<p>There are a number of misconceptions with this question.</p>

<p>First off, the "gravitational pull" (which I'm interpreting as gravitational force) by the Milky Way on the Earth is seven orders of magnitude smaller than that exerted by the Sun.</p>

<p>Secondly, gravitational force is the wrong metric. The Newtonian gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth. Yet all but a tiny, tiny minority of professionals who deal with the solar system will say that the Moon does indeed orbit the Earth.</p>

<p>Thirdly, it's not correct to look at "orbit" as being a mutually exclusive concept. That the Moon does indeed orbit the Earth doesn't mean that it doesn't orbit the Sun. It does. The Moon not only orbits the Earth and Sun, but it also orbits the Milky Way, the Local Group, and the Local Supercluster.</p>

<p><br></p>

<p>So what is the right way of looking at orbits amongst a hierarchy of masses? One issue that needs to be addressed is what the word "orbit" means. Note that people do write and talk about parabolic and hyperbolic "orbits". I'm assuming that "orbit" in the context of this question means a bounded trajectory, which would rule out calling a hyperbolic or parabolic trajectory an "orbit".</p>

<p>That leads to the first of three characteristics of whether object A can be said to be orbiting object B: Object A can be said to be orbiting object B only if object A is gravitationally bound to object B. The appropriate metric here is energy, at least for cosmologically small distances. (Note: The distance between the Milky Way and the Andromeda Galaxy is very small in a cosmological sense.) From the perspective of a frame centered at object B, the total mechanical energy of object A needs to be negative he mechanical energy to be able to say that object A is (at least temporarily) orbiting object B.</p>

<p>Bounded orbits remain bounded forever in the two body point mass problem. That may not be the case when other larger objects are involved. This leads to a second criterion: Object A can be said to be orbiting object B only if the gravitational influences of larger objects on object A are but small perturbative effects that don't result in instabilities. The appropriate metric here is an energy-based sphere of influence. The two most widely used spheres are the Hill sphere and the Laplace sphere. For example, because the Moon is gravitationally bound to the Earth and because the Moon is well inside the Earth's Hill and Laplace spheres with respect to the Sun, it is appropriate to say that the Moon does indeed orbit the Earth.</p>

<p>What about the Sun and the Milky Way (and even larger objects such as the Local Group)? Can the Moon also be said to orbit those? The answer is yes. One way to look at it is to ask what would happen if the Earth suddenly disappeared. Would the Moon's path around the Sun change much? The answer is no. The Moon would still be orbiting the Sun at about one astronomical unit. The same applies to the Earth with respect to the Milky Way. The Earth's trajectory about the galaxy would not change by much at all if the rest of the solar system magically disappeared. This point of view however leads to some troubling issues. Planets (and stars and galaxies) don't magically disappear. Moreover, if Saturn somehow did magically disappear, its rings and all but its very outermost moons would be on hyperbolic trajectories with respect to the Sun. A way around this magic is to say that if object A orbits object B, and object B orbits object C, then object A also orbits object C. "Orbit" is not a mutually exclusive concept.</p>

CC BY-SA 3.0
2015-01-24T16:22:57.357Z
2015-01-24T16:22:57.357Z
David Hammen
<p>The stars do orbit the center every 250 Million years according to a <a href="https://en.wikipedia.org/wiki/Galactic_year" rel="nofollow noreferrer">galactic year</a>, That's 20-25 times since the formation of the solar system. </p>

<p>The complex shape of the galaxy, with a disk, bulge, bar and arms adds many non concentric gravity poles to the equation. Easiest to imagine is a wave of movement relative to the disk. It would be nice to see a gravity map of a galaxy in 3D to see that it is not a 1 body at the center kind of gravity map. </p>

<p>Some (many? n%) stars can result from the pressure front of supernovae that happen inside star nurseries, which gives them erratic vectors compared to the disk and the arms. Perhaps if there wasn't the constant energy of supernovae affecting the birth and movement of objects in the galaxy, there would be rings as seen for Saturn. </p>

<p>There's a parallel with mapping rocks in the solar system. If you measure the denser zones to try to figure out where every rock was moving, you'd find your predictions soon become chaos of spirals and local deflection, rather than simple trajectories of a predictable fashion. </p>

CC BY-SA 4.0
2018-05-30T02:13:58.213Z
you say "The complex shape of the universe with a disk, bulge, bar and arms ..." Did you mean to say galaxy instead of universe?
Jim421616
CC BY-SA 4.0
2018-05-29T08:33:33.790Z
2018-05-30T22:46:13.657Z
2018-05-30T22:46:13.657Z
bandybabboon
<p>The so called <a href="https://en.wikipedia.org/wiki/Inverse-square_law" rel="nofollow noreferrer">inverse square law</a> makes gravity dominate locally. The Moon orbits Earth, Earth orbits the Sun, and <a href="https://en.wikipedia.org/wiki/Galactic_year" rel="nofollow noreferrer">the Sun orbits the Galactic center</a>! Because the distance from Earth to the Moon is 1/500 the distance from Earth to the Sun, it is Earth's gravitational field which dominates the orbit of the Moon.</p>

