would be the transient effect as it happened?
Huge tides. Gradual change of Earth from an oblate to a prolate shape:
collapse of the equatorial bulge, slow rise of bulges facing and facing
away from the superjovian. Flooding and slow reappearance of continents.
Violent and extensive vulcanism.
> What would happen climatically ... climactically ... what would happen to the
> weather? ;-)
When the volcanoes let out lots of CO2 and sulphate the temperature
would rise and the rain would turn acid. There would be mass
extinctions. The length of the day would stretch out to hundreds of
hours, the diurnal temperature range would also rise, catabatic winds
and local thunderstorms would become severe. Eventually the Coriolis
effect would practically cease and there would be no more cyclones ot
anticyclones, typhoons, hurricanes, etc. Winds would get weaker. The
longitudinal winds (Roaring Forties, Trade Winds, Jet Streams) would
cease.
Regards,
Brett Evill
Isaac Kuo
>>Clearly the superJovian will be big enough that we'd expect to see
the Earth's
>>rotation eventually phase-lock. (If this isn't clear, then assume
that too.)
>>What would be the transient effect as it happened?
>Huge tides. Gradual change of Earth from an oblate to a prolate shape:
>collapse of the equatorial bulge, slow rise of bulges facing and facing
>away from the superjovian. Flooding and slow reappearance of
continents.
>Violent and extensive vulcanism.
Surely there would be lots of earthquakes as well, right?
And with lots of underwater earthquakes, lots of tsunamis.
--
_____ Isaac Kuo mec...@yahoo.com ICQ 29055726
__|_)o(_|__
/___________\
\=\)-----(/=/
Sent via Deja.com
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Charles R Martin
>
> Charles R Martin wrote:
>
> > Clearly the superJovian will be big enough that we'd expect to see the Earth's
> > rotation eventually phase-lock. (If this isn't clear, then assume that too.)
> >
> > How long would this take?
>
> That would depend on how close Earth's orbit was to the superjovian.
And on the mass of the superjovian, I assume. But those are parameters. I'm
more curious about the kinds and magnitudes of the various forces. My first
intuition suggests that things like tidal drag will be quite small in
comparison to the angular momentum of the Earth's rotation.
Charles R Martin
>
> In article <3A3F69...@tyndale.apana.org.au>,
> b.e...@tyndale.apana.org.au wrote:
> >Charles R Martin wrote:
>
> >>Clearly the superJovian will be big enough that we'd expect to see
> the Earth's
> >>rotation eventually phase-lock. (If this isn't clear, then assume
> that too.)
>
> >>What would be the transient effect as it happened?
>
> >Huge tides. Gradual change of Earth from an oblate to a prolate shape:
> >collapse of the equatorial bulge, slow rise of bulges facing and facing
> >away from the superjovian. Flooding and slow reappearance of
> continents.
> >Violent and extensive vulcanism.
>
> Surely there would be lots of earthquakes as well, right?
> And with lots of underwater earthquakes, lots of tsunamis.
This all seems very sudden and violent, which would only happen close to
Roche's Limit (pretty well by definition.) The disappearing Moon part does
indeed (in my planned scenario) happen suddenly -- the Earth wouldn't be
deposited near the Roche Limit at the other end, by design.
Brian Davis
> How long would this take?
You know, I'm *beginning* to realize that I've spent altogether too
much of my time learning about this subject. The rough timescale for
tidal locking is:
t = 16 rho omega a^6 (Q/k2) / ( 45 G M^2 )
rho, Q, k2 = variable describing the body that's slowing (Earth)
a = orbital distance
M = mass of body that's doing the slowing (Jupiter, in this case)
omega = 2 pi / P
That a^6 term is a real killer (literally). For example, if you plopped
the Earth into Io's orbit, the characteristic timescale is measured in
*decades*, not millennia (and that's assuming you have to tidally brake
Earth to a complete stop; it won't be, in the example given it would
only decelerate until it's day was around 80% longer).
Note that I've done this several different ways now, and keep coming
out with these absurdly short timescales. Either this is far more
dramatic than I thought, or I've (once again) made some serious errors somewhere.
> What would be the transient effect as it happened?
