Distance between stars in M42

[batema (Mark)]

Hi,

I am just reading about M42 being approx 1300Ly away and 24Ly across. Ciould someone tell me how far apart the stars in the trapezium are???

Mark

[Kevnool (Kev)]

have a look here.

http://www.cityastronomy.com/trapezium-lg.jpg

Cheers Kev.

[ngcles]

Hi Mark,

Without doing the arithmetic myself, Wikipedia provides what seems like a sensible guesstimate of 1.5 ly.

http://en.wikipedia.org/wiki/Trapezium_(astronomy)

Trig will give the answer I think assuming the four main stars all lie at the same distance from Earth, but I can't find my trig and log tables at the present time.


Best

Les D

[batema (Mark)]

Thank you
Mark

[[1ponders] (Paul)]

I haven't read it all but you may find more info here, Mark

http://en.wikipedia.org/wiki/Orion_nebula

[Enchilada]

Using the distance to the Orion Nebula as 389 parsecs or 1270 +/- 76 light-years, means that the trig. parallax is 2.571 mas (milliarcsec.)

If the principal stars in the Trapezium are; (based on the relative position of STF 748A at 6.55V magnitude)

Star ....... ". mag Spec Class

STF 748 AB 08.9 7.49 O7 B1V
STF 748 AC 13.1 5.06 O7
STF 748 AD 21.5 6.38 B0.5V
STF 748 AE 04.5 11.1 O7

Then the projected distances in AU and light years between the components from A (Mabs=-1.4 Mass=4.5 Solar Masses) is;

B= 3462 AU (0.055 ly.) Mabs =-1.4; 3.6 Solar Masses
C= 5095 AU (0.081 ly.) Mabs =-2.9; 6.4 Solar Masses
D= 8362 AU (0.132 ly.) Mabs =-1.6; 4.7 Solar Masses
E= 1750 AU (0.028 ly.) Mabs =+3.2; 1.5 Solar Masses

In physical size, therefore for the Trapezium is about 10,000 AU across.
The largest estimated minimum orbital period of these stars is about 250,000 years. (about 2.5 times that of Alpha Centauri AB to Proxima) This augers well with the fractional movements so far observed with these stars.

You could compare these values with the older ones described in the dated "Burnham's Celestial Handbook" Vol.2 "Orion" on Theta Orionis.

Note: If value were between 1400 to 1600 light years, these projected separations would be slightly larger.

For more information and perspective read Andrew James' "Southern Astronomical Delights" on the History of the Orion Nebula at;

http://homepage.mac.com/andjames/Page204.htm

Hope this helps.

NOTE: I have given open permission for this text to be used in an article "New View of Young, High-Mass Binary Star at Heart of Orion" at Universe Today on this without any accreditation.

[Jen]

thats sounds very technical all those numbers

[Enchilada]

Originally Posted by Jen

thats sounds very technical all those numbers

Your right. The Trapezium in the Orion Nebula is about 700 times larger than the solar system...

[Jen]

Originally Posted by Enchilada

Your right. The Trapezium in the Orion Nebula is about 700 times larger than the solar system...

[Ric]

Space is big, really big.

[ngcles]

Hi Enchilada & All,

Originally Posted by Enchilada

Using the distance to the Orion Nebula as 389 parsecs or 1270 +/- 76 light-years, means that the trig. parallax is 2.571 mas (milliarcsec.) ...

Well done mate ... ) I'm still looking for my trig tables.

Just so we're all a little wiser in the future (actually its so I can remember my high-school maths) (note to Jen: I expect you to commit this to memory for a later quiz) ... plugging that figure of 2.571 into a trig calculation to get a true separation in AU occurs ... precisely how? Can you show the working for me??

I've had to do this once before for an article I wrote (can't remember which one) and took instruction from a current 3rd year university level mathematics student working at Syd Obs and had it checked by Nick Lomb, and I seem to recall it was the Cosine rule -- yes??


Best,

the 'mathematically challenged' Les D

[Enchilada]

Originally Posted by ngcles

Can you show the working for me??

I do it from a complicated spreadsheet that I created for my double star work. I just plug in the parallax in mas (milliarcseconds) with the given errors, and it spits out the results.
However, the nuts and bolts of the calculations you require are as follows;

Distance is calculated as;

d (pc) = 1 / p x 1000

Where p = parallax (in milliarcseconds)

If the parallax is in arcsecs, it is just ;

d (pc) = 1/ p

To convert parsecs (pc.) to light years (ly) it is multiplied by 3.2616
(To convert light years (ly) to parsecs (pc.)it is multiplied by 0.30660)

To find the true separation (TS) of stars, in suitable units is a little more complicated. Easiest is to do;

TS = sin (Sep/3600) x d (in parsec)

The using the calculated true separation (TSep), then convert it into the units you want by;

TD (light-years) = TS x 3.2616

TD (AU) = TS x 63240

TD (km) = TS x 9460000000000 (ten zeros)
[or 9.46x10^12]

Where;
Sep = Separation (arcsec)
d = distance in parsecs
TS = True Separation
TD = True Distance

Note: Finding the sine (sin) of very small angles is subject to significant errors due to the smallness of the angle. I use another long-winded method of calculating to deduce this more exactly doing it electronically. However, in simple calculators I find for most uses it is ample if you are seeking values that are fairly close.

[Enchilada]

An odd question to IIS folk.
Without doing detailed calculation - like the complicated mess I've just posted before this, really how big is the bowl of M42 - with the Trapezium at its heart?
Just a bit curious...
Cheers