Re: [meteorite-list] More fun with GR

[Matson, Rob D. via Meteorite-list]

Hi Doug -- you are very close to the correct altitude of ~3167 km (~1.4965 * 
earth equatorial
radius). I'm now working through the math to figure out the latitude on earth 
where you
age the slowest. ;-)  --Rob

-Original Message-
From: MexicoDoug [mailto:mexicod...@aol.com] 
Sent: Thursday, July 21, 2016 9:57 AM
To: falco...@sbcglobal.net; Matson, Rob D.
Cc: meteorite-list@meteoritecentral.com
Subject: Re: [meteorite-list] More fun with GR

Hi Rob and the other meteoroidal travelers,

I'd say a good mean altitude for government work would be about half of Earth's 
radius, and that ought to smooth out any technicalities to gain an 
understanding of the magnitudes which is what is interestng about the new 
question.

A shortcut to calculate that is to set the free fall velocity (no atmosphere) 
equal to the orbital (tangential) velocity; it avoids the calculus by using the 
velocity derived from the drop in potential from orbit altitude to surface 
level.

v^2 = GM/r'  (orbital)
v^2 = 2GM/r -2GM/r' (gravitational)

If you solve for the altitude simultaneously, r'-r, you get the altitude of 
half again Earth's diameter easily.

Unless there are more Golgafrinchans lurking somewhere in the thread history!

That is a Medium Earth Orbit.  In a perfect universe, 3189 km altitude.  
Nothing special orbit wise, unless you are temporally centric in which case it 
could be called a temporally synchronous orbit, which clearly the universe is 
notvery concerned about as we are ;-)

Kindest wishes
Doug








-Original Message-
From: James Beauchamp <falco...@sbcglobal.net>
To: Matson, Rob D. <robert.d.mat...@leidos.com>
Cc: MexicoDoug <mexicod...@aol.com>; meteorite-list 
<meteorite-list@meteoritecentral.com>
Sent: Thu, Jul 21, 2016 10:31 am
Subject: Re: [meteorite-list] More fun with GR

For the satellite, it varies according to the gravity field it flies over.

Technically none exists because the gravity field is never constant.  It 
dithers.

Sent from my iPhone

On Jul 21, 2016, at 2:01 AM, Matson, Rob D. via Meteorite-list 
<meteorite-list@meteoritecentral.com> wrote:

Hi Doug,

I think you would have come up with the correct answer if I had given a more 
precise value for the clock slow down relative to a stationary clock in deep 
space:  it should be 0.69693 parts per billion relative to a clock at sea-level 
on the earth's equator, or 60.2 microseconds per day. It is no accident that 
the distant rock's velocity would need to be
11.19 km/sec for its clock to remain synchronized with one on the earth's 
equator. That value should be very familiar to meteorite folks. :-)

Here's a harder, but related problem:  at what altitude must a satellite in a 
circular orbit fly for its clock to run at the same speed as a clock on the 
earth's equator?

Another interesting GR factoid:  the core of the earth is actually
2 1/2 years younger than the crust (ignoring convection in the core, plate 
tectonics, etc.) If the earth is modeled as having constant density, the 
calculation works out to about 1 1/2 years younger, but of course earth is much 
denser at the core, resulting in even greater time dilation there.  --Rob 

From: MexicoDoug [mexicod...@aol.com]
Sent: Wednesday, July 20, 2016 4:03 PM
To: Matson, Rob D.; meteorite-list@meteoritecentral.com
Subject: Re: [meteorite-list] age of meteorites

Rob and all,

> For instance, even at solar system escape velocity at earth's distance 
> from the sun (42 km/sec)

What is...The ultimate question of life and the answer to everything?

> Extra-credit question for the mathematically
> inclined:  at what velocity relative to the earth would a meteoroid 
> have to travel to have its clock stay in sync with a clock at the 
> earth's surface?  :-)

Given the figure you mention of 0.6 ppb (52 microseconds per day faster) this 
question asks be nullified, maybe 10 km/s velocity relative to earth?

