18 views

Skip to first unread message

Oct 30, 2002, 7:17:46 PM10/30/02

to

I'm addressing the scenario - what would the Earth be like if the Moon were

10x closer to the earth than it actually is - [with the earth and moon

tidally locked].

10x closer to the earth than it actually is - [with the earth and moon

tidally locked].

For the earth with the current values of precessional constant, I get an

estimate of 24,000 days (65 years roughly) for an object with a relatively

low inclanation (cos theta approximately equal to 1) to precess completely

(through 360 degrees or 2*pi radians).

Because the precession rate is so slow, I would guess that the relative rate

of orbital plane re-alignment of the moon as compared to the rate at which

the tidal lock occurs would be slowed by the same factor (i.e. 24,000).

However, since the tidal lock process will occur 1,000,000 times faster with

the moon 10x closer, I think it's fairly likely that the orbital plane

normalization process will occur assuming that the age of the Earth is the

same as it is now.

This is important because as the orbital plane precesses, it will drag

fairly large tides with it.

Based on

http://scienceworld.wolfram.com/physics/Tide.html

I'm estimating the tides as 500 meters or so. [The analysis above is for

the equator-pole total tidal height difference, but I think the analysis

should work just as well for the height difference at two points 90 degrees

distant on the equator - the potential terms are identical.]

Because these tides are relatively small, I don't think there would be a

significant impact on the moments of inertia/precessional constant.

Nov 1, 2002, 3:34:36 PM11/1/02

to

"pervect" wrote:

> I'm addressing the scenario - what would the Earth be like if the Moon were

> 10x closer to the earth than it actually is - [with the earth and moon

> tidally locked].

OK, you know the other post by me that will unfortunately be

appearing? The one where I say such a system is impossible? This is me

eatting crow 30 sec after hitting "send". I thought you said "100

times" closer, at which point the high rotational velocity of the

Earth would have ripped it apart. Moving the Moon in to a distance of

38,440 km gives you a month of 0.86 days, or 20.7 hours, plenty slow

enough.

> For the earth with the current values of precessional constant, I get an

> estimate of 24,000 days (65 years roughly) for an object with a relatively

> low inclanation (cos theta approximately equal to 1) to precess completely

> (through 360 degrees or 2*pi radians).

First, what do you mean "current value of the precessionl

constant"? Second, the Earth's precession of the obliquity depends on

both the distance to the Moon (&, complicating things, the Sun), and

also one the rotational flattening of the Earth. In other words, I'm

not sure this is a good approximation.

> Because the precession rate is so slow, I would guess that the relative rate

> of orbital plane re-alignment of the moon as compared to the rate at which

> the tidal lock occurs would be slowed by the same factor (i.e. 24,000).

Huh? That's a hellishly rapid precessional rate! Are you

calculating the lunar orbital precession? Or the Earth's precession of

obliquity? And how are you calculating it?

> However, since the tidal lock process will occur 1,000,000 times faster with

> the moon 10x closer, I think it's fairly likely that the orbital plane

> normalization process will occur assuming that the age of the Earth is the

> same as it is now.

Generally, yes. You also have to consider what controls the

precession of the orbital plane of the Moon - for our Moon, it's not

the Earth, but the Sun, while for closer objects (Io, for instance),

it's the non-spherodial planet.

If the Moon were ten times closer, it's somewhat questionable if

Solar or terrestrial influences would dominate here - it's within a

factor of two of the critical limit (although inside it, so I

*suspect* the Earth would dominate, and you'd have the Moon's orbit

precess with the Earth's obliquity)

> I'm estimating the tides as 500 meters or so. [The analysis above is for

> the equator-pole total tidal height difference, but I think the analysis

> should work just as well for the height difference at two points 90 degrees

> distant on the equator - the potential terms are identical.]

Actually (surprisingly) it doesn't quite. Tidal deformation of a

body is a three-axis ellipsoid, because the polar point is not

rotating (much), while the points on the equator leading and trailing

in it's orbital motion *are* experiencing an additional rotation.

