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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – Alberto Martínez 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! :) $\endgroup$ – Alberto Martínez 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) $\endgroup$ – Carl Witthoft 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
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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.

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  • $\begingroup$ thanks for your long and understandable explanation, really appreciate it, so there is no real formula to caluclate it then? $\endgroup$ – Alberto Martínez 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
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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.

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Max is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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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.

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.

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

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]

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.

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Philip Moseman is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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  • $\begingroup$ 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 at 5:37

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