A Space & astronomy forum. SpaceBanter.com

If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below.

Go Back   Home » SpaceBanter.com forum » Space Science » History
Site Map Home Authors List Search Today's Posts Mark Forums Read Web Partners

Is the moon leaving, or are we shrinking by 38 mm/year



 
 
Thread Tools Display Modes
  #1  
Old November 6th 03, 09:44 PM
OM
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

On Thu, 6 Nov 2003 15:59:41 -0500, "Ken Glover"
wrote:

"Brad Guth" wrote in message
. com...

[snip all]

Leave the Lunar Space Elevator idea for another time. Start work
immediately on a GUTH/English translator...


....Personally, I theorize that a GUTH/English translator can be easily
constructed using one single part - a 2x4.


OM

--

"No ******* ever won a war by dying for | http://www.io.com/~o_m
his country. He won it by making the other | Sergeant-At-Arms
poor dumb ******* die for his country." | Human O-Ring Society

- General George S. Patton, Jr
Ads
  #2  
Old November 11th 03, 02:25 AM
Ned Pike
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

In ,
OM [email protected]_blessed_lady_mary_of_the_holy_NASA_researc h_facility.org spewed:
On Thu, 6 Nov 2003 15:59:41 -0500, "Ken Glover"
wrote:

"Brad Guth" wrote in message
om...

[snip all]

Leave the Lunar Space Elevator idea for another time. Start work
immediately on a GUTH/English translator...


...Personally, I theorize that a GUTH/English translator can be easily
constructed using one single part - a 2x4.


OM


LART, LART, LART


  #3  
Old November 15th 03, 09:58 PM
OM
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

On 15 Nov 2003 12:21:36 -0800, (lensman1955)
wrote:

Neil Adams (of Batman fame) actually has a theory that Continential
Drift isn't real and the Earth is actually expanding!


....Actually, that's Neal Adams, and nobody takes him seriously since
he quit drawing the Bat-books and it took him something like 12 years
to get _Ms. Mystek_ #2 out after #1 hit the stands.

OM

--

"No ******* ever won a war by dying for |
http://www.io.com/~o_m
his country. He won it by making the other | Sergeant-At-Arms
poor dumb ******* die for his country." | Human O-Ring Society

- General George S. Patton, Jr
  #4  
Old November 16th 03, 02:57 PM
lensman1955
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

OM [email protected]_blessed_lady_mary_of_the_holy_NASA_researc h_facility.org wrote in message . ..
On 15 Nov 2003 12:21:36 -0800, (lensman1955)
wrote:

Neil Adams (of Batman fame) actually has a theory that Continential
Drift isn't real and the Earth is actually expanding!


...Actually, that's Neal Adams, and nobody takes him seriously since
he quit drawing the Bat-books and it took him something like 12 years
to get _Ms. Mystek_ #2 out after #1 hit the stands.


I'm not saying it is something to be taken seriously... just brought it up.
  #5  
Old November 17th 03, 06:43 PM
Marvin
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

So again, how much kinetic energy, and/or in terms of thrust (ISP
or kg/s in terms of fuel/energy consumption), would it take for a
satellite the size and mass of our moon, to be accelerated so as to
escape Earth by a rate of 38 mm/year?


Thanks so much for all the feedback. I'll try a little more input into
my three remaining brain cells and see if the amount of recession
energy can be realized.


The amount of energy needed to raise the moon's orbit by that 3.8cm/year is
easy.. Just plug into basic orbital mechanics, out pops an answer of some
1.25 * 10^20 joules per year, easier to visualise as a constant power input
of some 4 terawatt (4 * 10^12 watt)

The problem is that this is indeed the *resultant* of all forces
applicable. Just how big are the other forces working on the moon,like
frictional drag against the not-quite-empty vacuum of local space.(keep in
mind the moon orbits earth at about 1km/sec, and the earth whizzes around
the sun at about 30km/sec)?





  #6  
Old November 17th 03, 09:52 PM
Brad Guth
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

Marvin wrote in message ...
So again, how much kinetic energy, and/or in terms of thrust (ISP
or kg/s in terms of fuel/energy consumption), would it take for a
satellite the size and mass of our moon, to be accelerated so as to
escape Earth by a rate of 38 mm/year?


Thanks so much for all the feedback. I'll try a little more input into
my three remaining brain cells and see if the amount of recession
energy can be realized.


The amount of energy needed to raise the moon's orbit by that 3.8cm/year is
easy.. Just plug into basic orbital mechanics, out pops an answer of some
1.25 * 10^20 joules per year, easier to visualise as a constant power input
of some 4 terawatt (4 * 10^12 watt)

The problem is that this is indeed the *resultant* of all forces
applicable. Just how big are the other forces working on the moon,like
frictional drag against the not-quite-empty vacuum of local space.(keep in
mind the moon orbits earth at about 1km/sec, and the earth whizzes around
the sun at about 30km/sec)?