Well, to slow the Earth from a 24 hour day to a 42 hour day (Io's
tidally locked period) will produce 1.4e+29 J. Ouch. That's enough
energy to raise the bulk temperature of the Earth (all 6e+24 kg of it)
by 10-20 degrees C, but by far most of it will be produced in the
surface layers... And then you have to dissipate all this, over much
longer timescales...
Conservatively, I'd say you'll have a near global magma ocean for
hundreds to thousands of years. When & if it's ever inhabitable,
geothermal heat is going to be a popular power source...
If you're wondering about tidal waves in the very early stages of the
problem (i.e.- while there's still somebody there to see them), they'd
be big. Once again for Earth in place of Io, *simply* scaling implies
that the tidal bulges would be around 20,000 times as high as currently.
Titanic earthquakes as nearly every crustal stress is released in a
period of a few days, followed by a bulge of water circling the planet
every 56 hours.
> What would happen to the weather? ;-)
I doubt in the short term you need to worry about it <grin>. Are you
asking what would transiently happen to the weather, or how would the
weather be different after an equilibrium of some type is reached?
Also note that you can come out with a "no time at all for tidal
locking" solution to your problem - just dump the Earth into a close
orbit with a period that matches it's rotational period already. But
that's cheating <grin>. And you'd still have major problems with earthquakes.
--
Brian Davis
Geoffrey A. Landis
Brian Davis wrote:
> Charles R Martin wrote:
>
> > How long would this take?
>
> You know, I'm *beginning* to realize that I've spent altogether too
> much of my time learning about this subject. The rough timescale for
> tidal locking is:
>
> t = 16 rho omega a^6 (Q/k2) / ( 45 G M^2 )
> rho, Q, k2 = variable describing the body that's slowing (Earth)
> a = orbital distance
> M = mass of body that's doing the slowing (Jupiter, in this case)
> omega = 2 pi / P
That's an odd way to phrase it. I would put the equation in terms of *either* P
*or* a, but not both, since omega = SQRT(GM/a^2). The fact that you have both in
your equation suggests to me an error.
> That a^6 term is a real killer (literally). For example, if you plopped
> the Earth into Io's orbit, the characteristic timescale is measured in
> *decades*, not millennia (and that's assuming you have to tidally brake
> Earth to a complete stop; it won't be, in the example given it would
> only decelerate until it's day was around 80% longer).
a time scale is a time scale.
> > What would be the transient effect as it happened?
>
> Well, to slow the Earth from a 24 hour day to a 42 hour day (Io's
> tidally locked period) will produce 1.4e+29 J. Ouch. That's enough
> energy to raise the bulk temperature of the Earth (all 6e+24 kg of it)
> by 10-20 degrees C, but by far most of it will be produced in the
> surface layers...
No, the power has to radiate out through the surface (obviously), but it is actually
produced in the viscoelastic layers-- most likely the mantle.
> And then you have to dissipate all this, over much
> longer timescales...
> Conservatively, I'd say you'll have a near global magma ocean for
> hundreds to thousands of years. When & if it's ever inhabitable,
> geothermal heat is going to be a popular power source...
> If you're wondering about tidal waves in the very early stages of the
> problem (i.e.- while there's still somebody there to see them), they'd
> be big. Once again for Earth in place of Io, *simply* scaling implies
> that the tidal bulges would be around 20,000 times as high as currently.
Note that a true tidal wave (a wave caused by tides) is not the same as a tsunami
(often popularly but erroneously called a "tidal wave"). In this case, I expect you
would have both.
> Titanic earthquakes as nearly every crustal stress is released in a
> period of a few days, followed by a bulge of water circling the planet
> every 56 hours.
Why 56?
> ...
--
Geoffrey A. Landis
author of MARS CROSSING, now available at bookstores everywhere
http://www.sff.net/people/geoffrey.landis
Brett Evill
>
> > >>What would be the transient effect as it happened?
> >
> > >Huge tides. Gradual change of Earth from an oblate to a prolate shape:
> > >collapse of the equatorial bulge, slow rise of bulges facing and facing
> > >away from the superjovian.
> This all seems very sudden and violent, which would only happen close to
> Roche's Limit (pretty well by definition.)
It might not be sudden, but it would happen. The effect would be
greatest near Roche's limit, but even arbitrarily far from Roche's limit
the equatorial bulge will collapse when Earth's rotation slows towards a
stop. And old current crust just won't fit on Earth's new shape. Hence
earthquakes and vulcanism.
