Sun <==> Alpha Centauri gravity interactions

"AA Institute" wrote in message
m...
So if the transverse velocity of Alpha Cen is 5.0 AUs/yr and the
radial velocity is -5.5 AUs/yr, does this mean that in 50,000 years
(272,000 AUs current distance / 5.5 AUs radial velocity) Alpha
Centauri is going to be very close to us?! Probably not, since due to
gravitational interaction with the Sun, Alpha Centauri might describe
a 'curved' trajectory as opposed to a linear one.
I've not checked your figures but assuming them to be correct: since the
transverse velocity is of the same order as the radial velocity, then by the
time the radial velocity 'would' have closed the distance between Alpha
Centauri and the Sun, the transverse velocity would have carried it just as
far at right angles and it will end up a similar distance away. The closest
approach would then be about 0.7 times the current distance.

It would be so much easier to visualise the whole thing in a 3D diagram.
There are programs available for plotting just such things in 3D. I remember
mentioning Mathcad not too long ago! You can even allow a term for the
gravitational interaction between the stars and convince yourself that it
has little effect. I'd do it for you except I have more interesting projects
I would rather spend my time working on (no offence meant).

Also referring to memory, which, as I always remind everyone, is very dodgy,
I have a vague recollection that when the velocities of nearby stars are
compared, the stars essentially fall into two groups. Stars in our group
move pretty much in the same direction and speed as the Sun, while the other
group of stars travel in a direction and speed that is common to them and
different from ours. I believe there were other factors such as age and
composition that distinguished the two groups? I apologise if this is not
the case, however, like most things, I cannot remember my source.
Grim

Grimble Gromble wrote:

Also referring to memory, which, as I always remind everyone, is very dodgy,
I have a vague recollection that when the velocities of nearby stars are
compared, the stars essentially fall into two groups. Stars in our group
move pretty much in the same direction and speed as the Sun, while the other
group of stars travel in a direction and speed that is common to them and
different from ours. I believe there were other factors such as age and
composition that distinguished the two groups? I apologise if this is not
the case, however, like most things, I cannot remember my source.

You're probably thinking of the "Population I" _vs_ "Population II"
classification. The former stars, including our Sun, are part of the
galactic disk, having been born from its clouds of gas and dust, and
orbit the galactic centre pretty much in a plane. The latter group,
mostly older stars that are evolving out of the main sequence, form a
spherical 'halo' around the Galaxy, with orbits that tend to
intersect the disk at steep angles, and make up most of the globular
clusters. Arcturus (Alpha Boötis), a fairly nearby orange giant, is
one of the most prominent examples of a Population II star, and
because of the high inclination of its path through the galactic disk
it exhibits the largest proper motion of any first-magnitude star,
cutting across the 'stream' in which the Sun and its contemporaries
are moving.

--
Odysseus

"Grimble Gromble" wrote in message ...
"AA Institute" wrote in message
m...
So if the transverse velocity of Alpha Cen is 5.0 AUs/yr and the
radial velocity is -5.5 AUs/yr, does this mean that in 50,000 years
(272,000 AUs current distance / 5.5 AUs radial velocity) Alpha
Centauri is going to be very close to us?! Probably not, since due to
gravitational interaction with the Sun, Alpha Centauri might describe
a 'curved' trajectory as opposed to a linear one.
I've not checked your figures but assuming them to be correct: since the
transverse velocity is of the same order as the radial velocity, then by the
time the radial velocity 'would' have closed the distance between Alpha
Centauri and the Sun, the transverse velocity would have carried it just as
far at right angles and it will end up a similar distance away. The closest
approach would then be about 0.7 times the current distance.

Thanks Grim, silly me for not seeing the wood for the trees...
So the closest approach point for Alpha Centauri (around 3 LY) is
still yet to come? Hoooorrrraaayyy!!!

This could be the ideal interstellar *launch window* for the Aster-Com
starship. A future generation of Earth might face the challenging
choice of either taking this window of opportunity or declining the
offer in anticipation of another star passing by the Sun. But that's
gonna be a long, long time coming...

It would be so much easier to visualise the whole thing in a 3D diagram.
There are programs available for plotting just such things in 3D. I remember
mentioning Mathcad not too long ago! You can even allow a term for the
gravitational interaction between the stars and convince yourself that it
has little effect. I'd do it for you except I have more interesting projects
I would rather spend my time working on (no offence meant).

That's fair comment. Hey, what's the big deal with a 50,000 year
voyage inside some hollowed out gigantic boulder rolling across in the
deep, dark ocean of space toward some unknown destination
pre-programmed by your great great great grand parents? It sucks...

Also referring to memory, which, as I always remind everyone, is very dodgy,
I have a vague recollection that when the velocities of nearby stars are
compared, the stars essentially fall into two groups. Stars in our group
move pretty much in the same direction and speed as the Sun, while the other
group of stars travel in a direction and speed that is common to them and
different from ours. I believe there were other factors such as age and
composition that distinguished the two groups? I apologise if this is not
the case, however, like most things, I cannot remember my source.

No probs, really appreciate your thoughts.

One final question: interstellar navigation - how can I do it whilst
drifting in this great interstellar ocean where the shores reach out
to near eternity in every direction?

"In the extreme circumstance where no new bodies are found for meeting
projected resource requirements, the ship can turnaround and back
track towards previously charted bodies using emergency reserves. With
no magnetic fields, no bright planets, no "GPS" for relative
referencing, the minute positional shifts of nearby stars may be the
only method of interstellar navigation in the surrounding darkness of
3D space."

How can I precisely chart the *absolute* positions and ship-relative
velocities of icy comets encountered on a forward pass... then try to
re-intercept them on a reverse pass, having turned my ship around?

Abdul

"AA Institute" wrote in message
om...
How can I precisely chart the *absolute* positions and ship-relative
velocities of icy comets encountered on a forward pass... then try to
re-intercept them on a reverse pass, having turned my ship around?
That isn't necessary. Just plant a small transmitter on the comet; that will
give you directional information. If the transmissions and receptions are
accurately timed (pulsars make excellent clocks available to all) then the
distance to the comet can be easily calculated.
Grim

Wasn't it AA Institute who wrote:
Mike Williams wrote in message news:Q9MYRBANsrQBFwP
...
Wasn't it AA Institute who wrote:

Could it be that Alpha Centauri (A+B+C) and the Sun are
gravitationally *locked* together and share a common proper motion
around the galaxy?

To be gravitationally locked, their relative velocity would need to be
less than the escape velocity of one from the other. A quick calculation
shows the relevant escape velocity to be about 81 metres/second at this
distance. The radial component of the relative velocity is about 26400
metres per second, so they're not gravitationally locked.

According to a formula I found in my spherical astronomy notes for
proper motion, the 'transverse velocity' (component of total velocity
projected *across* our line of sight) is given by:

v = 4.74 * (proper motion / parallax) km/sec, so for Alpha Centauri, v
= 4.74 * (3.7 / 0.74) = 23.7 km/sec = 5.0 AUs per year. Translating
the star's given radial velocity of -24.6 km/sec to AUs per year =
-5.5 AUs/year

Are you certain that your values for "proper motion" and "parallax" have
the correct units for the equation you're using? I use a more direct
method and get a vastly different answer.

I started with the fact that the proper motion is RA: -7.54775
acsecs/year, Dec: +0.48180 arcsecs/year and the distance is 4.3 light
years.

A light year is 9.46e15 metres.
-7.54775 arcsecs/year of RA is -0.000549399 radians/year
0.48180 arcsecs/year of Dec is 2.33583e-06 radians/year
(Note a complete circle is 24h of RA but 360d of Dec)

The transverse motions are Distance * sin(Angle), giving
-2.23484e+13 and 9.5017e+10 metres/year. Divide by the number of seconds
in a year and combine the two velocities by Pythagoras and I get the
transverse motion to be 710 km/sec = 150 AU/year.

--
Mike Williams
Gentleman of Leisure

Mike Williams wrote in message

Are you certain that your values for "proper motion" and "parallax" have
the correct units for the equation you're using? I use a more direct
method and get a vastly different answer.

I started with the fact that the proper motion is RA: -7.54775
acsecs/year, Dec: +0.48180 arcsecs/year and the distance is 4.3 light
years.

A light year is 9.46e15 metres.
-7.54775 arcsecs/year of RA is -0.000549399 radians/year
0.48180 arcsecs/year of Dec is 2.33583e-06 radians/year
(Note a complete circle is 24h of RA but 360d of Dec)

The transverse motions are Distance * sin(Angle), giving
-2.23484e+13 and 9.5017e+10 metres/year. Divide by the number of seconds
in a year and combine the two velocities by Pythagoras and I get the
transverse motion to be 710 km/sec = 150 AU/year.

My equation is from page 250, "Spherical Astronomy" by W.M. Smart (a
very old book from the 1960s). On page 251, he gives an example using
the star Capella, where the annual proper motion is 0.439 arc sec,
parallax 0.075 arc sec, giving a transverse velocity of 27.7 km/sec.

On that basis, I think I've got it right... unless the 3.7 arc
sec/year total proper motion figure for Alpha Centauri I'm using is
wrong?

Considering also the Sun moves through space at roughly 20 km/sec, I
think your number is a bit on the high side.

Abdul

Mike Williams wrote:

Are you certain that your values for "proper motion" and "parallax"
have the correct units for the equation you're using? I use a more
direct method and get a vastly different answer.

There are two errors in your calculation, both of which inflate the
component of motion in right ascension.

-7.54775 arcsecs/year of RA is -0.000549399 radians/year
(Note a complete circle is 24h of RA but 360d of Dec)

It isn't necessary to convert -7.54775 from hours:minutes:seconds to
degrees:minutes:seconds. It's already expressed as arcseconds in the
d:m:s system.

But you do have to multiply it by the cosine of Alpha Centauri's
declination. To see why, consider the surface of the Earth. Degrees
latitude (north-south) always correspond to a surface distance of about
110 km, but the surface distance for a degree of longitude depends on
the latitude. It's 110 km at the equator, where cos(lat) = 1, but
smaller than 110 km by the factor cos(lat) at other latitudes.

The declination of Alpha Centauri is -60° 50', and cos(-60° 50') is
about 0.487.

So your figure for radians/year in RA is too big by a factor of about
30: (360 / 24) * (1 / cos(-60° 50')).

The formula Abdul used is pretty standard, and simpler to apply. You
can divide by the parallax (in arcseconds) or multiply by the distance
(in parsecs).

4.74 is just a constant of proportionality that converts between AU/year
and km/s. An object at a distance of 1 parsec with a proper motion of 1
arcsecond/year has a transverse motion of 1 AU/year, or 150 million
km/year, or 4.74 km/s.

- Ernie http://home.comcast.net/~erniew

Ernie Wright wrote:

The formula Abdul used is pretty standard, and simpler to apply. You
can divide by the parallax (in arcseconds) or multiply by the distance
(in parsecs).

4.74 is just a constant of proportionality that converts between AU/year
and km/s. An object at a distance of 1 parsec with a proper motion of 1
arcsecond/year has a transverse motion of 1 AU/year, or 150 million
km/year, or 4.74 km/s.

Oh, so that's where the 4.74 came from, I wasn't sure of its origins.

So to depart toward Alpha Centauri on a hypothetical voyage, one has
to leave the ecliptic plane of our solar system going south towards
-60° 50' declination.
Is there an easy calculation to work out how many degrees that
direction is off the ecliptic plane of our solar system?

Abdul