Generally,

r_1 = ( 1 + (14/9) h gamma ) r radius along primary-satellite line

r_2 = ( 1 - (4/9) h gamma ) r radius perpendicular to r_1 and

rotation axis

r_3 = ( 1 - (10/9) h gamma ) r radius along rotation axis

where,

r = volumetric radius ("mean radius")

gamma = equilibrium tidal height

= (3/4) (M/m) (r/a)^3

M = mass raising the tides

m = mass experiencing the tides

a = seperation distance between the two

h = correction factor for self-gravitation and rigidity

= 0.4167 for Earth

= 0.0333 for Moon

= 5/2 for an ideal self-gravitating liquid (zero strength) body

> Because these tides are relatively small, I don't think there would

> be a significant impact on the moments of inertia/precessional constant.

Probably true, but speeding up the Earth's rotation would certainly

alter it's precessional behavior (increase the flattening). The

flattening is proportional to omega^2 (that's the angular rotation

velocity), so flattening will increase to roughly 144% of it's current

value, which will increase the second-order moment (J_2) similarly, so

it throws you off by *at least* that much.

OK, gang, what did I drasticly screw up in *this* post? sigh...

--

Brian Davis

Nov 1, 2002, 7:59:05 PM11/1/02

to

"Brian Davis" <brd...@iusb.edu> wrote in message

news:f5ead130.02110...@posting.google.com...

> > For the earth with the current values of precessional constant, I get an

> > estimate of 24,000 days (65 years roughly) for an object with a

relatively

> > low inclanation (cos theta approximately equal to 1) to precess

completely

> > (through 360 degrees or 2*pi radians).

>

> First, what do you mean "current value of the precessionl

> constant"? Second, the Earth's precession of the obliquity depends on

> both the distance to the Moon (&, complicating things, the Sun), and

> also one the rotational flattening of the Earth. In other words, I'm

> not sure this is a good approximation.

I think this is a reasonable approximation (though not exact). But I'm

definitely open to second opinions as to things I might have missed.

As far as the background goes:

http://scienceworld.wolfram.com/physics/PrecessionalConstant.html

gives H, the precession constant, as 305

My other text [Goldstein, Classical Mechanics] doesn't give H, but gives a

precession rate of 700 orbits for a hypothetical body orbiting at a r=R at

an orbital inclination of 30 degrees, which winds up in the same ballpark

[

The formula I used for the precessional rates is from Goldstein

rate-of-precession = .5 * H * (R/r)^2 cos(theta)

where

1/rate-of-precession = number of orbits for plane to precess through 360

degrees

H = I3-I1/I3 (Goldstein's notation) = precessional constant (dimensionless)

R = radius of earth

r = radius of orbit of orbiting body

theta = inclanation of orbit of orbiting body

> > Because the precession rate is so slow, I would guess that the relative

rate

> > of orbital plane re-alignment of the moon as compared to the rate at

which

> > the tidal lock occurs would be slowed by the same factor (i.e. 24,000).

>

> Huh? That's a hellishly rapid precessional rate! Are you

> calculating the lunar orbital precession? Or the Earth's precession of

> obliquity? And how are you calculating it?

The formulas are above, I just put in r/R = 6 (that's with the moon 10x

closer, it would be r/R = 60 for the current moon's orbit). That's the

precession due to the Earth's quadropole moment.

It's rapid because I moved the moon in 10x closer, basically, it would be

100x slower with the moon at it's current distance.

> Generally, yes. You also have to consider what controls the

> precession of the orbital plane of the Moon - for our Moon, it's not

> the Earth, but the Sun, while for closer objects (Io, for instance),

> it's the non-spherodial planet.

> If the Moon were ten times closer, it's somewhat questionable if

> Solar or terrestrial influences would dominate here - it's within a

> factor of two of the critical limit (although inside it, so I

> *suspect* the Earth would dominate, and you'd have the Moon's orbit

> precess with the Earth's obliquity)

This is one of the things I didn't calculate and don't have a good handle

on - what exactly - well approxmiately is good enough - would happen to the

dynamics of the new earth-moon system due to the sun and other bodies in the

solar system.

I don't mind being off by less than 2:1 :-)

Well, these certainly look like much better formulas for the tides. What is

the source?

Is "h" related to the "Love numbers" I've seen mention of?

Anyway, I didn't intend to change the earth's day very much, I thought that

changing the distance by a factor of 10 was simpler than changing it by the

exact amount needed to make the day 24 hours.