Regards,
Brett Evill
Charles R Martin
>
> Charles R Martin wrote:
>
> > How long would this take?
>
> You know, I'm *beginning* to realize that I've spent altogether too
> much of my time learning about this subject.
Hey, it's a dirty job but somebody has to do it.
> The rough timescale for
> tidal locking is:
>
> t = 16 rho omega a^6 (Q/k2) / ( 45 G M^2 )
> rho, Q, k2 = variable describing the body that's slowing (Earth)
> a = orbital distance
> M = mass of body that's doing the slowing (Jupiter, in this case)
> omega = 2 pi / P
>
> That a^6 term is a real killer (literally). For example, if you plopped
> the Earth into Io's orbit, the characteristic timescale is measured in
> *decades*, not millennia (and that's assuming you have to tidally brake
> Earth to a complete stop; it won't be, in the example given it would
> only decelerate until it's day was around 80% longer).
> Note that I've done this several different ways now, and keep coming
> out with these absurdly short timescales. Either this is far more
> dramatic than I thought, or I've (once again) made some serious errors somewhere.
But hold on -- I'm not sure about what a few of the parameters are, but if we
hold rho, Q, k2, M and P constant, so that we vary only the orbital distance,
doesn't this mean that the time scale *increases* as the sixth power of the
radius? So pick an orbit relatively slightly further out and the time scale
increases *greatly*.
>
> > What would be the transient effect as it happened?
>
> Well, to slow the Earth from a 24 hour day to a 42 hour day (Io's
> tidally locked period) will produce 1.4e+29 J. Ouch. That's enough
> energy to raise the bulk temperature of the Earth (all 6e+24 kg of it)
> by 10-20 degrees C, but by far most of it will be produced in the
> surface layers... And then you have to dissipate all this, over much
> longer timescales...
Wow, cool.
> Conservatively, I'd say you'll have a near global magma ocean for
> hundreds to thousands of years. When & if it's ever inhabitable,
> geothermal heat is going to be a popular power source...
> If you're wondering about tidal waves in the very early stages of the
> problem (i.e.- while there's still somebody there to see them), they'd
> be big. Once again for Earth in place of Io, *simply* scaling implies
> that the tidal bulges would be around 20,000 times as high as currently.
> Titanic earthquakes as nearly every crustal stress is released in a
> period of a few days, followed by a bulge of water circling the planet
> every 56 hours.
Far out. I like this better and better.
>
> > What would happen to the weather? ;-)
>
> I doubt in the short term you need to worry about it <grin>. Are you
> asking what would transiently happen to the weather, or how would the
> weather be different after an equilibrium of some type is reached?
>
> Also note that you can come out with a "no time at all for tidal
> locking" solution to your problem - just dump the Earth into a close
> orbit with a period that matches it's rotational period already. But
> that's cheating <grin>. And you'd still have major problems with earthquakes.
Yup, that's an interesting possibility... or into a more distant orbit that's
an appropriate whole-number multiple. Earthquakes are one of the effects
that'd have to be dealt with somehow. Preferably fictionally.
>
> --
> Brian Davis
Charles R Martin
Right, understood. But I was really referring i9n my comment to the stuff up
to that point, not just your comment. While it would unquestionably shift
over time, at least one version here is talking about decades -- and I find
myself wondering if that wouldn't be even more disruptive than we'd been
thinking....
Brian Davis
>> The rough timescale for tidal locking is:
>>
>> t = 16 rho omega a^6 (Q/k2) / ( 45 G M^2 )
>> rho, Q, k2 = variable describing the body that's slowing...
>> a = orbital distance
>> M = mass of body that's doing the slowing...
>> omega = 2 pi / P
>
> That's an odd way to phrase it. I would put the equation in terms
> of *either* P *or* a, but not both, since omega = SQRT(GM/a^2).
Well, given that we're talking about tidal braking of an initially
free rotation state, omega(rotation) does not equal omega(revolution)
just yet: it does only after tidal lock is attained. I should have been
much more specific in my variable definitions, but omega is the moon's
rotational angular rate, not orbital angular rate. Does that make more sense?
> The fact that you have both in your equation suggests to me an error.
The fact that I get such a short timescale suggests to my an error,
but I've used this same equation to get rough estimates on bodies from
Mercury to Triton.
