NationStates Jolt Archive


Geek... overflowing..... urrkkk....

Neo Bretonnia
05-12-2007, 20:53
So I've been reading a Dune novel and kinda immersing myself in the whole Dune mythos, when I stumbled upon the tidbit that the planet Arrakis (Dune) orbits the star Canopus.

So I went over to Celestia and lo and behold, Canopus is a yellow/white supergiant star about 312 LY from here almost directly over the Earth's South Pole.

So I wondered... If there WERE a habitable planet like Arrakis orbiting Canopus, just how far from the sun would the planet have to be?

So looking at Wikipedia under the heading "Habitable Zone" I discovered that the ideal orbit for a planet around a star is the square root of the ratio of its brightness to our own sun's, in AUs.

So d=Sqrt(Luminosity Star/Luminosity Our Sun)

The luminosity of Canopus is 13,800xOur Sun

d=Sqrt(13,800/1)
d=Sqrt(13,800)
d=117 AUs

That's pretty far... Pluto orbits our Sun at just under 40 AUs average. I took a look at it in Celestia at 117 AUs distance and while it appears smaller in the sky than our Sun does from here, it would be about as bright.

That led me to wonder... How long would an Arrakian year be???


T=2(pi)Sqrt(a^3/GM)

Where

T= Orbital Period in Seconds
a = orbital radius (approx) (km)
G = Gravitational Constant
M = Mass of the Star (kg)

Well we know the orbital raduis of Arrakis is about 117 AUs. In km that's

1 AU = 149.6×10^6 km

so 149.6×10^6 * 117 = 175 x 10^8 km.

G = 6.67 x 10 ^ -8 by definition

The Mass of Canopus is about 8 x Our Sun, which has a mass of 1.98 x 10^30 kg which comes out to about 1.6 x 10 ^31 kg. Plugging all that in:

T=2(pi)Sqrt((175 x 10^8 )^3/6.67 x 10 ^-8 * 1.6 x 10 ^31))
T=2(pi)Sqrt((175x10^8 )^3/1.1 x 10^23)
T=2(pi)Sqrt(5.26 x 10 ^ 30/1.1 x 10 ^23)
T=2(pi)47818181.9
T= 300298181 seconds

or


9.5 Earth Years...

So either that number is grossly inaccurate (considering Pluto's orbit around the Earth is in hundreds of years) Or the planet would truly fly through space at an alarming speed due to the extremely high mass of Canopus.

Hmm...

Orbital Circumference = 117.00*3.14 = 367 AUs.

38.67 AUs traveled per Earth Year.
or 3557785263 Miles per Earth Year or
406,139 Miles per hour.

Earth moves at about 10,502 Miles per hour in its orbit (around a much less massive star and at a much closer distance.)

I dunno what do you guys think?
Tagmatium
05-12-2007, 20:56
Mmmmm... sciency.
Gay Valley
05-12-2007, 20:59
I think you should possibly try going out. Or atleast opening your curtains and letting some light in.
Tagmatium
05-12-2007, 21:00
I think you should possibly try going out. Or atleast opening your curtains and letting some light in.
S/he could be doing this after being out all day.

I just don't understand it as my brain works with more historical stuff.
Cannot think of a name
05-12-2007, 21:02
http://www.lechatnoirboutique.com/prodimages/Coffee%20Mug%20-%20Far%20Side%20My%20Brain%20is%20Full-sm.jpg
The Turkic Shahdom
05-12-2007, 21:04
I doubt it is wrong, the mass of the planet is many times more than pluto's combined with the large and assuming that you have cruched all the right numbers you should be right
Isidoor
05-12-2007, 21:06
that's probably the geekiest thing I've seen this year
Whereyouthinkyougoing
05-12-2007, 21:07
that's probably the geekiest thing I've seen this year

And in December that's really saying something.
ElJefe
05-12-2007, 21:08
your units are inconsistent.

G = 6.67300 × 10-11 m3 kg-1 s-2, so you need to have 'a' in m, not km.

I get 446.5 earth years, assuming the rest of your values are correct.
Free Soviets
05-12-2007, 21:17
that's probably the geekiest thing I've seen this year

i think you spoke too soon.

your units are inconsistent.

G = 6.67300 × 10-11 m3 kg-1 s-2, so you need to have 'a' in m, not km.

I get 446.5 earth years, assuming the rest of your values are correct.
ElJefe
05-12-2007, 21:18
and then, circumference = 2*pi*r = 6.8e10 miles, over 4 million hours = 17000 mph.
Neo Bretonnia
05-12-2007, 21:29
your units are inconsistent.

G = 6.67300 × 10-11 m3 kg-1 s-2, so you need to have 'a' in m, not km.

I get 446.5 earth years, assuming the rest of your values are correct.

and then, circumference = 2*pi*r = 6.8e10 miles, over 4 million hours = 17000 mph.

Those values seem much more reasonable. Thanks!
Setto Land
06-12-2007, 00:18
it may be geeky, but if i could work that kind of stuff out i would.
Call to power
06-12-2007, 00:59
http://www.myconfinedspace.com/wp-content/uploads/2007/03/lame.gif

I dub thee the nerd thread
Saige Dragon
06-12-2007, 01:52
I could comment, but I won't comment because in a thread full of geeks I'd stand out as not so geeky.