| MadSci Network: Astronomy |
Kevin,
All objects that orbit about stars, do so in partnership with the star.
In other words, a planet and its star orbit about a common center of
gravity. That is why a star with a massive planet appears to wobble
slightly as the planet moves from one side of the star to the other in its
orbit. To calculate how much the sun would apparently wobble, one needs
the distance between the centers of the Sun and Jupiter. One also needs
the relative masses of the two bodies.
Jupiter is 778,300,000 km from the Sun. Assume this distance is from their respective centers of mass. The Sun is 1,000 times more massive than Jupiter, having a mass of 2 x 1030 kg compared to Jupiter's mass of 2 x 1027 kg.
Immagine that these two objects are connected by a solid, massless rod and you want to balance them on a knife edge. The point at which they balance is the point around which the Sun will wobble due to Jupiter's mass.
The mathematics follow:
MSun x DSun = MJup x DJup
Where M is the mass of the object and D is the
length of the rod from the center of the object to the balancing knife
edge. Using the values above, the equation becomes
1000 x DSun = 1 x DJupor
DJup is therefore 1000 times the distance of
DSun. Numerically we have that 1/1000 of 778,300,000 km
is 778,300 km. The radius of the Sun is 695,000 km.
According to these calculations, the Sun wobbles about a point just outside its surface during the course of Jupiter's 12 Earth-year trip around the Sun.
The effect of other planets on the Sun's wobble would be smaller, proportionate to their masses and distances from the Sun.
I hope this helps.
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