>> Well, to slow the Earth from a 24 hour day to a 42 hour day (Io's
>> tidally locked period) will produce 1.4e+29 J. Ouch. That's enough
>> energy to raise the bulk temperature of the Earth (all 6e+24 kg of
>> it) by 10-20 degrees C, but by far most of it will be produced in
>> the surface layers...
>
> No, the power has to radiate out through the surface (obviously),
> but it is actually produced in the viscoelastic layers-- most
> likely the mantle.
True. My point was that it would not be released in the liquid core
(& I think the solid core would be largely unaffected as well),
concentrating the heating in the "outer layers" (ie- distinctly
non-uniform, even though my rough BotE consequence assumes a uniform
deposition of energy.
>> If you're wondering about tidal waves...
>
> Note that a true tidal wave (a wave caused by tides) is not the
> same as a tsunami (often popularly but erroneously called a "tidal
> wave"). In this case, I expect you would have both.
Probably, and thanks for mentioning the difference. I was speaking of
the shape of a forced tidal bulges as they would circumnavigate the
globe, not a classic shallow-water ocean wave as typified by a tsunami.
>> a bulge of water circling the planet every 56 hours.
>
> Why 56?
Did I get that wrong? Initially the transplanted Earth will have a
roughly 24 hour sidereal rotation rate, but Jupiter will have an
apparent rate of something very different, (synodic period? I confess I
get this things confused too often, especially when I just rough it out
which I did in this case). Given the Earth's orbital period around
Jupiter (1.77 days), I thought Jupiter would appear to move around with
a period of 1/(1/24 - 1/42) = 56 hours (OK, I'm also assuming some
things about the direction of rotation... should have thought more
before posting).
> author of MARS CROSSING, now available at bookstores everywhere
...which I've read some good things about, but alas, I can't buy it
till after the holidays (it was "on the list", and I think someone else
may have bought it for me). Looking forward to it.
--
Brian Davis
Brian Davis
>> The rough timescale for tidal locking is:
Both you and Geoffrey have rightly commented on my poorly-defined
variables, so let me re-state this. BTW, I'm cribbing this formula (very
slightly modified) from somewhere else, namely Burn's chapter in
"Satellites" (U of Az Press), edited by Burns & Mathews:
T = 16 rho omega a^6 (Q/k2) / ( 45 G M^2 )
rho = density of body being despun [kg/m^3]
omega = inital rotation rate of body being despun [rad/s]
= 2 pi / P, where P is the inital rotation rate [s]
a = semi-major axis of orbit [m]
Q/k2 = dissipation function divided by the 2nd order Love #
M = mass of body doing the despinning [kg]
I hope this is clearer - I've taken the formula for the despinning
timescale out of Burn's chapter and modified it very slightly.
> if we hold rho, Q, k2, M and P constant, so that we vary only the
> orbital distance, doesn't this mean that the time scale *increases*
> as the sixth power of the radius? So pick an orbit relatively
> slightly further out and the time scale increases *greatly*.
Exactly so, although the amount you have to despin the body also
increase slightly as well. But even for more remote orbits, the
timescales are mighty short: dropping the Earth in Callisto's orbit
yields a timescale around 340,000 years, still a very short time to
dissipate a lot of energy (about 45% more energy then for the "Earth
replaces Io" case).
> Far out. I like this better and better.
Yeah, well, I suspect that's because you're not one of the charecters
in the storyline <grin>.
--
Brian Davis
Geoffrey A. Landis
Brian Davis wrote:
> Well, given that we're talking about tidal braking of an initially
> free rotation state, omega(rotation) does not equal omega(revolution)
> just yet
Oops-- I was commenting on the wrong problem! This makes most of what I said
completely irrelevant. My apologies.
Can you comment on where you got the a^6 dependence? I get a different result
for the braking due to the slosh of water in ocean basins and that due to the
stretching of the planet, but my back-of-the-braincell estimate gives me a^4 and
a^3 respectively.
