Sun <==> Alpha Centauri gravity interactions

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

"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