# Effect of Sun movement on Planetary orbits?

1. Mar 16, 2007

### OSalcido

What is the effect (if any) of the sun's movement around the galactic center in relation to planetary orbits?

2. Mar 17, 2007

### mathman

The planetary orbits are computed relative to the sun. The motion of the sun within the galaxy translates into an added motion of all the planets within the galaxy. In other words, the entire solar system moves as a unit.

3. Mar 18, 2007

### tony873004

There is something called the galactic tide. The side of the solar system facing the galactic center is pulled harder than the side facing away. At the distance of the planets, its negligable and can be ignored. It won't cause more than a few meters of difference in the postions of the planets. But at greater distances, it makes a difference. The Oort Cloud is probably not spherical due to the galactic tide. The Sun's Hill Sphere is limited to about 1-2 light years due to the pull from the galactic center. So the Oort Cloud is probably more of a football (American) shape.

4. Mar 18, 2007

### Garth

We also have to remember the galactic gravitational field is not that of a central mass with negligible mass test particles orbiting it, but the gravitating mass itself is in orbit, they are the stars and gas of the galactic disk, together with a massive dark halo in which the galaxy is situated, which gives it a flat orbiting velocity/radius profile.

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

Garth

5. Mar 18, 2007

### tony873004

If I recall correctly, the galactic center can be approximated by a point mass containing all the mass of the galaxy interior to your position. Everything beyond you cancels out. I believe this is the reason for the flat velocity profile.

6. Mar 19, 2007

### Garth

Not true, such a scenario would produce a
$$V(r) \propto \frac{1}{\sqrt{r}}$$ velocity profile.

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

$$\rho(r) = \frac{C}{(a^2 + r^2)}$$

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

Garth

Last edited: Mar 19, 2007
7. Mar 19, 2007

### haiha

May i ask if all the stars and planets in the Galaxy rotate at the same angular velocity or the ones closer to the centre will have higher angular velocity?

8. Mar 19, 2007

### Garth

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

9. Mar 19, 2007

### tony873004

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, $$v(r)= \sqrt{\frac{GM}{r}}$$, G stays constant, so it can be simplified to $$v(r)\propto \sqrt{\frac{M}{r}}$$. 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 $$v(r)= \sqrt{\frac{GM}{r}}$$, 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.

10. Mar 19, 2007

### Garth

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
$$M(r) \propto r$$

This requires: If the mass is in a flat disc

$$\rho (r) \propto \frac{1}{r}$$

and if the mass is distributed spherically

$$\rho (r) \propto \frac{1}{r^2}$$,

Garth

Last edited: Mar 19, 2007