--
Geoffrey A. Landis
author of MARS CROSSING, now available at bookstores everywhere
http://www.sff.net/people/geoffrey.landis
Charles R Martin
>
> Charles R Martin wrote:
>
> >> The rough timescale for tidal locking is:
>
> Both you and Geoffrey have rightly commented on my poorly-defined
> variables, so let me re-state this. BTW, I'm cribbing this formula (very
> slightly modified) from somewhere else, namely Burn's chapter in
> "Satellites" (U of Az Press), edited by Burns & Mathews:
>
> T = 16 rho omega a^6 (Q/k2) / ( 45 G M^2 )
> rho = density of body being despun [kg/m^3]
> omega = inital rotation rate of body being despun [rad/s]
> = 2 pi / P, where P is the inital rotation rate [s]
> a = semi-major axis of orbit [m]
> Q/k2 = dissipation function divided by the 2nd order Love #
> M = mass of body doing the despinning [kg]
(Just by the by, what the hell's a "second order Love number?")
(I know, I should buy the book.)
>
> I hope this is clearer - I've taken the formula for the despinning
> timescale out of Burn's chapter and modified it very slightly.
>
> > if we hold rho, Q, k2, M and P constant, so that we vary only the
> > orbital distance, doesn't this mean that the time scale *increases*
> > as the sixth power of the radius? So pick an orbit relatively
> > slightly further out and the time scale increases *greatly*.
>
> Exactly so, although the amount you have to despin the body also
> increase slightly as well. But even for more remote orbits, the
> timescales are mighty short: dropping the Earth in Callisto's orbit
> yields a timescale around 340,000 years, still a very short time to
> dissipate a lot of energy (about 45% more energy then for the "Earth
> replaces Io" case).
Hmmm... but we're still talking 1e24 J (ROM) and using your figures because
I'm too lazy to redo it, we're changing from 1e23 J/yr to 1e18 J/yr ... or
from 2 degrees/yr to 0.00002 degrees per year. That seems *significantly*
less dramatic.
>
> > Far out. I like this better and better.
>
> Yeah, well, I suspect that's because you're not one of the charecters
> in the storyline <grin>.
Bite your tongue -- I'm *all* the characters in my storyline.
(It's just that I can *leave*.)
Brian Davis
> Can you comment on where you got the a^6 dependence?
Straight from the book <grin>! Seriously, I did *not* derive these
equations myself, so I've got only a very rough understanding of the
details. But, roughly speaking, the torque exerted on the body is
inversely proportional to the cube of the distance (think of a dipole
field, interaction drops as a^-3), while the "dipole strength" (actual
equlibrium height of the tidal bulges) also goes as a^-3 (closer the
bodies, the higher the tidal deformation). The combined effect is a a^6 dependence.
At least I *think* that's how you get that dependance.
> I get a different result for the braking due to the slosh of water
> in ocean basins and that due to the stretching of the planet,...
> a^4 and a^3 respectively.
I can see the a^3 (tidal height dependance on distance), but not the
a^4; what do you see as a qualitatively different for liquid vs. solid tides?
--
Brian Davis
Geoffrey A. Landis
Brian Davis wrote:
> Geoffrey A. Landis wrote:
>
> > Can you comment on where you got the a^6 dependence?
OK, I think I understand the a^6 dependence now.
The differential gravitational force is proportional to a^3 (tidal force law), so
essentially the water is pouring into a basin of depth a^3. The energy per unit mass
of water pouring into the basin is proportional to a^3, and the amount of water is
*also* proportional to a^3, since you can pour more water into a deeper well.
(for stretching a solid planet, I still think it only goes as a^3, since you stretch
the whole planet-- you don't stretch more planet if you stretch harder)
ameen . net
"Geoffrey A. Landis" <Geoffrey...@sff.net> spake from on high
claiming:
Ben Wilson (a.k.a. Ameen, Last of the Dausha)
____________________________
-"Ever heard of Aristotle . . . Plato . . . Socrates?!"
-"Yes."
-"Morons!"
Brian Davis
> The differential gravitational force is proportional to a^3 (tidal
> force law), so essentially the water is pouring into a basin of
> depth a^3...
Why would the potential energy of the oceans change at all as the
tidal bulge circulates around the globe? There are frictional or viscous
losses, but I'm not sure I see that in your analysis, and energy
conversion to heat is not a major driving factor; torque is. The water
surface is an equipotential, and while the equipotential rises in one
area, it falls in another - no net potential energy change during a
tidal cycle, *unless* you consider viscous loss mechanisms. To put it
another way, the net effect of a constant tidal force is not to change
an energy, but to deform the body in question, increasing the size of
the higher-order terms in the gravitational multipole expansion.
