<p>James K gave a good answer to this, but I just want to add that if the Sun was unmoving relative to the center of the Milky-way, it would fall towards the center along with your object.   The sun orbits within the Milky way with tangential velocity about 230 km/s.</p>

<p>It's rather pointless, but we can compare mass and distance and find the sweet spot where the gravitational pulls are equal on your object to the sun and towards the center of the Milky-way.</p>

<p>The mass of the center of the Milky-way, where, the center is everything that draws your object towards it, not just the big black hole at the center, precise numbers are impossible, but a rough <a href="https://www.universetoday.com/22790/mass-of-the-milky-way/" rel="nofollow noreferrer">estimate of 200 billion solar masses</a> can be used.  That might be low, but close enough.</p>

<p>The sun is roughly 26,000 light years from the center.   Using the inverse square rule, the distance would be the square root of 200 billion, or about 450,000, and divide the distance (26,000 light years) with that number and you get about 1/17th of 1 light year.    That's the sweet spot where the two gravitational tugs would be equal.</p>

<p>This demonstrates the uselessness of gravitational strength at astronomical distances.   Take the Moon for example.   It's under greater gravitational tug from the Sun than the Earth, but it still orbits the Earth (or, you could say it orbits both), but stronger gravity doesn't govern the hill sphere.   It's fun for calculating "how long would it take to hit the surface if I dropped it from . . . 1 light year", but beyond the occasional fun calculation, there's no real practical use.</p>

<p>With 3 bodies the equation gets much more exotic, especially if one is orbiting the other and you add a 3rd and you want to know which way it would fall.  The truth is, it would fall towards both, slowly, but if it got close enough to the sun it could get a gravitational kick, sending it further away for a time.   Total energy is always conserved, but calculating the movement of that 1 smaller object with both the sun and a uniform massive center is a more complicated.</p>

CC BY-SA 3.0
2017-12-18T04:36:10.067Z
2017-12-19T02:44:28.023Z
2017-12-19T02:44:28.023Z
userLTK
<p>The <a href="https://en.wikipedia.org/wiki/Equations_for_a_falling_body#Examples" rel="nofollow noreferrer">full equation</a> for the time for an object to drop is 
<span class="math-container">$$t =  \frac{ \arccos \Big( \sqrt{ \frac{x}{r} }\Big)  + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu }  }  \, r^{3/2},$$</span>
where <span class="math-container">$x$</span> is the radius of the sun, <span class="math-container">$r$</span> is distance of the object, and <span class="math-container">$\mu=GM=1.327\times10^{20}$</span>. (in SI units, so you will need to convert your distance to metres, and get a time in seconds.)</p>

<p>As <span class="math-container">$x&lt;&lt;r$</span> you can assume <span class="math-container">$\frac{x}{r}=0$</span> so the formula simplifies to 
<span class="math-container">$$\frac{\pi r^{3/2}}{2\sqrt{2\mu}}$$</span></p>

<p>For an object released at 0.1,0.5 or 1 light year that is 3000000 years, thirty million years or a hundred million years.</p>

<p>For the speed on impact, you can assume that the body will reach very close to the escape velocity of the sun, which is about 620 km/s. It matters little whether the body is released from 0.1, 0.5 or 1.0 lightyears, because the body will pick up speed only when it approaches the sun.</p>

<p>The last part of you question is unclear. If the body is at rest relative to the sun, and the various gravitational affects of other stars are ignored, then it will fall towards the sun. If on the otherhand the body is at rest relative to the centre of the galaxy, then it is moving extremely fast relative to the sun, much faster than the sun's escape velocity. It won't fall anywhere near to the sun, instead the sun will deflect it slightly as it falls towards the centre of the galaxy.</p>

CC BY-SA 3.0
2017-12-18T12:45:31.610Z
+1, but I think your free-fall times are a factor $\sqrt{1000}$ too small. Perhaps a g/kg error?
pela
CC BY-SA 3.0
2017-12-18T21:11:12.173Z
@pela  very likely,   I will edit.
James K
CC BY-SA 3.0
2017-12-18T23:55:57.553Z
Nicely done, @JamesK. I should've realised that it'd be close to escape v, but I'm in a permanent state of sleep deprivation & elderly confusion.
Yes, of course it'd fall to the sun, quite so -- although if we introduce the real-life bodies in orbit, esp. gas giants, then it'd likely miss by not very much, as suggested by TLK.
Phizzle
CC BY-SA 4.0
2017-12-17T21:22:49.410Z
2019-07-13T01:10:36.267Z
2019-07-13T01:10:36.267Z
James K
<p>A hypothetical situation which will help me with a real-life problem. 
Several stationary objects are placed at great distances from our Sun -- say 0.1 light year, 0.5 ly, 1.0 ly. Assuming no other influences; 
What velocity would each have when it reached the Sun?
How long would each take to reach the Sun?
Assuming the object-Sun path was perpendicular to the ecliptic &amp; the Sun had its actual motion, would the objects hit the Sun, or (roughly) by how much would they miss it?</p>

