# Galaxy rotation and Kepler law

by pixel01
Tags: galaxy, kepler, rotation
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 P: 691 According to Kepler third law, the ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. If I can apply this to the rotaion of galaxy, meaning stars in inner part will orbit much faster than the outer ones. But it seems not. Anyonoe please explain to me this. Thanks.
 Sci Advisor PF Gold P: 1,542 That's because the Sun is a fixed mass at the center of our solar system. But as you get further from the galactic core, there are more stars interior to you, adding to the mass that you would use to compute your orbital velocity. Technically, this happens in the solar system. Earth orbits the combined mass of the Sun, Mercury and Venus. So it orbits a little faster than it would if Mercury and Venus did not exist. But since Mercury and Venus are insignificant compared to the Sun, their effect is negligible. But in the galaxy, the additional mass interior to you as you move out is not negligible.
P: 185
 Quote by tony873004 That's because the Sun is a fixed mass at the center of our solar system. But as you get further from the galactic core, there are more stars interior to you, adding to the mass that you would use to compute your orbital velocity.
This is interesting in that I have been researching this topic throughout the day today. It seems that in fact after a certain distance from a galactic core, orbital velocities of stars become fairly constant in apparent contradiction of your statement.

The reason appears to be the presence of vast amounts of dark matter surrounding galaxies; the gravitational effect being the explanation for the faster-than-expected orbital velocities of outer systems. My reading suggests that the ratio of dark matter to luminescent matter is in the ballpark of 100:1.

 P: 10 Galaxy rotation and Kepler law As you know Kepler's laws are true for simple garvitional fields , but talking about galaxies and stars in it , the gravity isn't the same as the gravity between two "very small" objects. To sum up what Tony873004 and WhyIsItSo said , in situations of this kind the force is: -Kr instead of -K/r2! just for Mg use 4$$\pi$$r2$$\rho$$ to get what I'm saying Thanks a lot!

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