I'm trying to see how far can our star reaches with its gravity. I'm asking if anyone could give info as to what's our star's limit or the furthest object found in our solar system.


The Sun's gravity extends infinitely, but eventually solar objects would be unstable due to the influence of other stars. The minor planet "Sedna" has an orbit which takes it nearly 1000 AU (0.016 light years) from the sun at its furthest point (but now it is a lot closer)

It is also thought that billions of comets must orbit in the outer part of the solar system, out to 50000AU, or 0.8 light years, (or possibly further) forming the Oort Cloud. However, at such distances, they could not be directly observed. This marks the greatest distance at which orbiting solar system bodies can be found.

  • $\begingroup$ And for example, is there any limit of mass and distance? For example, jupiter can't orbit at 1000AU or it will fly away, or Saturn cant orbit at 2000AU or it fly away, is there any mass-distance orbit relation? $\endgroup$ May 29 '17 at 8:07
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    $\begingroup$ @AlbertoMartínez No, the mass of a planet has no influence on its orbit. The only thing you need to create a stable orbit is the right velocity. If you want you could place jupiter 1km aboves sun surfaces if you make sure jupiter has the right velocity $\endgroup$
    – RononDex
    May 29 '17 at 8:27
  • $\begingroup$ @RononDex thanks for the info, appreciate it! :) $\endgroup$ May 29 '17 at 10:43
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    $\begingroup$ The word "infinitely" doesn't really have meaning here. Even gravitational waves propagate, so despite the simple 1/R^2 model, one can claim that the Sun's field reaches no farther than the speed of light times the age of the Sun. (yes, I know that's inaccurate since the sun didn't pop into existence) $\endgroup$ May 30 '17 at 13:24
  • $\begingroup$ Corrected, it was meant to be 0.015 ly and 50000 AU $\endgroup$
    – James K
    May 30 '17 at 16:37

There's no straight forward answer. In the solar-system, which is well ordered, objects that are in stable orbits, and not too elliptical, have well defined spheres of influence. Planet 9, if/when it's discovered, will probably have the largest sphere of influence for known solar-system objects. Currently, Neptune has the largest.

If the stars near the sun were static relative to each other, the Sun's sphere of influence could be calculated and it would probably extend between 2 and 3 light years. But because the stars are not static, the sphere of influence is constantly changing and stars (probably) exchange outer, loosely orbiting debris fairly frequently.

The Oort cloud by this article is thought to extend to almost 2 light years, so that's one possible answer to your question. If you want to know the most distant aphelion of an object currently orbiting the sun, James Ks answer is good, but I think the outer most aphelion is a bit further than the 0.8 light years that he suggests. At least 2 light-years, possibly even 3. The problem is, an orbit that distant, such an object has a good chance of being deflected before it reaches it's perihelion, a journey that takes over 10 million years. Orbits that distant are likely not very stable. A lot depends on how close other stars get to our sun. A star that passes too close would likely throw everything in the vicinity that it passes through out of wack.

See chart and Wikipedia.

The tiny Scholz's star is thought to have passed within 1 light year of our sun about 70,000 years ago. Stars passing that close are quite rare, but, from the link above

A star is expected to pass through the Oort Cloud every 100,000 years or so. An approach as close or closer than 52,000 AU is expected to occur about every 9 million years.

This does make defining an outermost orbit somewhat difficult, as the most distant orbits take millions of years to reach their closest point from their most distant point, and they run a pretty good chance of being perturbed within a single orbit. Stars likely play Frisbee with their outer-most orbiting objects all the time. Picking an outermost stable orbit is impossible.

A curious sidebar on Scholz, is that, it may have sent a bunch of outer comets and oort cloud objects heading towards the inner solar system. We wont find out how many for another 2 million years or so. That's how long it will take any objects that were sent towards the inner solar-system to reach it.

  • $\begingroup$ thanks for your long and understandable explanation, really appreciate it, so there is no real formula to caluclate it then? $\endgroup$ May 29 '17 at 10:47
  • $\begingroup$ @AlbertoMartínez In theory, an object could be determined to be orbiting the sun up to perhaps about 2 light years based on tangental velocity, but there's no guarantee it would complete a single orbit before being orbitally disturbed. That far out the gravity from nearby stars is pretty close to our sun's gravity. It's very much a grey area, so to my knowledge, no, there's no formula and no clear distance to say this is where orbits stop being stable. $\endgroup$
    – userLTK
    May 29 '17 at 13:20

Since the effect on space-time curvature (gravity) of the Sun propagates through space at the speed of light, a observer beyond the Suns Cosmological horizon, or it's age in light years away, will never be able to feel it.

