[TML] The proper year length of a super jovian

[Garry Ward]

----- Original Message ----- 
From: "Knapp" <magick.crow at gmail.com>
To: "The Traveller Mailing List" <tml at travellercentral.com>
Sent: Sunday, February 03, 2008 12:46 PM
Subject: [TML] The proper year length of a super jovian


>This makes the length of year for a planet; based on WG.pgf, page 15, 2.1.1
>"where a is the orbital distance in AU and m is the mass of the
>primary, in solar masses. If the planet has significant mass compared
>to the
>star (like a large superjovian and a small red star or brown dwarf),
>add the masses together. Result is in Earth standard years."

>My question for all you space experts is what is meant by
>"Significant" here? I am now using, significant is when the the mass
>of the super is divided by the mass of the star and it is over 60%
>I am thinking that this changes the year because the barycenter is no
>longer in the center of the planet. Is that right? Also how would this
>effect other planets with them?

>Thoughts anyone?
Don't know about barycenters, but as the masses go up, the gravitational 
pull goes up which means that you need to go faster sideways the balance the 
speed towards so it takes less time to cover the circumfernence at a given 
orbital radius.

As I understand the forumla for orbital period, SQRT(distance in au^3 
divided by the sum of the masses in solar masses). If the mass of the 
orbiting object is small, it can be ignored. Or, simply convert it into 
solar masses and add them any way. After all, if the orbiting object is only 
.00001 solar masses, it really won't affect the final number that much.

If it is .6, then it will.

1^3 / 1 = 1 year
1^3/1.0001 = .9999499 or an hour or so less than a year.
1^3/1.1 =  .95 year or about 9 days less.
1^3/1.01 = .99550371 or about 2 days less than a year.
1^3/1.001 = .9995003 or abount 4 hours less than a year.

By and large, I'd say just add the masses anyway.

Garry




>Thanks,
>Douglas E Knapp


-- 
Ein Leben ohne Mops ist möglich,
doch völlig sinnlos.
-Loriot
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