<p>And it is not a single thing which attracts mass to orbit the galaxy. It is the sum of mass of everything in it. The center is of course the galactic center of mass.</p>

CC BY-SA 3.0
2015-01-24T13:21:48.047Z
Yes, the Earth is about 500 times nearer to the Moon (much closer to 400 actually), but the Sun is also 333,000 times more massive than the Earth making the Sun-Moon gravitational attraction over twice as strong. So it's actually the Sun's gravitational field which dominates the the orbit of the Moon.
David H
CC BY-SA 3.0
2015-01-24T14:36:17.233Z
@DavidH Are you trying to make things difficult here ;-) So what is your answer?
LocalFluff
CC BY-SA 3.0
2015-01-24T14:49:53.433Z
DavidH is correct. The gravitational force exerted by the Sun on the Moon is more than double that of the gravitational force exerted by the  Earth on the Moon.
David Hammen
CC BY-SA 3.0
2015-01-24T15:15:09.553Z
@DavidHammen Makes sense. As per your answer [here](http://physics.stackexchange.com/questions/148656/why-dont-stars-have-keplerian-orbits), is the inverse-square law even applicable in this case (Sun v. galactic center)? I would think not.
HDE 226868
CC BY-SA 3.0
2015-01-25T00:05:54.003Z
Sun, earth, moon, solar system all orbit center of galaxy with a period of about 225 million years: http://en.wikipedia.org/wiki/Galactic_year
Wayfaring Stranger
CC BY-SA 3.0
2015-01-25T09:18:20.387Z
If you want to see what the centre of the galaxy looks like, you can visit https://www.universeguide.com/fact/galaxy for what it looks like.  The galactic centre is in the constellation of https://www.universeguide.com/constellation/sagittarius
MiscellaneousUser
CC BY-SA 3.0
2016-11-12T08:29:27.840Z
The Earth and the Moon orbit the Sun at the same speed and distance **on average**. This means we can *almost* ignore the effects of the Sun on the Earth/Moon system and just consider the gravity between the two local objects. If the Sun was to suddenly disappear, the Earth and Moon would continue to orbit each other, and their barycentre would travel in approximately a straight line, whereas at the moment, the barycentre travels in an approximate circle around the Sun. In both cases, the Earth and Moon orbit the barycentre of the Earth/Moon system, which is inside the Earth.
CJ Dennis
CC BY-SA 4.0
2015-01-24T13:08:56.453Z
2018-05-23T07:24:00.030Z
2018-05-23T07:24:00.030Z
LocalFluff
<p>At the center of the milky way, there must be some giant mass that keeps the galaxy from just floating away from itself. So, obviously, we feel the immense gravitational pull of whatever this is. But, on a smaller scale we can see this with the earth and the sun. The sun keeps us from spinning into oblivion with its gravitational pull. The question I am asking is, why do we orbit the sun, but not the center of the galaxy, which has a much stronger gravitational pull?</p>

<gravity>
Why don't we orbit the center of our galaxy?
user5400
# Why don't we orbit the center of our galaxy?

(Asked by: username_0 on 2015-01-24)

At the center of the milky way, there must be some giant mass that keeps the galaxy from just floating away from itself. So, obviously, we feel the immense gravitational pull of whatever this is. But, on a smaller scale we can see this with the earth and the sun. The sun keeps us from spinning into oblivion with its gravitational pull. The question I am asking is, why do we orbit the sun, but not the center of the galaxy, which has a much stronger gravitational pull?

**Tags:** gravity


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## 3 answers

### Answer by username_4 on 2015-01-24. Score: 6 | Accepted answer

The so called [inverse square law](https://en.wikipedia.org/wiki/Inverse-square_law) makes gravity dominate locally. The Moon orbits Earth, Earth orbits the Sun, and [the Sun orbits the Galactic center](https://en.wikipedia.org/wiki/Galactic_year)! Because the distance from Earth to the Moon is 1/500 the distance from Earth to the Sun, it is Earth's gravitational field which dominates the orbit of the Moon.

And it is not a single thing which attracts mass to orbit the galaxy. It is the sum of mass of everything in it. The center is of course the galactic center of mass.