Terrific feed back, at least 4 terawatts plus something for drag,
thanks much.

BTW; here's something I picked up off the net:

http://www.europhysicsnews.com/full/.../article3.html
Hydrodynamics of planetary nebulae: 10e10 atoms per m3.
and of much lower density, typically 10e7 atoms per m3.

Perhaps the friction aspect can be roughly derived, as there's got to
be some worth in that much surface being driven through at the 1.025
km/s.

The part about keeping up with Earth I'd think is a given, as pushing
ahead (into the solar wind) should be fully counteracted by the exact
opposit of traveling down-wind while being taken along by the momentum
of Earth's traveling about the sun.
  #7  
Old November 19th 03, 12:08 AM
Brad Guth
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

Obviously there's tidal gravitational forces continuously at work, as
otherwise something as massive as the Earth/Moon union would become
one.

Since I've learned from Marvin ) that lunar
recession is presumably taking roughly 4 terawatts/year, while I'm
thoroughly confused about calculating something other upon lunar drag
into similar terms of continuous energy requirement, I've decided to
do a little shock and awe reverse engineering, based upon yet another
of my village idiot guestimates of lunar drag consuming another 1e12
W, a terawatt worth of tidal forces that's having only to offset for
drag in order to sustain the lunar orbit speed at zero recession.

If we were to allot said 1e12 (terawatt) to represent the necessary
energy to overcome a lunar drag coefficient.

1e12/38e12 = 26.3 mw/m2

If we were to further suggest that the space weather environment in
which the moon travels about Earth is representing but 1/1000th that
viscosity and kinetic energy of the ISS orbit.

If we were use an ISS surface reference area of 1000 m2, as well as
the 1000 times greater atmosphere as multiplier factors:

26.3e-3 * 1e3 * 1e3 = 26.3 kw (in other words 35.8 hp)

Thus it's taking 26.3 kw worth of continuous energy applied in order
to sustain the likes of ISS in a fixed (non-recession) orbit, in other
words of overcoming friction as that based upon the consideration of
the ISS environment being of 1e3 greater viscosity than lunar space,
as well as for having to travel roughly 11 fold faster than the moon,
which may actually compute as an overall 11e3:1.

Obviously if lunar space were merely 100 times thinner, then the
reverse engineering calculations for ISS compensation drops to a mere
2.6 kw.

Perhaps 26 kw is simply too much continuous energy applied for ISS
but, I'd have to bet that 2.6 kw is not sufficient, thereby the
atmosphere and/or space weather difference may actually become 400~500
times thinner for that of the moon traveling through at 1.025 km/s.

Of course absolutely none of this is sufficiently correct but, at
least it gives myself something to go along with those 4 terawatts
worth of recession energy, making the overall tally for continuous
gravitational tidal forces per year applied onto the moon as
representing perhaps 5 terawatts.

4e12 + 1e12 = 5e12 W

If there's now 5e12 W to draw upon (disregarding solar PV conversions
and/or induced solar plasma electrons and/or EMF factors taken from
solar weather as well as from our Van Allen zone of death, all of
which should actually be worth quite a great deal, so much so that it
seems exactly the sort of tether dipole extraction potential that I'm
thinking can be safely applied into those massive counter-rotating
flywheels situated at the ME-L1 (gravity-well null).

Just for a little further shock and awe argument sake, lets say that
the additional energy input (besides the overall recession energy)
provides us with one additional terawatt resource, now all toll were
looking at 6 terawatts, whereas tapping into 50% of that energy yields
3 terawatts, enough energy to run 20 of those 100 GW laser cannons
plus another terawatt to spare. Obviously only 2 of those terawatts
are those being extracted from the recession energy.

5e12 + 1e12 = 6e12 * 0.5 = 3e12 W extracted

We certainly can't take away everything, as that could reverse things
by pulling the moon into Earth, a seriously bad sort of thing to be
doing. Although, if the bulk of energy taken is what's converted into
laser cannon energy focused upon relatively small portions of Earth
(not that Earth isn't already getting a little too hot under the
collar), say quadrant targets zones of as little as 1 km diameter
along with a 10 km safety buffer zone, at least some of this extracted
energy should return itself as a slight repulsion factor.

As we manage to run ourselves out of natural petroleum as well as
other natural resources and remain too dumbfounded to safely utilize
nuclear energy (like those smart ass French have been doing for
decades), and way too energy inefficient to rely upon wind and ocean
tidal resources (too polluted and otherwise too greenhouse for solar
PV), thereby having insufficient energy for exterminating the
remaining populations we don't happen to like; By properly using laser
cannons of either near UV or near IR, or perhaps both spectrums so
that at least humans aren't blinded while the least amount of thermal
energy is contributed upon Earth (this is actually where I'd favor the
near IR [750~800 nm] that shouldn't blind the nocturnal species), this
tactic being where those multiple 100 GW cannons can sort of light
your fire from afar, just might do the trick.

If those 20 or so 100 GW laser cannon beams are being efficiently
transmitted and converted, I'd tend to think the overall input/output
conversion that's obtained on Earth might eventually reach 25%, thus 2
terawatts of input becomes 500 GW, which isn't all that bad for the
hundreds of billions it may take to pull that one off.

Regards, Brad Guth http://guthvenus.tripod.com
  #8  
Old November 19th 03, 01:16 AM
[email protected] \(formerly\)
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

Dear Brad Guth:

"Brad Guth" wrote in message
om...
Obviously there's tidal gravitational forces continuously at work, as
otherwise something as massive as the Earth/Moon union would become
one.


A real sidelight, but I don't think they would "become one". I think that
in the existing system, there is enough angular momentum that there is no
way a single "partially liquid" body wouldn't spin out one or more lobes.
You might jam the two together, but I don't think they'd stay together.

David A. Smith


  #9  
Old November 19th 03, 09:57 PM
Brad Guth
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

\(formerly\)" [email protected] wrote in message news:[email protected]
Dear Brad Guth:

"Brad Guth" wrote in message
om...
Obviously there's tidal gravitational forces continuously at work, as
otherwise something as massive as the Earth/Moon union would become
one.


A real sidelight, but I don't think they would "become one". I think that
in the existing system, there is enough angular momentum that there is no
way a single "partially liquid" body wouldn't spin out one or more lobes.
You might jam the two together, but I don't think they'd stay together.

David A. Smith


That's certainly a good thing to know about.

Any better notion upon my village idiot efforts at obtaining the
kinetic energy requirement for just sustaining the lunar orbit?

My guestimate of adding in another terawatt of continuous energy, as a
26 mw/m2 seems somewhat reasonable, though what do I know?


Perhaps this following context is just the tip of this tether dipole
iceberg.

Obviously there's tidal gravitational forces continuously at work, as
otherwise something as massive as the Earth/Moon union would become
one.

Since I've learned from Marvin ) that lunar
recession is presumably taking roughly 4 terawatts/year, while I'm
thoroughly confused about calculating something other upon lunar drag
into similar terms of continuous energy requirement, I've decided to
do a little shock and awe reverse engineering, based upon yet another
of my village idiot guestimates of lunar drag consuming another 1e12
W, a terawatt worth of tidal forces that's having only to offset for
drag in order to sustain the lunar orbit speed at zero recession.

If we were to allot said 1e12 (terawatt) to represent the necessary
energy to overcome a lunar drag coefficient.

1e12/38e12 = 26.3 mw/m2

If we were to further suggest that the space weather environment in
which the moon travels about Earth is representing but 1/1000th that
viscosity and kinetic energy of the ISS orbit.

If we were use an ISS surface reference area of 1000 m2, as well as
the 1000 times greater atmosphere as multiplier factors:

26.3e-3 * 1e3 * 1e3 = 26.3 kw (in other words 35.8 hp)

Thus it's taking 26.3 kw worth of continuous energy applied in order
to sustain the likes of ISS in a fixed (non-recession) orbit, in other
words of overcoming friction as that based upon the consideration of
the ISS environment being of 1e3 greater viscosity than lunar space,
as well as for having to travel roughly 11 fold faster than the moon,
which may actually compute as an overall 11e3:1.

Obviously if lunar space were merely 100 times thinner, then the
reverse engineering calculations for ISS compensation drops to a mere
2.6 kw.

Perhaps 26 kw is simply too much continuous energy applied for ISS
but, I'd have to bet that 2.6 kw is not sufficient, thereby the
atmosphere and/or space weather difference may actually become 400~500
times thinner for that of the moon traveling through at 1.025 km/s.

Of course absolutely none of this is sufficiently correct but, at
least it gives myself something to go along with those 4 terawatts
worth of recession energy, making the overall tally for continuous
gravitational tidal forces per year applied onto the moon as
representing perhaps 5 terawatts.

4e12 + 1e12 = 5e12 W

If there's now 5e12 W to draw upon (disregarding solar PV conversions
and/or induced solar plasma electrons and/or EMF factors taken from
solar weather as well as from our Van Allen zone of death, all of
which should actually be worth quite a great deal, so much so that it
seems exactly the sort of tether dipole extraction potential that I'm
thinking can be safely applied into those massive counter-rotating
flywheels situated at the ME-L1 (gravity-well null).

Just for a little further shock and awe argument sake, lets say that
the additional energy input (besides the overall recession energy)
provides us with one additional terawatt resource, now all toll were
looking at 6 terawatts, whereas tapping into 50% of that energy yields
3 terawatts, enough energy to run 20 of those 100 GW laser cannons
plus another terawatt to spare. Obviously only 2 of those terawatts
are those being extracted from the recession energy.

5e12 + 1e12 = 6e12 * 0.5 = 3e12 W extracted

We certainly can't take away everything, as that could reverse things
by pulling the moon into Earth, a seriously bad sort of thing to be
doing. Although, if the bulk of energy taken is what's converted into
laser cannon energy focused upon relatively small portions of Earth
(not that Earth isn't already getting a little too hot under the
collar), say quadrant targets zones of as little as 1 km diameter
along with a 10 km safety buffer zone, at least some of this extracted
energy should return itself as a slight repulsion factor.

As we manage to run ourselves out of natural petroleum as well as most
other natural resources and remain too dumbfounded to safely utilize
nuclear energy (like those smart ass French have been doing for
decades), and way too energy inefficient to rely upon wind and ocean
tidal resources (too polluted and otherwise too greenhouse for solar
PV), thereby having insufficient energy for exterminating the
remaining populations we don't happen to like; By properly using laser
cannons of either near UV or near IR, or perhaps both spectrums so
that at least humans aren't blinded while the least amount of thermal
energy is contributed upon Earth (this is actually where I'd favor the
near IR [750~800 nm] that shouldn't blind the nocturnal species), this
tactic being where those multiple 100 GW cannons can sort of light
your fire from afar, just might do the trick.

If those 20 or so 100 GW laser cannon beams are being efficiently
transmitted and converted, I'd tend to think the overall input/output
conversion that's obtained on Earth might eventually reach 25%, thus 2
terawatts of input becomes 500 GW, which isn't all that bad for the
hundreds of billions it may take to pull that one off.


For the continuing entertainment sake, plus on behalf of those coming
into this topic without benefit of nearly three years worth of my
efforts, I've added another infomercial page (GV-LM-1) and edited upon
a couple of others.

If I were to be suggesting upon the sorts of wild and crazy things, as
if this is what makes your life worth living, especially if those were
to be of the sorts of horrifically spendy and somewhat lethal agendas
like the Mars or bust and ESE fiasco, in that case I've got lots other
to say about utilizing the moon as well as Venus.
http://guthvenus.tripod.com/gv-lm-1.htm
http://guthvenus.tripod.com/gv-cm-ccm-01.htm
http://guthvenus.tripod.com/space-radiation-103.htm
http://guthvenus.tripod.com/earth-moon-energy.htm
http://guthvenus.tripod.com/gv-lse-energy.htm
http://guthvenus.tripod.com/vl2-iss-joke.htm
http://guthvenus.tripod.com/gv-illumination.htm
http://guthvenus.tripod.com/moon-sar.htm
http://guthvenus.tripod.com/gv-town.htm
plus a few dozen other pages, with more on their way.
  #10  
Old November 20th 03, 12:14 AM
[email protected] \(formerly\)
external usenet poster
 
Posts: n/a
Default Is the moon leaving, or are we shrinking by 38 mm/year

Dear Brad Guth:

"Brad Guth" wrote in message
om...
\(formerly\)" [email protected] wrote in message

news:[email protected]
Dear Brad Guth:

"Brad Guth" wrote in message
om...
Obviously there's tidal gravitational forces continuously at work, as
otherwise something as massive as the Earth/Moon union would become
one.


A real sidelight, but I don't think they would "become one". I think

that
in the existing system, there is enough angular momentum that there is

no
way a single "partially liquid" body wouldn't spin out one or more

lobes.
You might jam the two together, but I don't think they'd stay together.


That's certainly a good thing to know about.

Any better notion upon my village idiot efforts at obtaining the
kinetic energy requirement for just sustaining the lunar orbit?


Zero additional energy should do OK. The solar wind boosts one side, and
retards the other. The earth would have to freeze solid for all eternity,
of course. All the way down to the core. Might as well tidally lock it as
well.

Otherwise, it'll just continue to gain angular momentum from the Earth.

Should pretty much stop the recession...

David A. Smith


 




Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Sedna, space probes?, colonies? what's next? TKalbfus Policy 265 July 13th 04 12:00 AM
The New NASA Mission Has Been Grossly Mischaracterized. Dan Hanson Policy 25 January 26th 04 07:42 PM
NEWS: The allure of an outpost on the Moon Kent Betts Space Shuttle 2 January 15th 04 12:56 AM
We choose to go to the Moon? Brian Gaff Space Shuttle 49 December 10th 03 10:14 AM


All times are GMT +1. The time now is 07:53 AM.


Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2019, Jelsoft Enterprises Ltd.
Copyright 2004-2019 SpaceBanter.com.
The comments are property of their posters.