[meteorite-list] More fun with GR

From: MexicoDoug <mexicodoug_at_meteoritecentral.com>
Date: Thu, 21 Jul 2016 12:56:40 -0400
Message-ID: <1560e63405f-ec9-12bf_at_webprd-m67.mail.aol.com>

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 <falcon99 at sbcglobal.net>
To: Matson, Rob D. <ROBERT.D.MATSON at leidos.com>
Cc: MexicoDoug <mexicodoug at aol.com>; meteorite-list <meteorite-list at 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 at 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 [mexicodoug at aol.com]
Sent: Wednesday, July 20, 2016 4:03 PM
To: Matson, Rob D.; meteorite-list at 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 at meteoritecentral.com>
To: meteorite-list <meteorite-list at 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:meteorite-list-bounces at 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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Received on Thu 21 Jul 2016 12:56:40 PM PDT


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