Sun <==> Alpha Centauri gravity interactions
AA Institute
Moore's) often show projections of *brightest* and *closest* stars in
the several thousand years before and after the present time. I
distinctly recall that Barnard's star is one that will become the
closest star to the Sun in so many thousands of years from now at
around 3.5(?) odd light years and then it will start receding away
from us again after that time.
Firstly, how does one go about making such future positional guesses
and secondly, how long has Alpha Centauri been in close proximity to
the Sun? Is there any projections as to how long Alpha Centauri will
stay this close?
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? I know we observe distinct radial velocity and
proper motions that Alpha Centauri has relative to the Sun, but they
are based on short term measurements in the current era... and I don't
expect you can simply *extrapolate* forwards/backwards in time simply
on the basis of their present values... or can you?
According to "Solstation" (my number one favourite site for local star
system research!):-
http://www.solstation.com/stars/alp-cent3.htm
the next nearest large star system to have any *significant*
interaction with both the Sun and Alpha Centauri(A+B+C) is Sirius
(A+B) - which is quite far removed at 8.6 LY away from the Sun and 9.5
LY away from from Alpa Cen.
This means the Sun and Alpha Centauri system are relatively isolated
in space, where it is conceivable that bodies orbiting far out around
each system are gravitationally perturbed in the manner in which I
illustrate here:-
http://uk.geocities.com/aa_spaceagent/restricted/interstellar-propulsion.html#midrange
Its important I think to study our nearest triple star system in
greater depth (if only it rose above my horizon... but then I can't do
a lot with my tiny 8-inch Newtonian!). How much Hubble or other
space/ground-based telescope time is devoted to Alpha Centauri,
compared to all other stellar astronomy, I wonder...
cheers
Abdul Ahad
Brian Tung
> Firstly, how does one go about making such future positional guesses
> and secondly, how long has Alpha Centauri been in close proximity to
> the Sun? Is there any projections as to how long Alpha Centauri will
> stay this close?
The gravitational interactions between stars are pretty small. Unless
one star is predicted to pass very close indeed--say, less than a tenth
of a light-year or so--the impact on their orbit around the galactic
center is minimal. Testament to this is the relative stability of the
galactic disc. You can map the radial and proper velocities of the
star in three dimensions and get reasonably accurate results into the
fairly distant future (say, about a million years) for the nearest stars.
> 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?
No. I forget the exact figures, but the relative velocity of the
alpha Centauri system is considerably more than permitted for a locked
system.
> I know we observe distinct radial velocity and
> proper motions that Alpha Centauri has relative to the Sun, but they
> are based on short term measurements in the current era... and I don't
> expect you can simply *extrapolate* forwards/backwards in time simply
> on the basis of their present values... or can you?
You sure can, because the gravitational interactions between individual
stars separated by light-years are quite weak.
> the next nearest large star system to have any *significant*
> interaction with both the Sun and Alpha Centauri(A+B+C) is Sirius
> (A+B) - which is quite far removed at 8.6 LY away from the Sun and 9.5
> LY away from from Alpa Cen.
>
> This means the Sun and Alpha Centauri system are relatively isolated
> in space...
How do you figure? The Sun and alpha Centauri are 4.3 light-years apart,
or half the distance from here to Sirius. Since Sirius has a bit more
than twice the mass of the Sun, and is more massive than the entire alpha
Centauri system, I don't think you can consider the Sun and alpha Centauri
much more isolated from Sirius than they are from each other.
The essential paucity of gravitational interaction is why the "stellar
rape" hypothesis of the solar system's formation had such a short life
in the middle of the 20th century. It's just too darned unlikely for
stars to pass that close to one another. To give you an idea of how far
apart the stars are from one another, consider that the Sun and alpha
Centauri A are about the same size (1.4 million km across) and are
separated by about 42 *trillion* km. That is, the distance between them
is about 30 million times their diameters.
To put it another way, if the Sun is a golf ball in San Francisco, then
alpha Centauri is another one (or two golf balls and a marble) in Los
Angeles, and Sirius is a slightly larger racquetball, accompanied by an
incredibly dense BB pellet, in Boise. Picture that for a moment, and
ask yourself how likely it is that they will interact with one another.
And remember, they move in three dimensions, not two.
> Its important I think...
Steady, Brian, steady...
> Its important I think to study our nearest triple star system in
> greater depth (if only it rose above my horizon... but then I can't do
> a lot with my tiny 8-inch Newtonian!). How much Hubble or other
> space/ground-based telescope time is devoted to Alpha Centauri,
> compared to all other stellar astronomy, I wonder...
