Sweep Out Size?

[bdaunno]

Obviously, the amount of gravity that is needed to make an object round is not enough gravity to say, sweep out an asteroid belt similar to what our moon did when first created.<br /><br />Is there a minimum size where this will always occur? Or does it depend moreso on the velocity / orbital location of the objects?

 

[doubletruncation]

There was an interesting article recently that discusses exactly this effect. The author's conclusion is that there is a five order of magnitude difference between the eight big planets in our solar system and other objects when you use a parameter that measures the ability of an object to dominate the other objects in the disk. The article is at:<br /><br />http://xxx.lanl.gov/abs/astro-ph/0608359<br /><br /> <div class="Discussion_UserSignature"> </div>

 

[mikeemmert]

Dear doubletruncation;<br /><br />Thank you so much for this paper. I have found it so compelling that I have saved it in my computer for future reference.<br /><br />For the casual readers of this forum, the author proposes that a planet be defined as being greater that 100 times the mass of all the objects in a similar orbit. He devises elegant models that show this. Actually, most planets are about 10,000 times the mass of objects in a similar orbit, so the 100 figure is arbitrary to include some "pathological" cases, which always occur in science. Neptune is about 8500 times the mass of Pluto or similar-sized 2003 UB313 (note: the author did not have the latest size measurement for Xena when he wrote the paper, but that does not change it's validity, in fact it increases it).<br /><br />One of the factors in determining whether an object should be a planet is a, semimajor axis. This makes Mercury a planet, but an object the size of Earth at the distance of the Kuiper belt would fall short of being a planet.<br /><br />This paper is recommended reading.