CC BY-SA 3.0
2017-12-17T20:28:08.633Z
Hint: You can use conservation of energy (potential energy + kinetic energy = const.) to get the answer for the first part.
AtmosphericPrisonEscape
CC BY-SA 3.0
2017-12-18T09:21:14.793Z
The word "stationary" is problematic. Stationary relative to what? The sun? The Galactic centre? The average local velocity of nearby stars?
James K
CC BY-SA 3.0
2017-12-18T22:59:35.950Z
Thanks @AtmosphericPrisonEscape, but it's far too many decades since I failed Physics.  i'd have to stare at that for a long time to remember what to do.
Phizzle
CC BY-SA 3.0
2017-12-18T23:03:46.580Z
And well caught, @JamesK.  I meant relative to the Sun, of course, treating it as a simple 2 body system (except for the galaxy through which the sytem is moving).
Phizzle
<the-sun><gravity><newtonian-gravity>
Gravitational attraction of Sun on distant object
Phizzle
# Gravitational attraction of Sun on distant object

(Asked by: username_0 on 2017-12-17)

A hypothetical situation which will help me with a real-life problem.
Several stationary objects are placed at great distances from our Sun -- say 0.1 light year, 0.5 ly, 1.0 ly. Assuming no other influences;
What velocity would each have when it reached the Sun?
How long would each take to reach the Sun?
Assuming the object-Sun path was perpendicular to the ecliptic & the Sun had its actual motion, would the objects hit the Sun, or (roughly) by how much would they miss it?

**Tags:** the-sun, gravity, newtonian-gravity

> **Comments:**
> **username_1**: Hint: You can use conservation of energy (potential energy + kinetic energy = const.) to get the answer for the first part.  
> **username_2**: The word "stationary" is problematic. Stationary relative to what? The sun? The Galactic centre? The average local velocity of nearby stars?  
> **username_0**: Thanks @username_1, but it's far too many decades since I failed Physics.  i'd have to stare at that for a long time to remember what to do.  
> **username_0**: And well caught, @JamesK.  I meant relative to the Sun, of course, treating it as a simple 2 body system (except for the galaxy through which the sytem is moving).  

---

## 2 answers

### Answer by username_2 on 2017-12-17. Score: 8 | Accepted answer

The [full equation](https://en.wikipedia.org/wiki/Equations_for_a_falling_body#Examples) for the time for an object to drop is
$$t = \frac{ \arccos \Big( \sqrt{ \frac{x}{r} }\Big) + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu } } \, r^{3/2},$$
where $x$ is the radius of the sun, $r$ is distance of the object, and $\mu=GM=1.327\times10^{20}$. (in SI units, so you will need to convert your distance to metres, and get a time in seconds.)

As $x< you can assume $\frac{x}{r}=0$ so the formula simplifies to
$$\frac{\pi r^{3/2}}{2\sqrt{2\mu}}$$

For an object released at 0.1,0.5 or 1 light year that is 3000000 years, thirty million years or a hundred million years.

For the speed on impact, you can assume that the body will reach very close to the escape velocity of the sun, which is about 620 km/s. It matters little whether the body is released from 0.1, 0.5 or 1.0 lightyears, because the body will pick up speed only when it approaches the sun.

The last part of you question is unclear. If the body is at rest relative to the sun, and the various gravitational affects of other stars are ignored, then it will fall towards the sun. If on the otherhand the body is at rest relative to the centre of the galaxy, then it is moving extremely fast relative to the sun, much faster than the sun's escape velocity. It won't fall anywhere near to the sun, instead the sun will deflect it slightly as it falls towards the centre of the galaxy.

> **Comments:**
> **username_4**: +1, but I think your free-fall times are a factor $\sqrt{1000}$ too small. Perhaps a g/kg error?  
> **username_2**: @username_4  very likely,   I will edit.  
> **username_0**: Nicely done, @JamesK. I should've realised that it'd be close to escape v, but I'm in a permanent state of sleep deprivation & elderly confusion. Yes, of course it'd fall to the sun, quite so -- although if we introduce the real-life bodies in orbit, esp. gas giants, then it'd likely miss by not very much, as suggested by TLK.  

---

### Answer by username_3 on 2017-12-18. Score: 3

username_2 gave a good answer to this, but I just want to add that if the Sun was unmoving relative to the center of the Milky-way, it would fall towards the center along with your object. The sun orbits within the Milky way with tangential velocity about 230 km/s.

It's rather pointless, but we can compare mass and distance and find the sweet spot where the gravitational pulls are equal on your object to the sun and towards the center of the Milky-way.

The mass of the center of the Milky-way, where, the center is everything that draws your object towards it, not just the big black hole at the center, precise numbers are impossible, but a rough [estimate of 200 billion solar masses](https://www.universetoday.com/22790/mass-of-the-milky-way/) can be used. That might be low, but close enough.

The sun is roughly 26,000 light years from the center. Using the inverse square rule, the distance would be the square root of 200 billion, or about 450,000, and divide the distance (26,000 light years) with that number and you get about 1/17th of 1 light year. That's the sweet spot where the two gravitational tugs would be equal.

This demonstrates the uselessness of gravitational strength at astronomical distances. Take the Moon for example. It's under greater gravitational tug from the Sun than the Earth, but it still orbits the Earth (or, you could say it orbits both), but stronger gravity doesn't govern the hill sphere. It's fun for calculating "how long would it take to hit the surface if I dropped it from . . . 1 light year", but beyond the occasional fun calculation, there's no real practical use.

With 3 bodies the equation gets much more exotic, especially if one is orbiting the other and you add a 3rd and you want to know which way it would fall. The truth is, it would fall towards both, slowly, but if it got close enough to the sun it could get a gravitational kick, sending it further away for a time. Total energy is always conserved, but calculating the movement of that 1 smaller object with both the sun and a uniform massive center is a more complicated.


---