The actual direction of the gravity propagation can be affected through gravitational lensing from other galaxies/stars, so there may be some "blind spots", but it is hard to say where they are.

  • $\begingroup$ +1 This is always important to point out! Light is so darn slow. $\endgroup$
    – uhoh
    Dec 2 '20 at 22:45

Take a simple case where we know the mass of two solar systems (M_1 and M_2) and the distance between their centers of gravity (x). We want to find the location between them where the two forces of gravity from each system cancel out. Where an object placed on one or the other side of that point would eventually fall into one or the other the star system.

To find the point of equilibrium between 2 systems, first we need the formula for the force of gravity:

F = GmM/R^2 (this is Newtonian gravity, so it is ultimately wrong but a fair approximation anyhow).

I am using 'M_1' for the mass of the first system, 'R_1' for the distance from the center of the first system to the placed object, and 'm' for the mass of the placed object.

F_1 is the force from the first system acting on the placed object:

F_1 = GmM_1/R_1^2

We do the same thing for the second system:

F_2 = GmM_2/R_2^2

And then we set the forces equal eachother to find the point where they will cancel out:

F_1 = F_2

At this exact point an object is pulled equally by the forces from both systems and will stay motionless. We can see that several variables cancel out, the mass of the object (m) and the gravitational constant (G), and we are left with:

M_1/R_1^2 = M_2/R_2^2

Since we know the distance between the systems (x) we can make a substitution using the formula:

R_1 + R_2 = x

enter image description here

What we are left with (after some algebra) is the infamous quadratic equation solution, where:

a = 1 - M_1/M_2

b = - 2*x

c = x^2

Lastly, we need the quadratic equation:

R_2 = [-b +/- sqrt(b^2 - 4ac)]/[2a]

enter image description here

Skipping the algebra, simply plug your 3 knowns into a, b, and c and apply these to the quadratic solution.

You can then find R_1 using the formula:

R_1 = x - R_2

The concept is similar to rain basins, these are space basins. In the case of a basin in space we merely create a point and draw a line perpendicular to an imaginary line connecting the 2 systems. In a more complex set of multiple systems we find the equilibrium points between all neighboring systems and extend the perpendicular basin lines to where they first touch one another.

enter image description here

Sample problem:

M_1 = 1 solar masses

M_2 = 2 solar masses

x = 100 au


R_2 = 58.5786437626905 au

  • $\begingroup$ This doesn't seem to answer this question at all. Is there a chance that you have posted this under the wrong question by accident? $\endgroup$
    – uhoh
    Nov 28 '20 at 5:37
  • $\begingroup$ A more direct answer to the question is extremely complex. If we assume the universe is correctly 78 billion light years across and in fact 13.5 billion years old then there is expansion to contend with. What is meant by influence? If we mean even the slightest influence then gravity travels at the speed of light so what influence the sun has deoends on when we say the sun in fact came into existence. $\endgroup$ Dec 2 '20 at 21:53
  • $\begingroup$ Nobody understands dark energy. We don't know what future expansion will look like. What the sun from the past is influencing now or will influence in the future is out of reachof the sun of right now; it will likely not reach those same objects. There is no concrete answer. Mind you, the top answer in this stack is relating the "influence" of the sun, meaning the furthest distance an object might be in orbit around it. My answer creates the rough/newtonian estimate for that distance, allowing someone to get their own idea based on locations of neighboring stars. $\endgroup$ Dec 2 '20 at 22:08
  • $\begingroup$ The 2nd top voted answer also answers the question by considering the furthest an object would orbit our sun, contesting the first answer, while giving various estimates. I don't see anyone using math or physics equations to answer the question about astronomical distance, which seems a fairly vague way to answer the question. $\endgroup$ Dec 2 '20 at 22:27
  • $\begingroup$ Consider 2 stars of the same mass. An object will orbit whichever star it is closer to. Now consider the likely possibility that the 2 suns have different masses. The orbital dividing line is no longer merely the halfway point between the 2 stars and it is not a simple proportion of the 2 masses. An object in orbit about one star will remain in orbit about that star until the influence of its neighbor captures that objecr. Beyond that distance the star will continue to have some influence, changing the shape of the orbit by pulling the orbiter toward itself. Where that ends has no equation. $\endgroup$ Dec 2 '20 at 22:45

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