A good relative velocity to hunt a flock of wild space geese coming to roost on 
Earth, wearing accurate Rolexes ...  But should the meteoroid transition to our 
gravity, the on-board Rolex might abandon its precision for a few spectacular 
minutes, and have an "error" of a couple of nanoseconds ;-)

Kindest wishes
Doug


-Original Message-
From: Matson, Rob D. via Meteorite-list <meteorite-list@meteoritecentral.com>
To: meteorite-list <meteorite-list@meteoritecentral.com>
Sent: Mon, Jul 18, 2016 6:43 pm
Subject: Re: [meteorite-list] age of meteorites

It's not a bad idea, Pete, but unfortunately the time dilation is  really 
minimal unless you get up to a substantial fraction of the speed of light. For 
instance, even at solar system escape velocity at earth's distance from the sun 
(42 km/sec), a meteoroid's clock would be running at about
10 parts per billion slower than that of a stationary rock. (Additional note: 
due to general relativity, a cloc

Re: [meteorite-list] More fun with GR

[MexicoDoug via Meteorite-list]

Hi Rob and the other meteoroidal travelers,

I'd say a good mean altitude for government work would be about half of Earth's 
radius, and that ought to smooth out any technicalities to gain an 
understanding of the magnitudes which is what is interestng about the new 
question.

A shortcut to calculate that is to set the free fall velocity (no atmosphere) 
equal to the orbital (tangential) velocity; it avoids the calculus by using the 
velocity derived from the drop in potential from orbit altitude to surface 
level.

v^2 = GM/r'  (orbital)
v^2 = 2GM/r -2GM/r' (gravitational)

If you solve for the altitude simultaneously, r'-r, you get the altitude of 
half again Earth's diameter easily.

Unless there are more Golgafrinchans lurking somewhere in the thread history!

That is a Medium Earth Orbit.  In a perfect universe, 3189 km altitude.  
Nothing special orbit wise, unless you are temporally centric in which case it 
could be called a temporally synchronous orbit, which clearly the universe is 
notvery concerned about as we are ;-)

Kindest wishes
Doug








-Original Message-
From: James Beauchamp <falco...@sbcglobal.net>
To: Matson, Rob D. <robert.d.mat...@leidos.com>
Cc: MexicoDoug <mexicod...@aol.com>; meteorite-list 
<meteorite-list@meteoritecentral.com>
Sent: Thu, Jul 21, 2016 10:31 am
Subject: Re: [meteorite-list] More fun with GR

For the satellite, it varies according to the gravity field it flies over.

Technically none exists because the gravity field is never constant.  It 
dithers.

Sent from my iPhone

On Jul 21, 2016, at 2:01 AM, Matson, Rob D. via Meteorite-list 
<meteorite-list@meteoritecentral.com> wrote:

Hi Doug,

I think you would have come up with the correct answer if I had given
a more precise value for the clock slow down relative to a stationary
clock in deep space:  it should be 0.69693 parts per billion relative to
a clock at sea-level on the earth's equator, or 60.2 microseconds per
day. It is no accident that the distant rock's velocity would need to be
11.19 km/sec for its clock to remain synchronized with one on the
earth's equator. That value should be very familiar to meteorite folks. :-)

Here's a harder, but related problem:  at what altitude must a
satellite in a circular orbit fly for its clock to run at the same speed
as a clock on the earth's equator?

Another interesting GR factoid:  the core of the earth is actually
2 1/2 years younger than the crust (ignoring convection in the core,
plate tectonics, etc.) If the earth is modeled as having constant density,
the calculation works out to about 1 1/2 years younger, but of course
earth is much denser at the core, resulting in even greater time dilation
there.  --Rob

From: MexicoDoug [mexicod...@aol.com]
Sent: Wednesday, July 20, 2016 4:03 PM
To: Matson, Rob D.; meteorite-list@meteoritecentral.com
Subject: Re: [meteorite-list] age of meteorites

Rob and all,

> For instance, even at solar system escape velocity
> at earth's distance from the sun (42 km/sec)

What is...The ultimate question of life and the answer to everything?

> Extra-credit question for the mathematically
> inclined:  at what velocity relative to the earth
> would a meteoroid have to travel to have its
> clock stay in sync with a clock at the earth's
> surface?  :-)

Given the figure you mention of 0.6 ppb (52 microseconds per day faster) this 
question asks be nullified, maybe 10 km/s velocity relative to earth?