Nov 2, 2002, 1:12:29 PM11/2/02

to

"pervect" <perv...@netscape.net> wrote in message

news:trFw9.65573$C53.3...@news2.west.cox.net...

> The formula I used for the precessional rates is from Goldstein

>

> rate-of-precession = .5 * H * (R/r)^2 cos(theta)

I thought I had better add (in afterthought) a note that this particular

formula has some asasumptions peculiar to the Earth, and isn't a general

formula.

The biggest one of these assumptions is that the moment of inertia (I3) of

the primary body is on the order of 1/3 * m R^2. An uniform sphere would

have a moment of inertia of 2/5 m R^2, but the earth isn't a uniform body.

While this won't necessarily make a major difference in the result, I

thought (in retrospect) that a note of warning was called for.

Nov 2, 2002, 11:01:50 PM11/2/02

to

In article <trFw9.65573$C53.3...@news2.west.cox.net>, pervect

<perv...@netscape.net> wrote:

<perv...@netscape.net> wrote:

> > First, what do you mean "current value of the precessionl

> > constant"? Second, the Earth's precession of the obliquity depends on

> > both the distance to the Moon (&, complicating things, the Sun), and

> > also one the rotational flattening of the Earth. In other words, I'm

> > not sure this is a good approximation.

>

> I think this is a reasonable approximation (though not exact). But I'm

> definitely open to second opinions as to things I might have missed.

>

> As far as the background goes:

>

> http://scienceworld.wolfram.com/physics/PrecessionalConstant.html

>

> gives H, the precession constant, as 305

>

That particular precession constant is the 'free precession' rate at

which a solid body without external forces precesses.

The more relevant precession rate is the 18.6 year precession period of

the plane of the Moon's orbit, driven by solar tidal forces. (Low

Earth orbit satellites' orbits precess due to the bulge of Earth's

equator, but that is unimportant at the Moon's altitude.) The plane of

the Moon's orbit about Earth is 5 degrees from that of the Earth's

around the Sun (the 'ecliptic', not the equator), but the orientation

of the offset revolves in with this 18.6 year period.

If the Moon were 10x closer, then the solar tidal precession torque

would be only 1/10 as strong, but the angular momentum of the system

would be 1/sqrt(10) as much, so the precession period would be 3x more,

or roughly 55-60 years.

--

David M. Palmer dmpa...@email.com (formerly @clark.net, @ematic.com)

Nov 3, 2002, 3:28:00 AM11/3/02

to

"David M. Palmer" <dmpa...@email.com> wrote in message

news:021120022101500306%dmpa...@email.com...

> That particular precession constant is the 'free precession' rate at

> which a solid body without external forces precesses.

While it's true that a body with I1=I2 does precess uniformly, the equations

I posted really were for what I said they were, the precession of the

orbital plane of a satellite due to the Earth's quadropole moment.

The Earth's axis of rotation does precess some - this Chandler Wobble,

though, is related to a subtly different constant:

omega = (I3-I1/I1)*w3 [I just looked this up]

vs I3-I1/I3 (the constant used in my equations).

Since it's from my textbook rather than a webpage, I don't have an online

reference for the formulas I posted, but if you have a copy of Goldstein's

"Classical Mechanics", you can look it up. Goldstein does not call this

constant "H", however, it's just I3-I1/I3 in his text.

> The more relevant precession rate is the 18.6 year precession period of

> the plane of the Moon's orbit, driven by solar tidal forces.

> Earth orbit satellites' orbits precess due to the bulge of Earth's

> equator, but that is unimportant at the Moon's altitude.) The plane of

> the Moon's orbit about Earth is 5 degrees from that of the Earth's

> around the Sun (the 'ecliptic', not the equator), but the orientation

> of the offset revolves in with this 18.6 year period.

>

>

> If the Moon were 10x closer, then the solar tidal precession torque

> would be only 1/10 as strong, but the angular momentum of the system

> would be 1/sqrt(10) as much, so the precession period would be 3x more,

> or roughly 55-60 years.

It looks like ignoring the solar effects is a serious problem, then. It's

definitely wrong for the current lunar position, and it appears to be

problematical even with the moon 10x closer.

