[TML] The proper year length of a super jovian

[Jerry W Barrington]

On 2/3/08 2:20 PM, "Garry Ward" <garry.e.ward at worldnet.att.net> wrote:

> 1^3 / 1 = 1 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.
> 1^3/1.0001 = .9999499 or an hour or so less than a year.
> 
> By and large, I'd say just add the masses anyway.

Likewise, I'd just add them all, but I'm a stickler for accuracy.
Especially if you're still looking at this for a computer program, why not?

If you don't want to worry about little ones, note the the 1000:1 ratio
means that ignoring the secondary's mass leave the result accurate to 3
significant digits (.999 vs. 1).  Same for other ratios, same accuracy as
the number of zeroes in the ratio (100:1 -> .99 vs. 1, etc)