> **Comments:**
> **username_5**: Yes, the Earth is about 500 times nearer to the Moon (much closer to 400 actually), but the Sun is also 333,000 times more massive than the Earth making the Sun-Moon gravitational attraction over twice as strong. So it's actually the Sun's gravitational field which dominates the the orbit of the Moon.  
> **username_4**: @DavidH Are you trying to make things difficult here ;-) So what is your answer?  
> **username_1**: DavidH is correct. The gravitational force exerted by the Sun on the Moon is more than double that of the gravitational force exerted by the  Earth on the Moon.  
> **username_6**: @DavidHammen Makes sense. As per your answer [here](http://physics.stackexchange.com/questions/148656/why-dont-stars-have-keplerian-orbits), is the inverse-square law even applicable in this case (Sun v. galactic center)? I would think not.  
> **username_7**: Sun, earth, moon, solar system all orbit center of galaxy with a period of about 225 million years: http://en.wikipedia.org/wiki/Galactic_year  
> **username_8**: If you want to see what the centre of the galaxy looks like, you can visit https://www.universeguide.com/fact/galaxy for what it looks like.  The galactic centre is in the constellation of https://www.universeguide.com/constellation/sagittarius  
> **username_9**: The Earth and the Moon orbit the Sun at the same speed and distance **on average**. This means we can *almost* ignore the effects of the Sun on the Earth/Moon system and just consider the gravity between the two local objects. If the Sun was to suddenly disappear, the Earth and Moon would continue to orbit each other, and their barycentre would travel in approximately a straight line, whereas at the moment, the barycentre travels in an approximate circle around the Sun. In both cases, the Earth and Moon orbit the barycentre of the Earth/Moon system, which is inside the Earth.  

---

### Answer by username_1 on 2015-01-24. Score: 8

There are a number of misconceptions with this question.

First off, the "gravitational pull" (which I'm interpreting as gravitational force) by the Milky Way on the Earth is seven orders of magnitude smaller than that exerted by the Sun.

Secondly, gravitational force is the wrong metric. The Newtonian gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth. Yet all but a tiny, tiny minority of professionals who deal with the solar system will say that the Moon does indeed orbit the Earth.

Thirdly, it's not correct to look at "orbit" as being a mutually exclusive concept. That the Moon does indeed orbit the Earth doesn't mean that it doesn't orbit the Sun. It does. The Moon not only orbits the Earth and Sun, but it also orbits the Milky Way, the Local Group, and the Local Supercluster.

So what is the right way of looking at orbits amongst a hierarchy of masses? One issue that needs to be addressed is what the word "orbit" means. Note that people do write and talk about parabolic and hyperbolic "orbits". I'm assuming that "orbit" in the context of this question means a bounded trajectory, which would rule out calling a hyperbolic or parabolic trajectory an "orbit".

That leads to the first of three characteristics of whether object A can be said to be orbiting object B: Object A can be said to be orbiting object B only if object A is gravitationally bound to object B. The appropriate metric here is energy, at least for cosmologically small distances. (Note: The distance between the Milky Way and the Andromeda Galaxy is very small in a cosmological sense.) From the perspective of a frame centered at object B, the total mechanical energy of object A needs to be negative he mechanical energy to be able to say that object A is (at least temporarily) orbiting object B.

Bounded orbits remain bounded forever in the two body point mass problem. That may not be the case when other larger objects are involved. This leads to a second criterion: Object A can be said to be orbiting object B only if the gravitational influences of larger objects on object A are but small perturbative effects that don't result in instabilities. The appropriate metric here is an energy-based sphere of influence. The two most widely used spheres are the Hill sphere and the Laplace sphere. For example, because the Moon is gravitationally bound to the Earth and because the Moon is well inside the Earth's Hill and Laplace spheres with respect to the Sun, it is appropriate to say that the Moon does indeed orbit the Earth.

What about the Sun and the Milky Way (and even larger objects such as the Local Group)? Can the Moon also be said to orbit those? The answer is yes. One way to look at it is to ask what would happen if the Earth suddenly disappeared. Would the Moon's path around the Sun change much? The answer is no. The Moon would still be orbiting the Sun at about one astronomical unit. The same applies to the Earth with respect to the Milky Way. The Earth's trajectory about the galaxy would not change by much at all if the rest of the solar system magically disappeared. This point of view however leads to some troubling issues. Planets (and stars and galaxies) don't magically disappear. Moreover, if Saturn somehow did magically disappear, its rings and all but its very outermost moons would be on hyperbolic trajectories with respect to the Sun. A way around this magic is to say that if object A orbits object B, and object B orbits object C, then object A also orbits object C. "Orbit" is not a mutually exclusive concept.


---

### Answer by username_2 on 2018-05-29. Score: 0

The stars do orbit the center every 250 Million years according to a [galactic year](https://en.wikipedia.org/wiki/Galactic_year), That's 20-25 times since the formation of the solar system.

The complex shape of the galaxy, with a disk, bulge, bar and arms adds many non concentric gravity poles to the equation. Easiest to imagine is a wave of movement relative to the disk. It would be nice to see a gravity map of a galaxy in 3D to see that it is not a 1 body at the center kind of gravity map.

Some (many? n%) stars can result from the pressure front of supernovae that happen inside star nurseries, which gives them erratic vectors compared to the disk and the arms. Perhaps if there wasn't the constant energy of supernovae affecting the birth and movement of objects in the galaxy, there would be rings as seen for Saturn.

There's a parallel with mapping rocks in the solar system. If you measure the denser zones to try to figure out where every rock was moving, you'd find your predictions soon become chaos of spirals and local deflection, rather than simple trajectories of a predictable fashion.

> **Comments:**
> **username_3**: you say "The complex shape of the universe with a disk, bulge, bar and arms ..." Did you mean to say galaxy instead of universe?  

---