A good relative velocity to hunt a flock of wild space geese coming to roost on 
Earth, wearing accurate Rolexes ...  But should the meteoroid transition to our 
gravity, the on-board Rolex might abandon its precision for a few spectacular 
minutes, and have an "error" of a couple of nanoseconds ;-)

Kindest wishes
Doug


-Original Message-
From: Matson, Rob D. via Meteorite-list <meteorite-list@meteoritecentral.com>
To: meteorite-list <meteorite-list@meteoritecentral.com>
Sent: Mon, Jul 18, 2016 6:43 pm
Subject: Re: [meteorite-list] age of meteorites

It's not a bad idea, Pete, but unfortunately the time dilation is  really 
minimal unless you get up
to a substantial fraction of the speed of light. For instance, even at solar 
system escape velocity
at earth's distance from the sun (42 km/sec), a meteoroid's clock would be 
running at about
10 parts per billion slower than that of a stationary rock. (Additional note: 
due to general relativity,
a clock on a meteoroid would be running about 0.6 parts per billion *faster* 
than a clock at the
earth's surface, but that is more than made up for by the time dilation due to 
special relativity.)

Extra-credit question for the mathematically inclined:  at what velocity 
relative to the earth
would a meteoroid have to travel to have its clock stay in sync with a clock at 
the earth's
surface?  :-) --Rob

-Original Message-
From: Meteorite-list [mailto:m

Re: [meteorite-list] More fun with GR

[James Beauchamp via Meteorite-list]

For the satellite, it varies according to the gravity field it flies over.

Technically none exists because the gravity field is never constant.  It 
dithers.

Sent from my iPhone

On Jul 21, 2016, at 2:01 AM, Matson, Rob D. via Meteorite-list 
 wrote:

Hi Doug,

I think you would have come up with the correct answer if I had given
a more precise value for the clock slow down relative to a stationary
clock in deep space:  it should be 0.69693 parts per billion relative to
a clock at sea-level on the earth's equator, or 60.2 microseconds per
day. It is no accident that the distant rock's velocity would need to be
11.19 km/sec for its clock to remain synchronized with one on the
earth's equator. That value should be very familiar to meteorite folks. :-)

Here's a harder, but related problem:  at what altitude must a
satellite in a circular orbit fly for its clock to run at the same speed
as a clock on the earth's equator?

Another interesting GR factoid:  the core of the earth is actually
2 1/2 years younger than the crust (ignoring convection in the core,
plate tectonics, etc.) If the earth is modeled as having constant density,
the calculation works out to about 1 1/2 years younger, but of course
earth is much denser at the core, resulting in even greater time dilation
there.  --Rob

From: MexicoDoug [mexicod...@aol.com]
Sent: Wednesday, July 20, 2016 4:03 PM
To: Matson, Rob D.; meteorite-list@meteoritecentral.com
Subject: Re: [meteorite-list] age of meteorites

Rob and all,

> For instance, even at solar system escape velocity
> at earth's distance from the sun (42 km/sec)

What is...The ultimate question of life and the answer to everything?

> Extra-credit question for the mathematically
> inclined:  at what velocity relative to the earth
> would a meteoroid have to travel to have its
> clock stay in sync with a clock at the earth's
> surface?  :-)

Given the figure you mention of 0.6 ppb (52 microseconds per day faster) this 
question asks be nullified, maybe 10 km/s velocity relative to earth?

A good relative velocity to hunt a flock of wild space geese coming to roost on 
Earth, wearing accurate Rolexes ...  But should the meteoroid transition to our 
gravity, the on-board Rolex might abandon its precision for a few spectacular 
minutes, and have an "error" of a couple of nanoseconds ;-)

Kindest wishes
Doug


-Original Message-
From: Matson, Rob D. via Meteorite-list 
To: meteorite-list 
Sent: Mon, Jul 18, 2016 6:43 pm
Subject: Re: [meteorite-list] age of meteorites

It's not a bad idea, Pete, but unfortunately the time dilation is  really 
minimal unless you get up
to a substantial fraction of the speed of light. For instance, even at solar 
system escape velocity
at earth's distance from the sun (42 km/sec), a meteoroid's clock would be 
running at about
10 parts per billion slower than that of a stationary rock. (Additional note: 
due to general relativity,
a clock on a meteoroid would be running about 0.6 parts per billion *faster* 
than a clock at the
earth's surface, but that is more than made up for by the time dilation due to 
special relativity.)

