Sun <==> Alpha Centauri gravity interactions

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

"Grimble Gromble" wrote in message ...
"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

Thanks! That's a sound idea, but I think we will be *certain* of a
nice trail of comets stretching all the way to Alpha, long before the
mission is due for launch... projected at round 2200 AD... by which
time we will have filled all the interim experience gaps necessary
with adventures on the Moon and Mars!

Now, I was going to turn this thing into a cracking BIG SCREEN MOVIE,
if only I could present my script to some 'movie' agency!

Grim, you wouldn't happen to know of any contacts in the movie
business by any chance would you? I'll share the royalties if it
helps...! LOL..

Abdul

"AA Institute" wrote in message
m...
Grim, you wouldn't happen to know of any contacts in the movie
business by any chance would you? I'll share the royalties if it
helps...! LOL..
I don't know anybody in anything.
Grim

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

BP wrote:
Right, okay that is the 26k year cycle. So, I guess really a stars
movement would not really be noticeable on the scale of a star chart.
I am referring to the Sky Atlas (Tirion,Sinnott)..

The shift in coordinates due to precession is of the order of 1 degree every
70 years. The star with the largest proper motion, ie measured across the
sky in a plane perpendicular to the line of site is Barnard's star, which
moves approx 1 degree every 350 years, or about 6 times less than the motion
due to precession. Most other stars would appear to move a far smaller
angle.
If you have a 1950 star chart, there are algorithms for calculating the
position at another date.


DaveL


Bp
The main reason for the changing Epochs is to account for precession
of the earth's axis. The axis shifts by about an arcminute every
year (plucked from memory, could be wrong). Because of this the
coordinates of objects on the celestial sphere changes with time, by
a small amount each year.


DaveL

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

"Grimble Gromble" wrote in message ...
"AA Institute" wrote in message
om...
Take our Sun for instance and the recent high levels of solar activity
we've seen, sending massive amounts of charged particles toward the
Earth. ...
My god; it's shooting at us!

So it is. Question is: would it shoot harder and more violently if the
Sun had a binary companion of similar size and mass orbiting it at,
say around where Saturn orbits the Sun? And if the orbit of that
secondary star was highly elliptical, would the output vary along the
orbital cycle in line with the distance separating the two stars?

Do gravitational interactions between two stars in a binary system
cause variability in the radiation output...through some kind of a
'tidal wave' inducement?

AAI

Saul Levy wrote in message . ..
snip
Space is plenty empty enough that most stars are isolated. Only
occasionally will a star come close enough to perturb outlying objects
in orbit around another star.

F= G M m / r^2, the standard equation for interactive force between
any two bodies in space, be they stars, planets, comets or any
combination between them.

A hypothetical comet of mass 1 x 10^5 kg orbiting the Sun at 50,000
AUs (7.5 x 10^15 metres) out in the direction of Alpha Centauri, would
be tugged with a force of 2.4 x 10^-7 Newton by the Sun (of mass 2 x
10^30 kg). The same comet would be perturbed by Alpha Centauri's
combined mass of 4 x 10^30 kg exerting a force of just 2.5 x 10^-8
Newton (from a distance of 222,000 AUs = 3.3 x 10^16 metres). Thus,
the Sun will be 10 times more forceful on the body, compared to Alpha
Centauri.

Now, you could argue well 9/10th versus 1/10th... is that sufficiently
strong a force to perturb the orbit of an outlying comet in the manner
I describe he-
http://uk.geocities.com/aa_spaceagen....html#midrange

Back in the 1840s, French astronomer Urbain Jean Joseph Le Verrier
made calculations along these lines to show that Uranus was being
perturbed by an unseen body, which subsequently enabled him to predict
the position of (then undiscovered) Neptune and guide astronomers at
the Berlin observatory to discover this new planet.

Now, Uranus of mass 8.7 x 10^25 kg orbits the Sun at a distance of
19.18 AUs on average (2.9 x 10^12 metres. The Sun exerts a force of
1.4 x 10^21 Newtons on Uranus.

At their closest points in their respective mean orbits around the
Sun, Uranus and Neptune (the latter of mass 1 x 10^26 kg) are
separated by 1.6 x 10^12 metres. The maximum perturbational force
between Uranus and Neptune during such (mean) close encounters in
their orbits is 2.2 x 10^17 Newtons.

Thus, the Sun's influence on Uranus is over 6,000 times as strong as
the mutual perturbing forces between it and Neptune. Yet with this
microscopic 1/6,000th as strong an influence, measurable positional
shifts in the motion of Uranus were used to deduce the discovery of
Neptune!

The power of gravity, no matter how weak, no matter from what distance
its influence, is not to be underestimated...

AAI