> for stretching a solid planet, I still think it only goes as a^3...
Well, I wish I could help more, but I've got to admit I've *not*
worked out the equations, just adapted those that I studied out of
"Satellites". You may very well be right.
--
Brian Davis
Geoffrey A. Landis
Brian Davis wrote:
> Geoffrey A. Landis wrote:
>
> > The differential gravitational force is proportional to a^3 (tidal
> > force law), so essentially the water is pouring into a basin of
> > depth a^3...
>
> Why would the potential energy of the oceans change at all as the
> tidal bulge circulates around the globe? There are frictional or viscous
> losses, but I'm not sure I see that in your analysis,
It's not explicit. I'm modelling tides as sloshing in a finite basin, not
as a bulge circulating freely around the planet, but the dependence is the
same either way. If you then assume a Q factor for frictional dissipation,
a constant fraction of the energy is taken out at every energy transfer.
> and energy
> conversion to heat is not a major driving factor; torque is.
You can model it as torque if you like, the modelling goes exactly the same
way. Torque is mass displaced from the center of the planet times
gravitational (tidal) acceleration times distance from the planet's
center. Mass is proportional to the height of water raised, which goes as
a^3, Acceleration is also proportional to a^3, and distance is constant
at the planet's radius (assuming tidal bulge is small compared to the
radius).
> The water
> surface is an equipotential, and while the equipotential rises in one
> area, it falls in another - no net potential energy change during a
> tidal cycle, *unless* you consider viscous loss mechanisms.
Sure. I didn't go that far in the analysis because I was only trying to
see where the a^6 factor came from, not trying to reproduce the whole
thing.
> To put it
> another way, the net effect of a constant tidal force is not to change
> an energy, but to deform the body in question, increasing the size of
> the higher-order terms in the gravitational multipole expansion.
>
> > for stretching a solid planet, I still think it only goes as a^3...
>
> Well, I wish I could help more, but I've got to admit I've *not*
> worked out the equations, just adapted those that I studied out of
> "Satellites". You may very well be right.
>
> --
> Brian Davis
--
pervect
"Geoffrey A. Landis" <Geoffrey...@sff.net> wrote in message
news:3A40C2E7...@sff.net...
>
>
> Brian Davis wrote:
>
> > Well, given that we're talking about tidal braking of an initially
> > free rotation state, omega(rotation) does not equal omega(revolution)
> > just yet
>
> Oops-- I was commenting on the wrong problem! This makes most of what I
said
> completely irrelevant. My apologies.
>
> Can you comment on where you got the a^6 dependence? I get a different
result
> for the braking due to the slosh of water in ocean basins and that due to
the
> stretching of the planet, but my back-of-the-braincell estimate gives me
a^4 and
> a^3 respectively.
My off-the-cuff analysis gives an a^6 dependence, but I'm not sure how good
it is. It's much more oriented to stretching than sloshing. Here's the
argument.
Let's start off with a crude model of the stretchable earth as a
spring-mass-damper system. We can characterize this system by a quality
factor Q that relates the amount of energy stored each cycle to the amount
of energy dissipated each cycle.
Now all we need to do is to find out what the proportionality constant is
for the energy stored each cycle. Well, the tidal force is 1/r^3 as you
point out, and the distance that the tides move is proportional to the
force, so it's also proportional to 1/r^3. The product, energy =
force*distance is therefore proportional to 1/r^6. (The force is averaged ,
which actually makes it .5/r^6, but that doesn't change the exponent).
Geoffrey A. Landis
pervect wrote:
> My off-the-cuff analysis gives an a^6 dependence, but I'm not sure how good
> it is. It's much more oriented to stretching than sloshing. Here's the
> argument.
>
> Let's start off with a crude model of the stretchable earth as a
> spring-mass-damper system. We can characterize this system by a quality
> factor Q that relates the amount of energy stored each cycle to the amount
> of energy dissipated each cycle.
>
> Now all we need to do is to find out what the proportionality constant is
> for the energy stored each cycle. Well, the tidal force is 1/r^3 as you
> point out, and the distance that the tides move is proportional to the
> force, so it's also proportional to 1/r^3. The product, energy =
> force*distance is therefore proportional to 1/r^6. (The force is averaged ,
> which actually makes it .5/r^6, but that doesn't change the exponent).
That line of reasoning sounds good to me. Thanks.