Extra-credit question for the mathematically inclined:  at what velocity 
relative to the earth
would a meteoroid have to travel to have its clock stay in sync with a clock at 
the earth's
surface?  :-)  --Rob

-Original Message-
From: Meteorite-list [mailto:meteorite-list-boun...@meteoritecentral.com] On 
Behalf Of Pete Shugar via Meteorite-list
Sent: Monday, July 18, 2016 3:12 PM
To: The List
Subject: [meteorite-list] age of meteorites

greetings to all,
my background is in electronics. everything deals with either C or C2.
Einstein states that nothing goes faster than the speed of light and that as 
you approach the speed of light, things get older slower.
So this meteorite in it's travels is going at a rate that is a subtantual 
percentage of the speed of light. Has anyone taken this into consideration when 
placing an age on the meteorite?
Just a thought to tickle the old brain cells!!
Pete Shugar
__

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[meteorite-list] More fun with GR

[Matson, Rob D. via Meteorite-list]

Hi Doug,

I think you would have come up with the correct answer if I had given
a more precise value for the clock slow down relative to a stationary
clock in deep space:  it should be 0.69693 parts per billion relative to
a clock at sea-level on the earth's equator, or 60.2 microseconds per
day. It is no accident that the distant rock's velocity would need to be
11.19 km/sec for its clock to remain synchronized with one on the
earth's equator. That value should be very familiar to meteorite folks. :-)

Here's a harder, but related problem:  at what altitude must a
satellite in a circular orbit fly for its clock to run at the same speed
as a clock on the earth's equator?

Another interesting GR factoid:  the core of the earth is actually
2 1/2 years younger than the crust (ignoring convection in the core,
plate tectonics, etc.) If the earth is modeled as having constant density,
the calculation works out to about 1 1/2 years younger, but of course
earth is much denser at the core, resulting in even greater time dilation
there.  --Rob

From: MexicoDoug [mexicod...@aol.com]
Sent: Wednesday, July 20, 2016 4:03 PM
To: Matson, Rob D.; meteorite-list@meteoritecentral.com
Subject: Re: [meteorite-list] age of meteorites

Rob and all,

>For instance, even at solar system escape velocity
>at earth's distance from the sun (42 km/sec)

What is...The ultimate question of life and the answer to everything?

>Extra-credit question for the mathematically
>inclined:  at what velocity relative to the earth
>would a meteoroid have to travel to have its
>clock stay in sync with a clock at the earth's
>surface?  :-)

Given the figure you mention of 0.6 ppb (52 microseconds per day faster) this 
question asks be nullified, maybe 10 km/s velocity relative to earth?

A good relative velocity to hunt a flock of wild space geese coming to roost on 
Earth, wearing accurate Rolexes ...  But should the meteoroid transition to our 
gravity, the on-board Rolex might abandon its precision for a few spectacular 
minutes, and have an "error" of a couple of nanoseconds ;-)

Kindest wishes
Doug


-Original Message-
From: Matson, Rob D. via Meteorite-list 
To: meteorite-list 
Sent: Mon, Jul 18, 2016 6:43 pm
Subject: Re: [meteorite-list] age of meteorites

It's not a bad idea, Pete, but unfortunately the time dilation is  really 
minimal unless you get up
to a substantial fraction of the speed of light. For instance, even at solar 
system escape velocity
at earth's distance from the sun (42 km/sec), a meteoroid's clock would be 
running at about
10 parts per billion slower than that of a stationary rock. (Additional note: 
due to general relativity,
a clock on a meteoroid would be running about 0.6 parts per billion *faster* 
than a clock at the
earth's surface, but that is more than made up for by the time dilation due to 
special relativity.)

Extra-credit question for the mathematically inclined:  at what velocity 
relative to the earth
would a meteoroid have to travel to have its clock stay in sync with a clock at 
the earth's
surface?  :-)  --Rob

-Original Message-
From: Meteorite-list [mailto:meteorite-list-boun...@meteoritecentral.com] On 
Behalf Of Pete Shugar via Meteorite-list
Sent: Monday, July 18, 2016 3:12 PM
To: The List
Subject: [meteorite-list] age of meteorites

greetings to all,
my background is in electronics. everything deals with either C or C2.
Einstein states that nothing goes faster than the speed of light and that as 
you approach the speed of light, things get older slower.
So this meteorite in it's travels is going at a rate that is a subtantual 
percentage of the speed of light. Has anyone taken this into consideration when 
placing an age on the meteorite?
Just a thought to tickle the old brain cells!!
Pete Shugar
__

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Archives at http://www.meteorite-list-archives.com
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