瞬時ケプラー軌道要素に基づいた解の表現

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  • 瞬時ケプラー軌道要素に基づいて、式(6)の解を表現することは便利です。
  • 相対運動の瞬時ケプラー軌道要素(e、i、b、τ、ω、φ)に関連する式(7)を使用すると、軌道が次のように与えられます。
  • b、e、I、τ、およびωは、相対運動の軌道の軌道要素であり、それぞれの半長軸、離心率、軌道傾斜角、近日点の経度、昇交点の経度です。
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和訳してください。 よろしくお願いします。

Thus, it is convenient to express a solution to Eq. (6) in terms of the instantaneous Keplerian orbital elements of the relative motion (e, i, b, τ, ω, φ): the orbit is given by x=b-ecos(t-τ), y=-3b(t-φ)/2+2esin(t-τ), z=isin(t-ω), ・・・・・・・(7) where b, e, I, τ, and ω are the semimajor axis, eccentricity, inclination, the longitude of perihelion, and that of ascending node of the relative motion, respectively.

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  • ddeana
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よって、相対運動(e, i, b, τ, ω, φ)の瞬間ケプラー軌道要素の観点から方程式(6)の解を説明したい。軌道は次のように与えられる。 x=b-ecos(t-τ), y=-3b(t-φ)/2+2esin(t-τ), z=isin(t-ω), ・・・・・・・(7) ここでは、b, e, I, τ, ωはそれぞれ、軌道長半径、離心率、軌道傾斜角、近日点経度、そして相対運動の昇交点のことである。

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