<p>Galactic orbits are not Keplerian: There is not a single massive centre, whose gravity attracts the sun, rather the whole disc, and dark matter halo which surrounds the galaxy. We can't use the inverse square law to calculate the force of gravity without knowing the distribution of mass in the galaxy.</p>

<p>Nevertheless, the Sun's orbit is roughly circular, so we can use kinematics to get some idea of the forces involved:  For circular motion $a=\frac{v^2}{r}$. The velocity of the sun is about <a href="http://hypertextbook.com/facts/2001/AngelaChan.shtml" rel="noreferrer">225000 m/s</a>, and we are at a radius of about 2.5e20 m from the centre. The formula above gives a very small centripetal acceleration of 2e-10 m/s²</p>

<p>However the sun is quite massive, 2e30 kg, so using $F=ma$, the force on the sun is of the order 4e20 N.  This is about 0.01 of the force acting on the Earth by the Sun. (3.6e22 N)</p>

CC BY-SA 3.0
2016-10-26T12:49:21.717Z
Just out of interest, if we model the galaxy as a disc with uniform density at a given radius (but varying with radius), does the mass distribution inside the sun's orbit, applied to $\frac{gMm}{r^2}$ come close to the estimate in your formulation?
Carl Witthoft
CC BY-SA 3.0
2016-10-26T13:41:13.293Z
You can *definitely* use the inverse square law to calculate the force of gravity. It's a bit harder than a standard keplerian system, but it can be done. You just need to figure out all the mass interior to the Sun's orbit which involves some integrating. If you couldn't do this, N-body simulations of galactic dynamics wouldn't work as that's basically what they do.
zephyr
CC BY-SA 3.0
2016-10-27T08:04:55.800Z
That's true, but as you say you would need to know the mass distribution inside and (since the galaxy is not spherically symmetric) outside the Sun's orbit. On the other hand, the kinematic approach above gives a solution without calculus, n-body dynamics. I'll edit to clarify.
James K
CC BY-SA 3.0
2016-10-27T08:20:30.287Z
@CarlWitthoft If you sum up what you can see, then no it doesn't. That's why dark matter is required. I believe the Sun's speed would be of order 2/3 what it is just using the inventory of baryonic matter. This deficit of course becomes much more serious at larger galactocentric radii.
ProfRob
CC BY-SA 3.0
2016-10-26T09:27:36.067Z
2016-10-27T08:28:28.817Z
2016-10-27T08:28:28.817Z
James K
<p>I know the Sun is orbiting around the Milky Way, but how strong is the attractive force between them (e.g. what is the order of magnitude in terms of newtons)?</p>

<orbit><the-sun><milky-way>
How strong is the force between the Sun and the centre of the Milky Way?
Gstestso
# How strong is the force between the Sun and the centre of the Milky Way?

(Asked by: username_0 on 2016-10-26)

I know the Sun is orbiting around the Milky Way, but how strong is the attractive force between them (e.g. what is the order of magnitude in terms of newtons)?

**Tags:** orbit, the-sun, milky-way


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## 1 answers

### Answer by username_1 on 2016-10-26. Score: 11 | Accepted answer

Galactic orbits are not Keplerian: There is not a single massive centre, whose gravity attracts the sun, rather the whole disc, and dark matter halo which surrounds the galaxy. We can't use the inverse square law to calculate the force of gravity without knowing the distribution of mass in the galaxy.

Nevertheless, the Sun's orbit is roughly circular, so we can use kinematics to get some idea of the forces involved: For circular motion $a=\frac{v^2}{r}$. The velocity of the sun is about [225000 m/s](http://hypertextbook.com/facts/2001/AngelaChan.shtml), and we are at a radius of about 2.5e20 m from the centre. The formula above gives a very small centripetal acceleration of 2e-10 m/s²

However the sun is quite massive, 2e30 kg, so using $F=ma$, the force on the sun is of the order 4e20 N. This is about 0.01 of the force acting on the Earth by the Sun. (3.6e22 N)

> **Comments:**
> **username_2**: Just out of interest, if we model the galaxy as a disc with uniform density at a given radius (but varying with radius), does the mass distribution inside the sun's orbit, applied to $\frac{gMm}{r^2}$ come close to the estimate in your formulation?  
> **username_3**: You can *definitely* use the inverse square law to calculate the force of gravity. It's a bit harder than a standard keplerian system, but it can be done. You just need to figure out all the mass interior to the Sun's orbit which involves some integrating. If you couldn't do this, N-body simulations of galactic dynamics wouldn't work as that's basically what they do.  
> **username_1**: That's true, but as you say you would need to know the mass distribution inside and (since the galaxy is not spherically symmetric) outside the Sun's orbit. On the other hand, the kinematic approach above gives a solution without calculus, n-body dynamics. I'll edit to clarify.  
> **username_4**: @username_2 If you sum up what you can see, then no it doesn't. That's why dark matter is required. I believe the Sun's speed would be of order 2/3 what it is just using the inventory of baryonic matter. This deficit of course becomes much more serious at larger galactocentric radii.  

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