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  #21  
Old May 13th 09, 03:25 PM posted to sci.astro.amateur
Dr J R Stockton[_29_]
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Default Celestial sphere image?

In sci.astro.amateur message
, Tue, 12 May 2009 18:27:52, Dave Typinski

posted:

A proof closer to home is to examine the relationship of satellite
orbit altitude and orbit period.

Geostationary satellites live at an orbit radius of 42,165 km, plus or
minus a couple kilometers. Given Earth's mass and that orbit radius
and Newton's laws of motion and gravitation, they orbit in one
sidereal day plus or minus a few seconds.

To do a full orbit in one solar day, they'd have to be either a)
higher by about 80 km or b) the Earth would have to be lighter by
about 3x10^19 metric tons.

Since they aren't higher and Earth isn't lighter, they don't orbit in
one solar day, but in a sidereal day. Since they're geostationary,
the Earth itself must complete one rotatation in one sidereal day, not
in one solar day. QED.


But where does the sufficiently-exact figure for the Earth's mass come
from? Does it not come from measurements of orbit times and sizes,
making your argument circular?

--
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  #22  
Old May 13th 09, 08:33 PM posted to sci.astro.amateur
Dave Typinski[_3_]
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Default Celestial sphere image?

Dr J R Stockton wrote:

In sci.astro.amateur message
, Tue, 12 May 2009 18:27:52, Dave Typinski

posted:

A proof closer to home is to examine the relationship of satellite
orbit altitude and orbit period.

Geostationary satellites live at an orbit radius of 42,165 km, plus or
minus a couple kilometers. Given Earth's mass and that orbit radius
and Newton's laws of motion and gravitation, they orbit in one
sidereal day plus or minus a few seconds.

To do a full orbit in one solar day, they'd have to be either a)
higher by about 80 km or b) the Earth would have to be lighter by
about 3x10^19 metric tons.

Since they aren't higher and Earth isn't lighter, they don't orbit in
one solar day, but in a sidereal day. Since they're geostationary,
the Earth itself must complete one rotatation in one sidereal day, not
in one solar day. QED.


But where does the sufficiently-exact figure for the Earth's mass come
from? Does it not come from measurements of orbit times and sizes,
making your argument circular?


I'm glad you brought that up, JR, because I was originally reticent to
post that message for that very reason.

So I did some looking, and lo and behold, it turned out that it isn't
circular resoning after all.

Yes, you calculate Earth's mass using satellite orbit data, but you
also use big G calculated in an entirely different fashion.

First, calculate big G by measuring the acceleration between test
objects in a torsion balance. This is the key step that breaks the
circle.

Second, measure the orbit altitude and orbit period of a satellite.

Third, using your new big G number, calculate the Earth mass required
to produce the measured orbit altitude and period.

Gundlach and Merkowitz at UWash did just that a while ago.

http://asd.gsfc.nasa.gov/Stephen.Merkowitz/G/Big_G.html

Their refinement of big G resulted in an uncertainty of 8x10^16 metric
tons in Earth's mass. That's three orders of magnitude smaller than
the accuracy required to show the difference between solar and
sidereal orbit periods.

After figuring that out, I felt somewhat more confortable posting my
explanation. ;-)
--
Dave
  #23  
Old May 14th 09, 12:53 AM posted to sci.astro.amateur
Quadibloc
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Default Celestial sphere image?

On May 13, 1:33*pm, Dave Typinski wrote:

Third, using your new big G number, calculate the Earth mass required
to produce the measured orbit altitude and period.


And then you compare it with a known value for Earth's mass,
determined by core samples of the Earth's crust, mantle, and core? I
didn't think we had the latter two yet.

Actually, though, you _are_ right that the reasoning is not circular,
but not for the reasons you've outlined.

Think of satellites in polar orbit.

John Savard
  #24  
Old May 14th 09, 02:43 AM posted to sci.astro.amateur
Dave Typinski[_3_]
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Default Celestial sphere image?

Quadibloc wrote:

On May 13, 1:33*pm, Dave Typinski wrote:

Third, using your new big G number, calculate the Earth mass required
to produce the measured orbit altitude and period.


And then you compare it with a known value for Earth's mass,


Why compare it to anything? If you have the local acceleration due to
gravity, and you have G, you can calculate the mass required to set up
a gravitational potential field necessary to cause the acceleration
you measured.

determined by core samples of the Earth's crust, mantle, and core? I
didn't think we had the latter two yet.


Inferred from seismic analyses, aren't they? Whether they're known or
not, I don't understand why this would matter. The distribution of
mass within the Earth will make a difference to the gravitational
potential field interior to Earth's surface. External to the surface,
it's a point mass with corrections for centripetal acceleration and
surface altiutude deviation from an ideal ellipsoid.

Or am I missing your point entirely?

Actually, though, you _are_ right that the reasoning is not circular,
but not for the reasons you've outlined.

Think of satellites in polar orbit.


You've stumped me, John. How does changing the orbit inclination
simplify things make the reasoning non-circular?
--
Dave
  #25  
Old May 14th 09, 06:43 AM posted to sci.astro.amateur
oriel36[_2_]
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Default Celestial sphere image?

On May 14, 2:43*am, Dave Typinski wrote:


determined by core samples of the Earth's crust, mantle, and core? I
didn't think we had the latter two yet.


Inferred from seismic analyses, aren't they? *Whether they're known or
not, I don't understand why this would matter. *The distribution of
mass within the Earth will make a difference to the gravitational
potential field interior to Earth's surface. *External to the surface,
it's a point mass with corrections for centripetal acceleration and
surface altiutude deviation from an ideal ellipsoid.


Dave


Point mass indeed !,dynamicists have been working with spherical
deviation and fluid dynamics in stars for years but can't make the
leap to applying it to the Earth's viscous interior even though
surface features such as the MAR contain signatures of rotational
dynamics and specifically differential rotation.

Oh ,that's right,you have trouble understanding my sentences so I am
going to make a special effort to translate it for you in the most
simple way - the Earth is round and rotating and because the Earth is
mainly a viscous rotating composition,it will not be pefectly round.

I would not want to overload your brain with the next step which is
applying the same dynamic to crustal evolution and motion which
replaces stationary Earth 'convection cells' (which require no
reference to rotation or rotational orientation).





  #26  
Old May 14th 09, 07:18 AM posted to sci.astro.amateur
Quadibloc
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Posts: 7,018
Default Celestial sphere image?

On May 13, 7:43*pm, Dave Typinski wrote:

You've stumped me, John. *How does changing the orbit inclination
simplify things make the reasoning non-circular?


You're right that we don't need the Earth's mass to know the
acceleration of gravity at the satellite's altitude. But that makes
the Earth's mass - and G - a red herring.

If the question is that we don't know whether the Earth's axial
rotation period is 24 hours or 23 hours, 56 minutes, and 4 seconds,
then one satellite in orbit at one altitude would be compatible with
either, given different values of the Earth's mass.

Comparing the satellite with the acceleration of gravity on the ground
will settle that. Comparing the periods of satellites at different
heights will settle that, since only in a unique frame of reference
will Kepler's square/cube law be followed.

A satellite in polar orbit is a nice case for comparison, because
there is agreement that the Earth does not rotate with a pole
somewhere on the Equator, so the period of such a satellite at a given
altitude would be the same as one in an equatorial orbit relative to a
nonrotating Earth, from which the period of the Earth's rotation is
simple to determine.

John Savard
  #27  
Old May 14th 09, 01:50 PM posted to sci.astro.amateur
Dave Typinski[_3_]
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Default Celestial sphere image?

Quadibloc wrote:

On May 13, 7:43*pm, Dave Typinski wrote:

You've stumped me, John. *How does changing the orbit inclination
simplify things make the reasoning non-circular?


You're right that we don't need the Earth's mass to know the
acceleration of gravity at the satellite's altitude. But that makes
the Earth's mass - and G - a red herring.


In this case, the mass of the Earth is calculated from big G and the
satellite's orbit parameters. Big G comes from analysis of a torsion
balance experiment and the orbit altitude and period are measured
directly. What's red herringish about that?

If the question is that we don't know whether the Earth's axial
rotation period is 24 hours or 23 hours, 56 minutes, and 4 seconds,
then one satellite in orbit at one altitude would be compatible with
either, given different values of the Earth's mass.


Okay, I think I see your objection now.

The satellite needn't be geostationary or polar. One just has to know
its orbit period and altitude.

You're right, of course: a polar satellite removes the possibly
confusing parameter of Earth's rotation from the orbit period
measurement. It crosses the equator when it crosses and that's that.

Comparing the satellite with the acceleration of gravity on the ground
will settle that. Comparing the periods of satellites at different
heights will settle that, since only in a unique frame of reference
will Kepler's square/cube law be followed.


Yep and yep.

A satellite in polar orbit is a nice case for comparison, because
there is agreement that the Earth does not rotate with a pole
somewhere on the Equator, so the period of such a satellite at a given
altitude would be the same as one in an equatorial orbit relative to a
nonrotating Earth, from which the period of the Earth's rotation is
simple to determine.


Yep.
--
Dave
 




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