Nov 3, 2002, 7:39:20 PM11/3/02

to

"pervect" wrote:

>> the Earth's precession of the obliquity depends on both the

>> distance to the Moon (&, complicating things, the Sun), and

>> also one the rotational flattening of the Earth.

>

> I think this is a reasonable approximation (though not exact).

I see - your precessional constant takes into account the

quadrapole moment of the Earth.

> My other text [Goldstein, Classical Mechanics] doesn't give H,

> but gives a precession rate of 700 orbits for a hypothetical

> body orbiting at a r=R at an orbital inclination of 30 degrees,

> which winds up in the same ballpark.

OK, so you're talking about the precession of the Moon's orbital

plane relative to the Earth's equator, yes? I would find it rather

unlikely that any tidally-locked bodies would *have* such a relative

inclination - tidal effects drive the inclination to zero if the

orbit is expanding (and if tides are contracting or collapsing the

orbit, there's other issues). So in the process of achiving tidal

lock, the inclination is reduced to zero (and if it's not, tidal

interactions continue reducing the inclination even for a

synchronous but inclined orbit I suspect).

Note that this holds for nearly all the satellites in the solar

system *except* the Moon, and (IMS) this is due partially to the

Moon being so large and far from the Earth (look up "cassini states"

sometime, and then maybe someone can explain them to me).

> The formula I used for the precessional rates is from Goldstein

I use a version of that one as well, but I (hopefully correctly)

put it into terms of more use to me in world-building:

d_CapOmega/dt = (-3/2) omega J2 (R/a)^2 Cos(i) / ( 1 - e^2 )^2

CapOmega = angular position of the ascending node

d_CapOmega/dt = precession of the line of nodes [radians/s]

omega = orbital angular velocity [radians]

J2 = second order moment of Earth (or whatever central planet)

R = radius of Earth (or whatever)

a = semi-major axis of satellite orbit (same units as R)

i = inclination of orbital plane relative to equatorial plane

e = eccentricity

I prefer this because (while more complicated), I can play with

the parameters far more (recalculate J2, for instance, from first

principles; it's harder for me at least to look up I1, I2, & I3).

The correction for eccentricity is from "Satellites", from the

excellent AZ Press series.

> rate-of-precession = .5 * H * (R/r)^2 cos(theta)

Same functional form as mine I suspect, although not directly

analogous due to mine being in radians/sec, and yours (Goldsteins)

delivering in terms of the orbital period.

>>> Because the precession rate is so slow,

>>

>> Huh? That's a hellishly rapid precessional rate!

>

> It's rapid because I moved the moon in 10x closer...

I agree - my point was that your state precession is very very

rapid, while you had just termed it "so slow". I'm unclear I guess

on the original interpretation.

>> If the Moon were ten times closer, it's somewhat questionable

>> if Solar or terrestrial influences would dominate here - it's

>> within a factor of two of the critical limit (although inside it,

>> so I *suspect* the Earth would dominate, and you'd have the

>> Moon's orbit precess with the Earth's obliquity)

>

> This is one of the things I didn't calculate and don't have a

> good handle on - what exactly - well approxmiately is good enough

> - would happen to the dynamics of the new earth-moon system due

> to the sun and other bodies in the solar system.

WhatI was refering to is that satellites can be seperated into two

groups - close ones who's orbital planes precess with the parent (Io,

Europa, Titan, etc.) and distant ones where the orbital plane

precession is dominated by solar effects, thus keeping them from

aligning with the planet's equatorial plane. The critical distance

(once again from the U of Az "Satellites" text) is:

a_crit = ( 2 J2 R^2 a^3 (M_planet/M_star) )^(1/5)

R = radius of planet

a = semi-major axis of planet's orbit

This isn't a hard limit, but a pretty good one. Inside that, the moon

will remain in an orbit with zero inclination to the planetary equator,

with the orbital plane precessing in lock-step with the planet's

obliquity. Outside it, solar effects precess the moon's orbit... and

I have to think on that one (it's easy enough to see how it works,

but I've not worked it out).

>> r_1 = ( 1 + (14/9) h gamma ) r radius along primary-satellite line

>> [etc.]

>> where,...

