View Full Version : Size of Solar System
2003-Dec-13, 01:41 PM
I am cinfused as to the estimated size of the Solar system. A recent story based on Voyager 1 now being about 90 AU's away astimates the heliopause, the edge of the solar sytem, at about 150 Au's away. However, in a recent report on the Oort cloud it is estimated that this reaches more than 1 light year away. Presumably the Oort cloud is in the Solar System as it is held by the Sun's gravity. So can anyone clarify what is the extent of the Solar System? David
2003-Dec-14, 01:23 AM
:rolleyes: Hmm my books saids, the extent of the solar system is approximately 39 astronomical units, which was about 5,850,000,000 km, but if we consider comets, then the solar system is a much larger body. Cometary orbits may be on the order of 100,000 astronomical units. This is about one and one-half light years, or roughly one-third the distance to the nearest stars. :lol:
2003-Dec-14, 06:07 PM
So does this mean that comets and the Oort Cloud, lie outside the solar system and heliopause? If so what holds the Oort cloud in place?
2003-Dec-14, 07:04 PM
try ur first question : Yes
"If so what holds the Oort cloud in place?"
I think its gravity that holds the Oort cloud in this place :lol: , but most likely its gravity ;)
2003-Dec-14, 07:07 PM
or u can visit this website for some information :lol:
2003-Dec-14, 08:45 PM
Yhanks for the website address. I read it and it says that the Oort cloud is weakly bound to the Sun and is also at the edge of the Sun's gravitational influience. So I take it that the Solar sysytem does extend to the Oort cloud and that the gravity that holds it in place is the gravity of the Sun as the I think that the cloud is reckoned to be around the whole Solar system and not just at one edge only. So I am satisfied in my own mind that the Solar system does in fact extend for more than one light year or over 100,000 AU's.
2003-Dec-14, 11:16 PM
The 39 A.U.s is the distance to the outer solar system, but not to it's edge.
Generally anything closer to the Sun then the asteroid belt is considered inner solar system. From the asteroid belt to near Neptune's orbit, (39 AU) is middle.
Neptune is considered as the beginning of the Kupier belt which is all in the outer solar system. The Kupier belt contains a lot more asteroids, the majority of the comets, Pluto, and part of the Orrt cloud.
The heliopause is not well understood but is often considered as the outermost boundry of the Sun's actual influence. Most of the comets are within the heliopause, which is entirely within the Orrt cloud.
I suppose the best way of putting it is the Orrt cloud isn't part of the solar system, but the solar system is part of the Orrt cloud, because the Orrt cloud contains the solar system, as well as comets, dust, and who knows what else.
2003-Dec-18, 06:14 AM
Technically speaking, the solar system is defined as a configuration where ALL member objects are gravitationally BOUNDED to the Sun. Thus, the Oort Cloud IS part of the solar system.
Reposting bits on Gravitational well of the Sun from another thread>>>
THE SUN'S GRAVITATIONAL POTENTIAL WELL: CAN OUR SUN SUPPORT A REMOTE COMPANION?
As we know, the potential of the Sun is calculated from Newtonian gravity, except very close to the Sun where general relativistic effects may be significant for sensitive experiments. By Newtonian theory, the potential of a mass M at a distance r is given by -GM/R where G is the gravitational constant. By definition, the potential is zero at r = infinity. Hence the potential well of the Sun extends to infinite distances. However, the effect of Sun's gravity is not unlimited due to the presence of other masses in our galaxy.
Thus the gravity of the Sun extends to the Oort cloud. It is postulated to extend for a couple of light years even. The gravity well of the Sun will end when its gravity at a certain point in space equals the gravity from another star. At any point further than that point, the gravity well of the other star will be deeper than that of the Sun so that a particle at that point will be in orbit around the other star rather than the Sun. For instance, imagine that the Alpha Centauri system has a single star of the mass of the Sun (in reality it comprises of three stars). Then, the Sun's gravity will extend exactly to half the distance to this star.
Hope this clarifies things!
2003-Dec-18, 09:56 AM
I was going to merge these two topics but decided just to put the URL to the other topic that basically asks the same question ...
And here it is - posted originally by GOURDHEAD (http://www.universetoday.com/forum/index.php?showtopic=1133)
2003-Dec-18, 07:30 PM
Fascinating to read the answers to this question and the others threads. I am taken by the concept that our Sun's gravity extends as far as the gravitational influence of the nearest star. On this basis the gravitational pull of three Proxima Centauri must be about the same as our Sun's as I see from Norton's Atlas that the three PC's absolute magnitudes are 15.5, 4.4 and 5.7 compared to our Sun's 4.8. So it does seem reasonable to suppose that the Oort cloud and our Solar system extend out to about 2 light years.
2003-Dec-19, 12:28 AM
Just one tiny correction polymath...absolute magnitudes have NOTHING to do with the gravitational potential of a star or system of stars. Absolute magnitudes relate to the "brightness" of a star at a standard distance of 10 pc (1 pc = 3.26 Light Years). Gravitational attraction on the other hand has more to do with the mass of the star or stars and distance under consideration.
Also just something useful to augment this thread...
The heliopause is indeed an edge of sorts (BUT it is NOT the equivalent of a gravitional edge): there is material in the solar system of various kinds, and one of these sources of materials is the solar wind. This consists of electrons and protons and so on, and goes out from the Sun to the rest of the solar system. The strength of the wind depends on the solar cycle, and the solar wind is also responsible for aurora, but that's a whole other problem. The heliopause is the boundary where the density of the material in the solar wind equals the density of the regular interstellar medium. Does that make sense? It's not really a gravity boundary, it's just a density boundary.
Hope this helps!
2003-Dec-21, 03:47 AM
From earlier in this thread: "The gravity well of the Sun will end when its gravity at a certain point in space equals the gravity from another star. At any point further than that point, the gravity well of the other star will be deeper than that of the Sun so that a particle at that point will be in orbit around the other star rather than the Sun."
I wonder a bit about that last statement a bit.
Let's extend that to the Earth / Moon / Sun system. (ie. replace the second star in the above with the Earth, and the "particle" with the moon). When I do the math, it looks to me like the sun's gravity exerts a larger force on the moon than the earth does.
Gravitational force is GMm/r^2.
Mass of the sun is 2.0 X 10^30 kg. (2 significant digits is more than enough for this.)
Call the mass of the moon L (it's going too fall out of the calculation at the end, and I'm too lazy to look it up).
Mass of the Earth is 6.0 X 10^24 kg.
Distance from Sun to Moon = 1.5 x 10^8 km
Distance from Earth to Moon = 3.9 x 10^5 km
Sun-Moon gravitational force = (G X L X 2.0 X 10^30) / ((1.5 X 10^8)^2)
= (G X L X 2.0 X 10^30) / (2.25 X 10^16)
= G X L X 9.0 X 10^13
Earth - Moon gravitational force = (G X L X 6.0 X 10^24) / ((3.9 X 10^5)^2)
= (G X L X 6.0 X 10^24) / (1.5 X 10^11)
= G X L X 4.0 X 10^13
So the sun attracts the moon with a force about twice as much as the earth. And according to logic in the statement I quoted above, the moon should be in orbit around the sun.
And yet the moon orbits the earth. Or does it? Maybe if we were less geo-centric, we'd recognize that the moon and earth both orbit the sun, just in a special co-orbital arangment. Would that make the moon a planet under the definition you proposed in that other thread <G>?
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