Earth, Sun and Vega.

Hello,
I'm trying to calculate something and I wonder if someone can
help me.
The Earth travels around the Sun and the Sun is travelling towards
Vega (not quite, so I'm told?). I'd like to know at what moment (which
day in the solar year) the Earth is moving in the same direction as
the Sun, towards Vega. Also, as the Earth's mouvement is not in the
same plane as that of the Sun, what is the angle(a) between these two
planes.
What I want to find out is the Earth's maximum absolute speed. This
would be 30km/s (the Earth's speed around the Sun) plus 20km/s (the
Sun's speed towards Vega) multiplied by cos(a). But these two speeds
only add up at a precise time of year, which I haven't managed to work
out.
Hoping to hear from you. Regards, Ken

On 4 Sep 2004 07:21:28 -0700, (kenneth
couesbouc) wrote:

I'm trying to calculate something and I wonder if someone can
help me.
The Earth travels around the Sun and the Sun is travelling towards
Vega (not quite, so I'm told?). I'd like to know at what moment (which
day in the solar year) the Earth is moving in the same direction as
the Sun, towards Vega. Also, as the Earth's mouvement is not in the
same plane as that of the Sun, what is the angle(a) between these two
planes.
What I want to find out is the Earth's maximum absolute speed. This
would be 30km/s (the Earth's speed around the Sun) plus 20km/s (the
Sun's speed towards Vega) multiplied by cos(a). But these two speeds
only add up at a precise time of year, which I haven't managed to work
out.

The answer the first question is "never" because the motion of the sun
through the galaxy is roughly in the direction of the constellation
Hercules, which is not in the plane of the ecliptic.

The answer to the second question is to add the vectors. Imagine two
arrows in space. The first arrow has a length (magnitude) equal to
the speed of the Sun and a direction the same as the direction of the
Sun's motion though the galaxy. The second arrow has a length equal
to the speed of the Earth and a direction the same as the direction of
the Earth's motion through the galaxy. Touch the tail of one arrow to
the head of the other and draw a new arrow from the free tail to the
free head. The new arrow is the composite of the two motions through
space. The new arrow is constantly changing.

In article , William Hamblen
wrote:

On 4 Sep 2004 07:21:28 -0700, (kenneth
couesbouc) wrote:

I'm trying to calculate something and I wonder if someone can
help me.
The Earth travels around the Sun and the Sun is travelling towards
Vega (not quite, so I'm told?). I'd like to know at what moment (which
day in the solar year) the Earth is moving in the same direction as
the Sun, towards Vega. Also, as the Earth's mouvement is not in the
same plane as that of the Sun, what is the angle(a) between these two
planes.
What I want to find out is the Earth's maximum absolute speed. This
would be 30km/s (the Earth's speed around the Sun) plus 20km/s (the
Sun's speed towards Vega) multiplied by cos(a). But these two speeds
only add up at a precise time of year, which I haven't managed to work
out.

The answer the first question is "never" because the motion of the sun
through the galaxy is roughly in the direction of the constellation
Hercules, which is not in the plane of the ecliptic.

That's right, our system will reach the vicinity of the present position
of the green star 99 Hercules in about 82 thousand years from now, but it
of course will be long gone by then.

This green star is presently some 317 trillion miles away, and it appears
to us as a 5.04 magnitude star, so it's not very impressive. ( But I
think it should be more famous than it is..



The answer to the second question is to add the vectors. Imagine two
arrows in space. The first arrow has a length (magnitude) equal to
the speed of the Sun and a direction the same as the direction of the
Sun's motion though the galaxy. The second arrow has a length equal
to the speed of the Earth and a direction the same as the direction of
the Earth's motion through the galaxy. Touch the tail of one arrow to
the head of the other and draw a new arrow from the free tail to the
free head. The new arrow is the composite of the two motions through
space. The new arrow is constantly changing.


Actually, if I understand what Kenneth wants, our Local Group of galaxies
and dwarfs are moving 1.5 million mph towards the Great Attractor, so any
local motions are quickly cancelled out. The G.A. is at least 56 galaxy
clusters spanning a volume some 1.2 billion light years in diameter,
behind the stars of Centaurus from our vantage point 50 LYs above the
Plane.

Defender

In article , William Hamblen
wrote:

On 4 Sep 2004 07:21:28 -0700, (kenneth
couesbouc) wrote:

I'm trying to calculate something and I wonder if someone can
help me.
The Earth travels around the Sun and the Sun is travelling towards
Vega (not quite, so I'm told?). I'd like to know at what moment (which
day in the solar year) the Earth is moving in the same direction as
the Sun, towards Vega. Also, as the Earth's mouvement is not in the
same plane as that of the Sun, what is the angle(a) between these two
planes.
What I want to find out is the Earth's maximum absolute speed. This
would be 30km/s (the Earth's speed around the Sun) plus 20km/s (the
Sun's speed towards Vega) multiplied by cos(a). But these two speeds
only add up at a precise time of year, which I haven't managed to work
out.

The answer the first question is "never" because the motion of the sun
through the galaxy is roughly in the direction of the constellation
Hercules, which is not in the plane of the ecliptic.

The answer to the second question is to add the vectors. Imagine two
arrows in space. The first arrow has a length (magnitude) equal to
the speed of the Sun and a direction the same as the direction of the
Sun's motion though the galaxy. The second arrow has a length equal
to the speed of the Earth and a direction the same as the direction of
the Earth's motion through the galaxy. Touch the tail of one arrow to
the head of the other and draw a new arrow from the free tail to the
free head. The new arrow is the composite of the two motions through
space. The new arrow is constantly changing.


William, I was trying to apply your vector explanation, but I soon found
it to be a very complicated problem because of the description of our
motions below;

The American astronomer Frank Bash has interpreted a great deal of data to
mean that the sun is in a roughly elliptical orbit where its far point
(apogalacticon) is seven per cent further away than its current location.
The sun is now moving inwards and has almost reached its nearest point to
the galactic center (perigalacticon) which
lies 99.5 per cent of the sun's current distance from the galactic center
and will reach that point in only 15 million years - less than a
month in terms of the cosmic year.
The sun is also moving upwards, out of the plane of the Milky Way, at a
speed of 7 km per sec. At this moment we are 50 LYs above the
mid-plane of the galaxy, and our motion is steadily carrying us further
away. Fortunately, the gravitational pull of the stars in the galactic
plane is slowing down the sun's escape. Bash calculates that in 14
million years the sun will reach its maximum height above the galactic
disc. From that 250 LY perch it will be pulled back towards the plane of
the galaxy. Then passing through, it will travel to a point 250 LYs below the
disc, only to oscillate upwards again to reach its present position 66
million years from now.

kenneth couesbouc wrote:
Hello,
I'm trying to calculate something and I wonder if someone can
help me.
The Earth travels around the Sun and the Sun is travelling towards
Vega (not quite, so I'm told?). I'd like to know at what moment (which
day in the solar year) the Earth is moving in the same direction as
the Sun, towards Vega. Also, as the Earth's mouvement is not in the
same plane as that of the Sun, what is the angle(a) between these two
planes.
What I want to find out is the Earth's maximum absolute speed. This
would be 30km/s (the Earth's speed around the Sun) plus 20km/s (the
Sun's speed towards Vega) multiplied by cos(a). But these two speeds
only add up at a precise time of year, which I haven't managed to work
out.
Hoping to hear from you. Regards, Ken

There is no absolute speed... Velocity is ALWAYS with respect to
a reference. You could estimate the Solar System's orbital speed
around the center [of mass] of our galaxy. You could do the same
for Vega. Doppler shift data will tell you the relative velocity
of the Vega with respect to the Earth. The average of Doppler
measurements of a year will give the relative velocity of Vega
with respect to the Sun.

http://scienceworld.wolfram.com/phys...lerEffect.html

kenneth couesbouc wrote:
Hello,
I'm trying to calculate something and I wonder if someone can
help me.
The Earth travels around the Sun and the Sun is travelling towards
Vega (not quite, so I'm told?). I'd like to know at what moment (which
day in the solar year) the Earth is moving in the same direction as
the Sun, towards Vega. Also, as the Earth's mouvement is not in the
same plane as that of the Sun, what is the angle(a) between these two
planes.
What I want to find out is the Earth's maximum absolute speed. This
would be 30km/s (the Earth's speed around the Sun) plus 20km/s (the
Sun's speed towards Vega) multiplied by cos(a). But these two speeds
only add up at a precise time of year, which I haven't managed to work
out.
Hoping to hear from you. Regards, Ken



The link below provides a musical and highly entertaining description of
the motion of the earth through space, and a lot of other velocities as
well. This may not answer your question completely, but there are many
in this group who know better than I, but they don't know about this
link. ;-)

http://www.resource-intl.com/THE_GALAXY_SONG.MP2

Enjoy!
CDSTYA
Uncle Bob


__________________________________________________ _____________________________
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In article , Robert English
wrote:


The link below provides a musical and highly entertaining description of
the motion of the earth through space, and a lot of other velocities as
well. This may not answer your question completely, but there are many
in this group who know better than I, but they don't know about this
link. ;-)

http://www.resource-intl.com/THE_GALAXY_SONG.MP2

Enjoy!
CDSTYA
Uncle Bob

Thanks for the song, but it doesn't mention our oscillation as we revolve. LOL

Anyone care to do the math to estimate the extra miles we must travel due
to our path up above the Plane 250 LYs and then back down 250 LYs?

According to globular positions (a 1982 study by Frenk and White at
Berkeley), infrared measurements and maser studies of Sagittarius B2
stars, our solar system is estimated to be only 25000 (plus or minus 2000) light
years from the center of rotation.

The globular position averaging is probably misleading.

Defender

"sheep defender" wrote
in message
...
In article , Robert
English
wrote:


The link below provides a musical and highly entertaining
description of
the motion of the earth through space, and a lot of other
velocities as
well. This may not answer your question completely, but
there are many
in this group who know better than I, but they don't know
about this
link. ;-)

http://www.resource-intl.com/THE_GALAXY_SONG.MP2

Enjoy!
CDSTYA
Uncle Bob

Thanks for the song, but it doesn't mention our
oscillation as we revolve. LOL

Anyone care to do the math to estimate the extra miles we
must travel due
to our path up above the Plane 250 LYs and then back down
250 LYs?

According to globular positions (a 1982 study by Frenk and
White at
Berkeley), infrared measurements and maser studies of
Sagittarius B2
stars, our solar system is estimated to be only 25000
(plus or minus 2000) light
years from the center of rotation.

The globular position averaging is probably misleading.

Defender

Just tilting a plate (rotation path) to coincide with the
galaxy plane would keep the same circumference?

In article , "Robin R. Wier"
wrote:

"sheep defender" wrote
in message
...
In article , Robert
English
wrote:


The link below provides a musical and highly entertaining
description of
the motion of the earth through space, and a lot of other
velocities as
well. This may not answer your question completely, but
there are many
in this group who know better than I, but they don't know
about this
link. ;-)

http://www.resource-intl.com/THE_GALAXY_SONG.MP2

Enjoy!
CDSTYA
Uncle Bob

Thanks for the song, but it doesn't mention our
oscillation as we revolve. LOL

Anyone care to do the math to estimate the extra miles we
must travel due
to our path up above the Plane 250 LYs and then back down
250 LYs?

According to globular positions (a 1982 study by Frenk and
White at
Berkeley), infrared measurements and maser studies of
Sagittarius B2
stars, our solar system is estimated to be only 25000
(plus or minus 2000) light
years from the center of rotation.

The globular position averaging is probably misleading.

Defender

Just tilting a plate (rotation path) to coincide with the
galaxy plane would keep the same circumference?

No, because our system moves inward and outward, in addition to upward and
downward sinusoidally as it orbits the Galaxy. See my previous post for
the observed amounts..

Defender

(sheep defender) wrote in message ...
In article , "Robin R. Wier"
wrote:

"sheep defender" wrote
in message
...
In article , Robert
English
wrote:


The link below provides a musical and highly entertaining
description of
the motion of the earth through space, and a lot of other
velocities as
well. This may not answer your question completely, but
there are many
in this group who know better than I, but they don't know
about this
link. ;-)

http://www.resource-intl.com/THE_GALAXY_SONG.MP2

Enjoy!
CDSTYA
Uncle Bob

Thanks for the song, but it doesn't mention our
oscillation as we revolve. LOL

Anyone care to do the math to estimate the extra miles we
must travel due
to our path up above the Plane 250 LYs and then back down
250 LYs?

According to globular positions (a 1982 study by Frenk and
White at
Berkeley), infrared measurements and maser studies of
Sagittarius B2
stars, our solar system is estimated to be only 25000
(plus or minus 2000) light
years from the center of rotation.

The globular position averaging is probably misleading.

Defender

Just tilting a plate (rotation path) to coincide with the
galaxy plane would keep the same circumference?

No, because our system moves inward and outward, in addition to upward and
downward sinusoidally as it orbits the Galaxy. See my previous post for
the observed amounts..

Defender

As the solar system (with a line drawn through the Sun as the center
of galactic orbital motion) moves towards a inner galactic orbital
trajectory,heliocentric orbital motion of the planets becomes less
elliptical,as it moves outwards in galactic orbital motion the
planetary heliocentric orbits become more elliptical hence ice ages
and the Milankovitch cycle.

It is quite an enormous task to remove the geocentric/heliocentric
orbital equivalency which Newton put in place to justify mean orbital
distances to the Sun and which now would prove to be an obstacle in
considerations of the influence of the solar system's galactic orbital
motion on heliocentric orbital motion.

"Cor. 2. And since these stars are liable to no sensible parallax from
the annual motion of the earth, they can have no force, because of
their immense distance, to produce any sensible effect in our system.
Not to mention that the fixed stars, every where promiscuously
dispersed in the heavens, by their contrary actions destroy their
mutual actions, by Prop. LXX, Book I." Principia

The 'universal laws of gravitation' is just a poorly constructed
ballistic theory,an astronomical cataloguer would not recognise it but
a true astronomer would.