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- Thread starter OSalcido
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mathman

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tony873004

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Garth

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It would be interesting to be able to observe how the galactic tide does perturb the Oort cloud.

Garth

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tony873004

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Garth

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Not true, such a scenario would produce a

[tex]V(r) \propto \frac{1}{\sqrt{r}} [/tex] velocity profile.

The Milky Way's rotation curve may be modelled in a first approximation by a density distribution of DM

[tex]\rho(r) = \frac{C}{(a^2 + r^2)}[/tex]

A more detailed treatment may be found in this eprint published today Disk galaxy rotation curves and dark matter distribution.

Garth

[tex]V(r) \propto \frac{1}{\sqrt{r}} [/tex] velocity profile.

The Milky Way's rotation curve may be modelled in a first approximation by a density distribution of DM

[tex]\rho(r) = \frac{C}{(a^2 + r^2)}[/tex]

A more detailed treatment may be found in this eprint published today Disk galaxy rotation curves and dark matter distribution.

Garth

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Garth

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In general over a range in which the velocity profile is 'flat' the further out stars orbit with less angular velocity around the galactic centre.

In a barred spiral the stars of the bar all rotate at the same angular velocity and the velocity profile linearly increases with distance from the centre.

Garth

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tony873004

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Not true, such a scenario would produce a

[tex]V(r) \propto \frac{1}{\sqrt{r}} [/tex] velocity profile.

This formula works well for the solar system, where ~99.8% of the solar system's mass is contained in the Sun. However, for a galaxy, the mass interior to your position, as you move outward, does not stay constant, so we can't simplify it with a 1. Starting with the formula for circular velocity, [tex]v(r)= \sqrt{\frac{GM}{r}}[/tex], G stays constant, so it can be simplified to [tex]v(r)\propto \sqrt{\frac{M}{r}}[/tex]. Unlike our solar system, as you move away from the center of the galaxy, the mass of the galaxy interior to your position significantly increases.

Even for our solar system, if you want an exact answer for an object's circular orbital velocity, using [tex]v(r)= \sqrt{\frac{GM}{r}}[/tex], you must add the masses of all the planets interior to it to the mass of the sun. This is especially true for Kuiper Belt objects, where your answers will be off by more than 2 meters/second if you use only the Sun's mass.

See question 3 on this page: http://corelli.sdsu.edu/courses/astro101_fall2006/readingquiz/Reading13_quiz.pdf

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Garth

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Okay, my error, I didn't read your "interior to your position" in the phrase "the galactic center can be approximated by a point mass containing all the mass of the galaxy interior to your position. "

For your explanation to hold good it means that for a flat rotation velocity profile

[tex]M(r) \propto r[/tex]

This requires: If the mass is in a flat disc

[tex]\rho (r) \propto \frac{1}{r}[/tex]

and if the mass is distributed spherically

[tex]\rho (r) \propto \frac{1}{r^2}[/tex],

Garth

For your explanation to hold good it means that for a flat rotation velocity profile

[tex]M(r) \propto r[/tex]

This requires: If the mass is in a flat disc

[tex]\rho (r) \propto \frac{1}{r}[/tex]

and if the mass is distributed spherically

[tex]\rho (r) \propto \frac{1}{r^2}[/tex],

Garth

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