Re: Two suns

Ilmari Karonen <usenet2@xxxxxxxxxxxxxx> wrote:
Tina Hall <Tina_Hall@xxxxxxxxxxx> kirjoitti 11.11.2007:
FennelGiraffe <sraarytvenssr@xxxxxxxxxxxxxxxxx> wrote:

[discussion of solar system with second sun in place of Jupiter]

On the other hand, if you put a sun-sized mass where Jupiter is, it
would orbit faster than Jupiter does (by a factor of the square
root of two, I think).

How many times bigger (massive, whatever matters) is the sun than
Jupiter?

Much bigger.

The time it takes for two bodies to orbit one another depends on
their distance and on the _sum_ of their masses. For the sun and
Jupiter, you can basically assume that the total mass is about the
same as the mass of the sun by itself. If you replace Jupiter with a
second sun, however, the total mass approximately doubles.

(Specifically, the orbital period is inversely proportional to the
square root of the sum of the masses, and directly proportional to the
square root of the cube of the distance, so doubling the total mass
indeed decreases the orbital period by the square root of two.

Ok, that's numbers I can work with (no 'funny' calculations).

The problem is just that I don't know the masses of any of these, and
for size I'm just looking at it from the planet's viewpoint, how does it
look. :)

(No doubt that can be calculated, too. I'm just at a loss right now for
how.)

"Distance" above strictly means the sum of the semi-major axes of the
orbits of each body, but for nearly circular orbits simply "distance"
will do.)

Which one is the semi-major axis?

Really, this is going in a completely unintuitive (for me)
direction. How about something straightforward?

Sun2 takes n planet-days a year at X orbit radius around Sun1, with
n and X shifting at <whatever ratio>. The X orbit radius in AU is
fine (multiples of planet distance to Sun1). If n (how long it takes
for a full circle) depends on its size (mass, whatever), then that's
just another variable. A Jupiter sun where it is is a small star,
fine, a sun sized sun where Jupiter is is (IIRC you said) a quarter
its size, also fine. What I don't know is the rate at which it
shifts, and how the length of its orbit shifts with that (or even
just its orbit at all).

See above. If you want a formula, try:

P = sqrt( d^3 / (M+m) )

where:

d = distance of orbiting body in AU
M = mass of sun in solar masses (M = 1 for our solar system)
m = mass of orbiting body in solar masses (approx. zero for
planets) P = orbital period in years

Ok. Thanks.

That's the "true" or sidereal orbital period. If you want the
synodic period, which is what FennelGiraffe seems to be using -- i.e.
the time it takes for the body to return to the same position with
respect to the sun and the earth [...]

The point at which they align again, somewhat further around the
primary? That's a bit unintuitive, awkward for me to work with.

Does that help?

Yes, thanks.

--
Tina
WIP: [Untitled]: 12427 words
WISuspension: Seasons & Elements trilogy | Magic Earth series
Posted to Usenet newsgroup rec.arts.sf.composition.

.

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