Re: Two suns

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.
"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.)


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

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 -- use the following formula:

S = 1 / abs( 1/P - 1/E )

where:

P = the orbital period of the body, as above
E = the orbital period of "earth" (E = 1 year for our solar system)
S = apparent/synodic period of the body as seen from "earth"

(P, E and S all have the same units.)

Note that, if you want to apply the first formula above to a planet
orbiting both components of a close binary star, you should set M
equal to the total mass of the component stars.

Does that help?

--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.
.

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