What you require is accurate astrometry, and this has all been done
previously--most recently by the Hipparcos mission.
Brian Tung <br...@isi.edu>
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt
Ernie Wright
> 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?
Brian's already answered this, but you don't have to take his word for
it. Your own page has the formulas you need to estimate the strength
of the gravitational interaction.
The gravitational force f between the sun and Alpha Centauri (assuming
I haven't goofed this up) is roughly
f = G M m / r^2
G = 6.7 * 10^-11 N m^2 / kg^2 Newton's Gravitational Constant
M = 2 * 10^30 kg mass of sun
m = 4 * 10^30 kg mass of Alpha Centauri
r = 4 * 10^16 m distance
f = 3.4 * 10^17 N
That looks like a lot, but it takes a lot of force to move a star. The
acceleration of the sun due to f is
a = f / M = 1.7 * 10^-13 m / s^2 = 0.00000000000017 m / s^2
Pretty small. If the sun were a car powered by the gravitational
attraction of Alpha Centauri, it would go from 0 to 60 (miles per hour)
in about 5 million years.
t = v / a
v = 60 mph = 27 m / s
a = 1.7 * 10^-13 m / s^2
t = 27 / (1.7 * 10^-13) s
Finally, if you substitute the Earth for Alpha Centauri,
m = 6 * 10^24 kg
f = 5.4 * 10^23 N
you find that the force between them is a million times stronger.
- Ernie http://home.comcast.net/~erniew
AA Institute
> Abdul Ahad wrote:
> > Firstly, how does one go about making such future positional guesses
> > and secondly, how long has Alpha Centauri been in close proximity to
> > the Sun? Is there any projections as to how long Alpha Centauri will
> > stay this close?
>
> alpha Centauri is another one (or two golf balls and a marble) in Los
> Angeles, and Sirius is a slightly larger racquetball, accompanied by an
> incredibly dense BB pellet, in Boise. Picture that for a moment, and
> ask yourself how likely it is that they will interact with one another.
> And remember, they move in three dimensions, not two.
>
distances in every day terms. Of course, the inverse square law
governing gravitational interactivity dictates that for a system such
as Sirius which is twice as far away from the Sun as Alpha Centauri,
the force of gravity will be only a quarter as strong (other things
like relative masses, being equal). So you'd expect the Sun and Alpha
Centauri to interact 4 times as strongly between one another, compared
to the strength with which each one interacts with the Sirius system.
So I would say, from a perspective of gravitational interactivity, the
Sirius system is relatively *far removed*... but you have every right
to disagree of course!
So if Alpha Centauri is not gravitationally *connected* with the Sun,
are you saying that its just another passing star system? Is there a
3D model available to show the projected distances separating Sun from
Alpha Centauri on a time-series basis like this:-
Epoch: Distance:
==============================================================
Now - 1 million years ?
Now - 500,000 years ?
Now - 100,000 years ?
Now 4.3 LY
Now + 1 million years ?
Now + 500,000 years ?
Now + 100,000 years ?
> > Its important I think...
>
> What you require is accurate astrometry, and this has all been done
> previously--most recently by the Hipparcos mission.
>
What about extrasolar planet detection efforts around Alpha Centauri,
to your knowledge?
Thanks for sharing your thoughts.
Abdul Ahad
Mike Williams
>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.
--
Mike Williams
Gentleman of Leisure
Brian Tung
> So I would say, from a perspective of gravitational interactivity, the
> Sirius system is relatively *far removed*... but you have every right
> to disagree of course!
And I do. I still think you are entranced by this idea of the Sun and
alpha Centauri as being partners in space, when they very clearly are
not. Since the gravitational influence of Sirius on alpha Centauri is
just about half that of the Sun's, I wouldn't call that at all far
removed. Further removed, yes, but somebody has to be closest. That
doesn't say anything about whether we're bound to each other. And, as
it happens, we are not.
> So if Alpha Centauri is not gravitationally *connected* with the Sun,
> are you saying that its just another passing star system?
Yes. We've said that a number of times, now.
> Is there a
> 3D model available to show the projected distances separating Sun from
> Alpha Centauri on a time-series basis like this:-
>
> Epoch: Distance:
> ==============================================================
> Now - 1 million years ?
> Now - 500,000 years ?
> Now - 100,000 years ?
> Now 4.3 LY
> Now + 1 million years ?
> Now + 500,000 years ?
> Now + 100,000 years ?
You can construct one very simply from available three-dimensional
velocity data. From the Doppler shift, one gets the velocity along the
line connecting us and alpha Centauri; from the proper motions, one
gets the velocity in the plane perpendicular to that line. Combine
that, and you can derive the values in the table above.
> What about extrasolar planet detection efforts around Alpha Centauri,
> to your knowledge?
To my knowledge, such efforts have been unsuccessful in detecting planets
around alpha Centauri A, B, or C. That doesn't mean that there aren't
any planets around any of those stars--only that if there are, they are
too small (or, just conceivably, in an orbital plane that is too close to
perpendicular to our line of sight) to have been detected yet.
AA Institute
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
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.
It would be so much easier to visualise the whole thing in a 3D
diagram.
Abdul
AA Institute
> a = f / M = 1.7 * 10^-13 m / s^2 = 0.00000000000017 m / s^2
>
> Pretty small. If the sun were a car powered by the gravitational
> attraction of Alpha Centauri, it would go from 0 to 60 (miles per hour)
> in about 5 million years.
>
Ernie,
That is a seriously tiny acceleration and for a massive body like the
Sun, I'd expect that sort of result but thanks for putting some hard
numbers to illustrate it all - makes it so much easier to visualise.
However, in relation to my interstellar journey "blueprint" (or
proposal to a far future generation!) I am concerned with
gravitational accelerations of tiny comets (effectively infinitesimal
*particles* of negligible mass in comparison with the mass of a star)
which will be perturbed (gravitationally accelerated) from a fraction
of the total Sun-Alpha Centauri distance.
Of course, looking across the other side of the interstellar "pond" we
find Proxima Centauri (with just 12% of Sun's mass) sharing a definite
common proper motion with Alpha Centauri A+B, and its placed at a huge
range of 13,000 AUs from the primary pair.
Its all going to be a *conjecture* sort of result I think...! But the
beauty of the "Aster-Com" starship concept is you can turn back at any
time you run into vaccuums with regards to resource availability on
comets/planetoids towards Alpha Centauri!
I wonder if its possible to see a mirage ahead from the control room
of a water-starved starship... LOL!!!
Abdul
Grimble Gromble
news:adbf5bc1.0409...@posting.google.com...
> 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.
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
Odysseus
>
> 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
Mike Williams
>f...@econym.demon.co.uk>...
>> 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.
AA Institute
> "AA Institute" <abdul...@ntlworld.com> wrote in message
> news:adbf5bc1.0409...@posting.google.com...
> > 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.
>
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
>
> Yes. We've said that a number of times, now.
Thanks Brian, that's awfully clear now.
>
> > What about extrasolar planet detection efforts around Alpha Centauri,
> > to your knowledge?
>
> To my knowledge, such efforts have been unsuccessful in detecting planets
> around alpha Centauri A, B, or C. That doesn't mean that there aren't
> any planets around any of those stars--only that if there are, they are
> too small (or, just conceivably, in an orbital plane that is too close to
> perpendicular to our line of sight) to have been detected yet.
>
I've heard it once mentioned that a planet orbiting one of the stars
of a binary system can receive large doses of radiation from two stars
as opposed to one and that 'flaring' between the two suns of the
system could be common place.
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. Now imagine if Alpha Centauri B was orbiting somewehere around
where Uranus or Neptune is in our solar system, how much of an effect
would the additional radiation from this secondary sun add to our
every day lives?
If solar type flaring occurs on Alpha Centauri A and B, I wonder if
the gravitational interactivity between the two stars would increase
this flaring disproportionately. In other words, if the two stars were
isolated and not confined in a binary setup then (radiation from A +
radiation from B) would be less than that where the two stars are
gravitationally bound as in the case of Alpha Centauri? Is it
possible that the invisible gravitational flux lines (if that's the
right way to put it?) could induce too much flaring and cause harm to
any life evolving around a planet orbiting either star?
If such flaring caused *minute* variability to the overall
brightnesses of each star, surely that would vary across the 80 year
orbital period of the system and we should perhaps see an increase
towards the periastron passage time when the stars are closest
together? I wonder if such tiny variations could in fact be tracked by
photometric equipment carried on space telescopes in orbit around
Earth like the Canadian MOST mission. etc...
Thanks,
Abdul
Grimble Gromble
> Take our Sun for instance and the recent high levels of solar activity
> we've seen, sending massive amounts of charged particles toward the
My god; it's shooting at us!
Grim
Grimble Gromble
> 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?
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
AA Institute
> 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
Ernie Wright
> 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
AA Institute
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
Grimble Gromble
> 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..
Grim
AA Institute
> 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
Ernie Wright
> -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/ScholarX/coords.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
AA Institute
> 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?
>
> 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
AA Institute
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