> > gamma = equilibrium tidal height

> > = (3/4) (M/m) (r/a)^3

> > h = correction factor for self-gravitation and rigidity

> > = 0.4167 for Earth

> > = 0.0333 for Moon

> > = 5/2 for an ideal self-gravitating liquid... body

>

> Well, these certainly look like much better formulas for the tides.

> What is the source?

U of AZ "Satellites" again (I think the chapter by Peale). Great

(if technical) worldbuilding source.

> Is "h" related to the "Love numbers" I've seen mention of?

Yes. h = (5/3) k2, where k2 is the 2nd order Love number. More

ugly math:

k2 = (3/2) / ( 1 + 19 mu / ( 2 g rho R ) )

mu = rigidity of body (in N/m^2; for Earth, roughly 1e+11)

g = surface gravity (m/s^2)

rho = mean density (kg/m^3)

R = radius (in meters)

If the body is stengthless (mu=0), k2 = 3/2.

>>> Because these tides are relatively small, I don't think there

>>> would be a significant impact on the moments of

>>> inertia/precessional constant.

>>

>> Probably true, but speeding up the Earth's rotation would

>> certainly alter it's precessional behavior (increase the flattening).

>

> I don't mind being off by less than 2:1 :-)

Agreed - but it's more fun (for a sufficiently demented definition

of the word "fun") to get the number right. And you can calculate

the flattening (or at least limits on it) as well as J2, given

the rotation period.

--

Brian Davis

Nov 3, 2002, 9:04:47 PM11/3/02

to

> OK, so you're talking about the precession of the Moon's orbital

> plane relative to the Earth's equator, yes?

Yep, you've got it.

> I would find it rather

> unlikely that any tidally-locked bodies would *have* such a relative

> inclination - tidal effects drive the inclination to zero if the

> orbit is expanding (and if tides are contracting or collapsing the

> orbit, there's other issues). So in the process of achiving tidal

> lock, the inclination is reduced to zero (and if it's not, tidal

> interactions continue reducing the inclination even for a

> synchronous but inclined orbit I suspect).

It was my guess as well that the inclination of the moon's orbit to the

earth's equator would probably be driven to zero, but I wasn't positive my

guess was right.

> I use a version of that one as well, but I (hopefully correctly)

> put it into terms of more use to me in world-building:

It looks to me like the same basic formula - note again that my formula

wasn't completely general, as it had as a built in assumption that

I3=1/3mR^2 (you have to read the fine print on these things! :-)).

<snip>

> What I was refering to is that satellites can be seperated into two

> groups - close ones who's orbital planes precess with the parent (Io,

> Europa, Titan, etc.) and distant ones where the orbital plane

> precession is dominated by solar effects, thus keeping them from

> aligning with the planet's equatorial plane. The critical distance

> (once again from the U of Az "Satellites" text) is:

>

> a_crit = ( 2 J2 R^2 a^3 (M_planet/M_star) )^(1/5)

> R = radius of planet

> a = semi-major axis of planet's orbit

Coolness. Well, if you tell me what J2 is for the earth, I can plug-n-chug

to double-check the numbers. Not that my calculations are necessarily more

reliable than yours :-) [I've made the occasional serious numerical blunder

here and there and it seems especially likely when I post.]

I can appreciate in general that J2 is a measure of the Earth's quadropole

moment, but right now I'd have to do some fancy footwork to get an actual

numerical value for it to plug into this equation.

I'll have to put that U of AZ book on my wishlist, too.

Nov 3, 2002, 9:26:20 PM11/3/02

to

pervect wrote:

> Coolness. Well, if you tell me what J2 is for the earth, I can

> plug-n-chug

> to double-check the numbers.

Planetary Fact Sheets says it's 1082.63 x 10^-6.

All hail PFS:

http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

--

Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/

__ San Jose, CA, USA / 37 20 N 121 53 W / &tSftDotIotE

/ \ Light ... more light!

\__/ (the last words of Goethe)

PyUID / http://www.alcyone.com/pyos/uid/

A module for generating "unique" IDs in Python.

Nov 4, 2002, 3:47:04 PM11/4/02

to

"Erik Max Francis" <m...@alcyone.com> wrote in message

news:3DC5DACC...@alcyone.com...

> Planetary Fact Sheets says it's 1082.63 x 10^-6.

>

> All hail PFS:

>

> http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

Thanks - I'm adding this to my bookmarks.

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu