Sun and moon distance questions - Feb. 3, 2014

SUN, MOON, AND EARTH DATA

Posted by E.D.G. February 3, 2014

A somewhat informal earthquake triggering forces research project is
getting underway by several researchers including me. The number of people
involved might gradually expand.

Attempts are being made to obtain several types of data and equations
such as the ones listed here. Those data can be found elsewhere or
generated. But it is probably best to ask for this information from people
who do this work on a regular basis so that there is some continuity with
the effort.


Data Already Available

We have already developed a computer program that provides us with:

--- The distance between the center of the sun and the center of the Earth
for any input time.

--- The distance between the center of the moon and the center of the Earth
for any input time.

--- The subsolar location for any input time. If a line is drawn between
the center of the sun and center of the Earth then what we regard as the
subsolar location is the latitude and longitude of the location where that
line crosses the surface of the Earth.

--- The sublunar location for any input time. If a line is drawn between
the center of the moon and center of the Earth then what we regard as the
sublunar location is the latitude and longitude of the location where that
line crosses the surface of the Earth.


The strength of the sun and moon gravities as they are felt at the
location of specific earthquake fault zones at specific times is needed.
And to help determine that we would like to know what the distance is
between the center of the sun and the fault zone and also the distance
between the center of the moon and the fault zone at specific times.


Question 1 - The average sun center to Earth center distance

To start with we would like to know what people regard as the value
of the distance between the center of the sun and the center of the Earth
for the time when the sun gravity pull on the Earth is considered to have a
value of 1 for calculation purposes.

With that information we can then determine using mass equations etc.
what the sun and moon gravity pulls on the Earth are relative to one another
at different times of the year.

I am aware that the sun - Earth and moon - Earth distances and their
subsolar and sublunar locations change throughout the year. And we could
actually select any sun - Earth distance we wanted to start with for these
types of calculations. But as I said, we would like to stay consistent with
other researchers on this.


Question 2 - What actual sun, moon, and Earth masses in kilograms (or
whatever) do people prefer to use for those equations


Question 3 - Sun and moon centers to earthquake fault zone distances

There might be some people who have the following information readily
available. If not then I know how to do the calculations myself. But it
would probably take a few days as I am not a math wizard.

As stated before, we have those sun - Earth distance, moon - Earth
distance, and subsolar and sublunar latitude and longitude data to work
with. And they are actually enough.

Since we want to know what the sun and moon gravity strengths are as
felt at specific earthquake fault zones at specific times, we first need to
know what the sun center to earthquake fault zone location and moon center
to earthquake fault zone location distances are for specific times. And
from those distance data and the right equations we can generate gravity
strength numbers.

If anyone knows where those types of distance equations can be found
I would be interested in learning that. Otherwise it will be a matter of
just working them out.

What I myself have been doing for this type of effort is to view the
Earth as being stationary in space. 0 latitude and longitude is directly
ahead. 90 N latitude is at the top of the image. 90 S latitude is at the
bottom. 90 W longitude and 0 latitude is at the left of the image. And 90
E longitude and 0 latitude is at the right.

Then in that imaginary image, the subsolar, sublunar, and earthquake
fault zone locations are pictured as points on the surface of the Earth.
And their x, y, and z components are calculated and merged with the original
sun and moon distance data so that new distances and various latitudes and
longitudes can be obtained.


Additional Comment Regard Earthquake Triggering

We already know why earthquakes occur. And data that I have
collected over the years indicate to me that at least some of our more
powerful earthquakes are being triggered at specific times by forces related
to the gravitational pulls of the sun and the moon on the Earth's crust.
However, whether or not this is really the case or if they are simply random
events is a debate that can be held at some other time. Right now we are
only interested in those sun, moon, and Earth mass, distance, and gravity
data.

Regards to all and thanks for any assistance.

E.D.G.

On Monday, February 3, 2014 7:02:50 AM UTC-8, E.D.G. wrote:
SUN, MOON, AND EARTH DATA



Posted by E.D.G. February 3, 2014



A somewhat informal earthquake triggering forces research project is

getting underway by several researchers including me. The number of people

involved might gradually expand.



Attempts are being made to obtain several types of data and equations

such as the ones listed here. Those data can be found elsewhere or

generated. But it is probably best to ask for this information from people

who do this work on a regular basis so that there is some continuity with

the effort.





Data Already Available



We have already developed a computer program that provides us with:



--- The distance between the center of the sun and the center of the Earth

for any input time.



--- The distance between the center of the moon and the center of the Earth

for any input time.



--- The subsolar location for any input time. If a line is drawn between

the center of the sun and center of the Earth then what we regard as the

subsolar location is the latitude and longitude of the location where that

line crosses the surface of the Earth.



--- The sublunar location for any input time. If a line is drawn between

the center of the moon and center of the Earth then what we regard as the

sublunar location is the latitude and longitude of the location where that

line crosses the surface of the Earth.





The strength of the sun and moon gravities as they are felt at the

location of specific earthquake fault zones at specific times is needed.

And to help determine that we would like to know what the distance is

between the center of the sun and the fault zone and also the distance

between the center of the moon and the fault zone at specific times.





Question 1 - The average sun center to Earth center distance



To start with we would like to know what people regard as the value

of the distance between the center of the sun and the center of the Earth

for the time when the sun gravity pull on the Earth is considered to have a

value of 1 for calculation purposes.



With that information we can then determine using mass equations etc.

what the sun and moon gravity pulls on the Earth are relative to one another

at different times of the year.



I am aware that the sun - Earth and moon - Earth distances and their

subsolar and sublunar locations change throughout the year. And we could

actually select any sun - Earth distance we wanted to start with for these

types of calculations. But as I said, we would like to stay consistent with

other researchers on this.





Question 2 - What actual sun, moon, and Earth masses in kilograms (or

whatever) do people prefer to use for those equations





Question 3 - Sun and moon centers to earthquake fault zone distances



There might be some people who have the following information readily

available. If not then I know how to do the calculations myself. But it

would probably take a few days as I am not a math wizard.



As stated before, we have those sun - Earth distance, moon - Earth

distance, and subsolar and sublunar latitude and longitude data to work

with. And they are actually enough.



Since we want to know what the sun and moon gravity strengths are as

felt at specific earthquake fault zones at specific times, we first need to

know what the sun center to earthquake fault zone location and moon center

to earthquake fault zone location distances are for specific times. And

from those distance data and the right equations we can generate gravity

strength numbers.



If anyone knows where those types of distance equations can be found

I would be interested in learning that. Otherwise it will be a matter of

just working them out.



What I myself have been doing for this type of effort is to view the

Earth as being stationary in space. 0 latitude and longitude is directly

ahead. 90 N latitude is at the top of the image. 90 S latitude is at the

bottom. 90 W longitude and 0 latitude is at the left of the image. And 90

E longitude and 0 latitude is at the right.



Then in that imaginary image, the subsolar, sublunar, and earthquake

fault zone locations are pictured as points on the surface of the Earth.

And their x, y, and z components are calculated and merged with the original

sun and moon distance data so that new distances and various latitudes and

longitudes can be obtained.





Additional Comment Regard Earthquake Triggering



We already know why earthquakes occur. And data that I have

collected over the years indicate to me that at least some of our more

powerful earthquakes are being triggered at specific times by forces related

to the gravitational pulls of the sun and the moon on the Earth's crust.

However, whether or not this is really the case or if they are simply random

events is a debate that can be held at some other time. Right now we are

only interested in those sun, moon, and Earth mass, distance, and gravity

data.



Regards to all and thanks for any assistance.


E.D.G.

What is the total volumetric sphere of tidal morphing energy applied by our moon?

In other words, if we were to fully simulate the tidal morphing of what our moon does to the whole volume-metric mass and lithosphere of our planet (not just of its oceans), how much applied energy would that require?

"Brad Guth" wrote in message
...

What is the total volumetric sphere of tidal morphing energy applied by
our moon?

In other words, if we were to fully simulate the tidal morphing of what
our moon does to the whole volume-metric mass and lithosphere of our
planet (not just of its oceans), how much applied energy would that
require?

Posted by E.D.G. February 5, 2014

There might be some people who have calculated that if I understand
your question correctly.

The process to determine how much energy is involved would I expect
be fairly complex.

As the moon rotates around the Earth it affects ocean height (tides).
And it affects solid mass (Solid Earth Tide).

And if that were all that was involved then changes in the speed of
the moon might be used to determine how much moon rotational energy was lost
each year.

However, the rotation of the Earth on its axis relative to the moon
is the main contributor to the ocean and Solid Earth tides. And that effect
actually causes the moon to spin faster in its orbit every year. Gravity
associated with the ocean tide bulges "drag" the moon along.

At some point in the very distant future, in theory, the moon is
supposed to rotate so fast that it will fly away into space. And we will
have to stop singing all of those songs that have the word "moon" in them
(humor intended).

That Earth rotation and tide effect would, I think, make any moon
energy calculations a lot more complex.

These are theories as I understand them.

On Wednesday, February 5, 2014 5:15:40 AM UTC-8, E.D.G. wrote:
"Brad Guth" wrote in message

...



What is the total volumetric sphere of tidal morphing energy applied by

our moon?



In other words, if we were to fully simulate the tidal morphing of what

our moon does to the whole volume-metric mass and lithosphere of our

planet (not just of its oceans), how much applied energy would that

require?



Posted by E.D.G. February 5, 2014



There might be some people who have calculated that if I understand

your question correctly.



The process to determine how much energy is involved would I expect

be fairly complex.



As the moon rotates around the Earth it affects ocean height (tides).

And it affects solid mass (Solid Earth Tide).



And if that were all that was involved then changes in the speed of

the moon might be used to determine how much moon rotational energy was lost

each year.



However, the rotation of the Earth on its axis relative to the moon

is the main contributor to the ocean and Solid Earth tides. And that effect

actually causes the moon to spin faster in its orbit every year. Gravity

associated with the ocean tide bulges "drag" the moon along.



At some point in the very distant future, in theory, the moon is

supposed to rotate so fast that it will fly away into space. And we will

have to stop singing all of those songs that have the word "moon" in them

(humor intended).



That Earth rotation and tide effect would, I think, make any moon
energy calculations a lot more complex.


These are theories as I understand them.

It would likely take more than 50 TW applied in order to do what our moon constantly does to morphing and/or modulating our whole planet (not just that of its oceans).

The accelerating of our moon away from Earth can also be calculated to within a kw of applied energy, because supposedly the mass of our moon is known to the exact thousand kg or tonne.

William Mook is actually very good at scientific calculations, and understands physics better than most. So, perhaps Mook can be of some use to this topic.

"Brad Guth" wrote in message
...

It would likely take more than 50 TW applied in order to do what our moon
constantly does to morphing and/or modulating our whole planet (not just
that of its oceans).

The accelerating of our moon away from Earth can also be calculated to
within a kw of applied energy, because supposedly the mass of our moon is
known to the exact thousand kg or tonne.

William Mook is actually very good at scientific calculations, and
understands physics better than most. So, perhaps Mook can be of some use
to this topic.

Posted by E.D.G. February 5, 2014

Without thinking too much about this, I suspect that the amount of
energy the moon contributes to the Earth at any point in time could be
calculated from knowing the distance to the moon and its orbital speed.

The moon is constantly accelerating towards the Earth. The ocean
tides are constantly increasing the moon's orbital speed by pulling it
forward. But they are not doing that as much as they might because the moon
already has an orbital speed.

Those things affect both the moon's orbital speed and its distance
from the Earth. So by knowing the the moon's true speed and distance and
comparing them with the theoretical speed and distance, one of the
differences calculated would be related to the amount of energy the moon is
not gaining versus what it should be gaining. And that amount would show
what its effects are on the Earth.

The net result is that the ocean tides are both indirectly affecting
the speed of the Earth and moving energy around on the Earth through their
gravitational interaction with the moon.

Again, that is a theoretical picture based on my understanding of
what takes place.

Someone, somewhere has probably done some calculations regarding the
energies involved.

In article ,
"E.D.G." writes:
However, the rotation of the Earth on its axis relative to the moon
is the main contributor to the ocean and Solid Earth tides. And that effect
actually causes the moon to spin faster in its orbit every year. Gravity
associated with the ocean tide bulges "drag" the moon along.

The basic explanation is correct, but the effect is to raise the
Moon's orbit. This means the Moon's speed in orbit decreases, and
orbit period increases.

At some point in the very distant future, in theory, the moon is
supposed to rotate so fast that it will fly away into space.

I don't think that ever happens. Once the Moon reaches synchronous
rotation (day length = month length), the tidal effect starts
decreasing the Moon's orbit height. My vague memory, though, is that
these effects have longer time scales than solar evolution, the
effects of which will probably overwhelm tidal effects.

For the original problem, you probably want to calculate tidal stress
-- differential of tidal force with position -- rather than tidal
force itself. I'd be surprised if this calculation has not been done
before.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA

In sci.astro message , Mon, 10 Feb 2014
22:19:03, Steve Willner posted:

At some point in the very distant future, in theory, the moon is
supposed to rotate so fast that it will fly away into space.

I don't think that ever happens. Once the Moon reaches synchronous
rotation (day length = month length), the tidal effect starts
decreasing the Moon's orbit height. My vague memory, though, is that
these effects have longer time scales than solar evolution, the
effects of which will probably overwhelm tidal effects.

I thought it was, considering only circular orbits,

(1) The moon's day is locked to the month, so that the front of the Moon
is always in front as we see it;

(2) As the Earth's day is less than the lunar month, the tides on the
Earth have the effect of pushing the Moon forwards, raising the orbit
and further lengthening the lunar month - neglecting everything else,
the Moon will escape to infinity in infinity^N time;

(3) As the Martian day is greater than the Martian months, the tides on
Mars have the effect of retarding the moons, so that they will in finite
times crash on the Martian equator, give or take a bit.

--
(c) John Stockton, nr London, UK. Mail via homepage. Turnpike v6.05 MIME.
Web http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms and links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Mail no News.

On Wednesday, February 5, 2014 12:35:02 PM UTC-8, E.D.G. wrote:
"Brad Guth" wrote in message

...



It would likely take more than 50 TW applied in order to do what our moon

constantly does to morphing and/or modulating our whole planet (not just

that of its oceans).



The accelerating of our moon away from Earth can also be calculated to

within a kw of applied energy, because supposedly the mass of our moon is

known to the exact thousand kg or tonne.



William Mook is actually very good at scientific calculations, and

understands physics better than most. So, perhaps Mook can be of some use

to this topic.



Posted by E.D.G. February 5, 2014



Without thinking too much about this, I suspect that the amount of

energy the moon contributes to the Earth at any point in time could be

calculated from knowing the distance to the moon and its orbital speed.



The moon is constantly accelerating towards the Earth. The ocean

tides are constantly increasing the moon's orbital speed by pulling it

forward. But they are not doing that as much as they might because the moon

already has an orbital speed.



Those things affect both the moon's orbital speed and its distance

from the Earth. So by knowing the the moon's true speed and distance and

comparing them with the theoretical speed and distance, one of the

differences calculated would be related to the amount of energy the moon is

not gaining versus what it should be gaining. And that amount would show

what its effects are on the Earth.



The net result is that the ocean tides are both indirectly affecting

the speed of the Earth and moving energy around on the Earth through their

gravitational interaction with the moon.



Again, that is a theoretical picture based on my understanding of

what takes place.


Someone, somewhere has probably done some calculations regarding the

energies involved.

Modulating the whole planet demands at least tenfold as much energy as for driving ocean tides.

In sci.astro message , Wed, 12 Feb 2014
08:20:26, Bill Owen posted:

On 02/11/14 13:54, Dr J R Stockton wrote:
In sci.astro message , Mon, 10 Feb 2014
22:19:03, Steve Willner posted:

At some point in the very distant future, in theory, the moon is
supposed to rotate so fast that it will fly away into space.

I don't think that ever happens. Once the Moon reaches synchronous
rotation (day length = month length), the tidal effect starts
decreasing the Moon's orbit height. My vague memory, though, is that
these effects have longer time scales than solar evolution, the
effects of which will probably overwhelm tidal effects.

I thought it was, considering only circular orbits,

(1) The moon's day is locked to the month, so that the front of the Moon
is always in front as we see it;

(2) As the Earth's day is less than the lunar month, the tides on the
Earth have the effect of pushing the Moon forwards, raising the orbit
and further lengthening the lunar month - neglecting everything else,
the Moon will escape to infinity in infinity^N time;

(3) As the Martian day is greater than the Martian months, the tides on
Mars have the effect of retarding the moons, so that they will in finite
times crash on the Martian equator, give or take a bit.


(3) is true for Phobos (7.66 hour period) but not for Deimos (30.35 hours).


Agreed. Phobos would crash in 40 million years or so, but it is
probably weak enough to break up first (like Roche Limit, but with
possible material strength). ISTM that if it breaks up into an even
ring, tidal effects will average out and the ring will not crash (so
fast) (perhaps). Deimos drifts outwards.

Thanks.

--
(c) John Stockton, nr London, UK. Mail via homepage. Turnpike v6.05 MIME.
Web http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms and links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.

In article id,
Dr J R Stockton writes:
I thought it was, considering only circular orbits,

(1) The moon's day is locked to the month, so that the front of the Moon
is always in front as we see it;

(2) As the Earth's day is less than the lunar month, the tides on the
Earth have the effect of pushing the Moon forwards, raising the orbit
and further lengthening the lunar month -

Yes, that's observed.

The other effect is that Earth's rotation is slowing, meaning the day
length is increasing. Once the day is longer than the month, the
tidal effect reverses, and the Moon's orbit starts shrinking.

neglecting everything else,
the Moon will escape to infinity in infinity^N time;

Because of the reversal, my understanding is that the Moon never
escapes. But as I wrote earlier, I _think_ all this takes longer
than the ~4 Gyr remaining main-sequence lifetime of the Sun. The
Sun's eventual evolution, in particular mass loss, will probably
control what really happens.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA