Sun <==> Alpha Centauri gravity interactions

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

AA Institute wrote:

-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?

Not quite as easy as the transverse motion calculation. What you'd be
doing is converting from equatorial coordinates to ecliptic coordinates.
The conversion involves a single rotation, but in three dimensions. The
axis of the rotation is the intersection of the equatorial and ecliptic
planes, and the amount of the rotation is the obliquity of the ecliptic
(the tilt of the Earth). You'd start with both RA and Dec (you need
both, since it's a 3D transformation) and end up with lamda and beta,
ecliptic longitude and latitude.

For more precise calculations, or calculations over long time frames,
you'd have to account for various effects that cause the two planes to
move with respect to each other. Nutation is a slight wobble of the
equatorial system caused by the Moon. The Earth's axis also precesses
slowly.

A search would probably turn up online calculators to do the conversion
for you, along with any number of pages on celestial coordinates, e.g.

http://www.seds.org/~spider/spider/S....html#ecliptic

Most star charting software is capable of displaying grid lines for
several coordinate systems, and some can provide the location of
specified objects in your choice of coordinate systems.

Approximate current ecliptic coordinates for Alpha Centauri are

beta = -42.6°
lamda = 239.5°

But don't take my word for it. I just did this on a calculator and may
have gotten it wrong, and I haven't told you the epoch or whether I
accounted for nutation. You'll get more satisfaction from understanding
and doing the calculation yourself.

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

Ernie Wright wrote in message ...
AA Institute wrote:

-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?

snip
Approximate current ecliptic coordinates for Alpha Centauri are

beta = -42.6°
lamda = 239.5°

But don't take my word for it. I just did this on a calculator and may
have gotten it wrong, and I haven't told you the epoch or whether I
accounted for nutation. You'll get more satisfaction from understanding
and doing the calculation yourself.

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

Thanks for confirming; yes I made it -42.6 degrees based on the
obliquity, R.A. & Dec, for the current epoch around 2000.0-ish.

The dynamics are quite complex and moving over time, hence if we're
launching a hypothetical starship in a future era the variables are
bound to be radically different by then. It just serves as an
illustration for now.

Abdul