散乱特性から予想される直接衝突は起こらない

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  • 散乱の特徴から予想される直接的な衝突は、bi≧5およびbi≦1.5の場合に起こりません。なぜなら、このような場合、粒子はヒル球に入ることができないからです。
  • 惑星と衝突する粒子の初期値は、図1(a)と(b)の上の直交軸に示されていますが、衝突するbiのすべての場合で直接衝突が必ずしも起こるわけではありません。
  • バンドが分解されると、図3(a)で見られるように、微細構造が現れます。さらに、分解されたバンドにはより高い階層の微細構造もあります。
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和訳お願いします。

初めまして。なんとなくは分かるのですが・・・ 和訳お願いできますでしょうか。 As is expected from the features of the scattering, the direct collision with the planet never occurs for the cases |bi|≧5 and |bi|≦1.5, because in these cases a particle cannot enter the Hill sphere. The initial values of bi, with which a particle collides with the planet are marked on the upper abscissa of Figs. 1(a), and (b). However, for all the cases where bi lies in the collision band, shown in Fig. 1(a), the direct collision does not always occur. When the band is decomposed, the fine structure appears as seen in Figs. 3(a). Furthermore, the decomposed band is found to have also a fine structure of a higher hierarchy. 以上です、宜しくお願いします。

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  • ddeana
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回答No.1

拡散の特性から想像しうるに、地球との直接衝突は |bi|≧5 and |bi|≦1.5の場合決して起りえない、なぜならこういった場合において粒子はヒル球に入ることができないからである。粒子が地球に衝突する場合のbiの初期値については図1(a)および (b)の上部横座標上に示したとおりである。しかしながら、図1(a)に示してある、bi が衝突バンド上に位置するすべての場合において、直接衝突は常に起らない。バンドが分解される時、図3(a)で確認できるように、微細構造(※1)が現れる。さらに、分解されたバンドもより高い階層の微細構造を有することも発見されている。 ※1:原子物理学においては、原子のスペクトル線に現れる微細な分裂を指す http://ja.wikipedia.org/wiki/%E5%BE%AE%E7%B4%B0%E6%A7%8B%E9%80%A0_%28%E5%8E%9F%E5%AD%90%E7%89%A9%E7%90%86%E5%AD%A6%29

stargazer1231
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