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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Lambert's problem ： ウィキペディア英語版 Lambert's problem In celestial mechanics Lambert's problem is concerned with the determination of an orbit from two position vectors and the time of flight, solved by Johann Heinrich Lambert. It has important applications in the areas of rendezvous, targeting, guidance, and preliminary orbit determination.〔E. R. Lancaster & R. C. Blanchard, (''A Unified Form of Lambert’s Theorem'' ), Goddard Space Flight Center, 1968〕 Suppose a body under the influence of a central gravitational force is observed to travel from point ''P1'' on its conic trajectory, to a point ''P2'' in a time ''T''. The time of flight is related to other variables by Lambert’s theorem, which states: :''The transfer time of a body moving between two points on a conic trajectory is a function only of the sum of the distances of the two points from the origin of the force, the linear distance between the points, and the semimajor axis of the conic.''〔James F. Jordon, (''The Application of Lambert’s Theorem to the Solution of Interplanetary Transfer Problems'' ), Jet Propulsion Laboratory, 1964〕 Stated another way, Lambert's problem is the boundary value problem for the differential equation :$\backslash ddot\; =\; -\backslash mu\; \backslash cdot\; \backslash frac\; \backslash \; \backslash$ of the two-body problem for which the Kepler orbit is the general solution. The precise formulation of Lambert's problem is as follows: Two different times $\backslash \; t\_1\; \backslash \; ,\backslash \; t\_2\backslash$ and two position vectors $\backslash bar\; r\_1\; =\; r\_1\; \_1\; ,\backslash \; \backslash bar\; r\_2\; =\; r\_2\; \_2\backslash$ are given. Find the solution $\backslash bar\; r(t)$ satisfying the differential equation above for which :$\backslash bar\; r(t\_1)=\backslash bar\; r\_1$ :$\backslash bar\; r(t\_2)=\backslash bar\; r\_2.$ ==Initial geometrical analysis== The three points ; $F\_1\; \backslash$ : The centre of attraction ; $P\_1\; \backslash$ : The point corresponding to vector $\backslash bar\; r\_1\backslash$ ; $P\_2\; \backslash$ : The point corresponding to vector $\backslash bar\; r\_2\backslash$ form a triangle in the plane defined by the vectors $\backslash bar\; r\_1\backslash$ and $\backslash bar\; r\_2\backslash$ as illustrated in figure 1. The distance between the points $P\_1\; \backslash$ and $P\_2\; \backslash$ is $2d\; \backslash$, the distance between the points $P\_1\; \backslash$ and $F\_1\; \backslash$ is $r\_1\; =\; r\_m-A\; \backslash$ and the distance between the points $P\_2\; \backslash$ and $F\_1\; \backslash$ is $r\_2\; =\; r\_m+A\; \backslash$. The value $A\; \backslash$ is positive or negative depending on which of the points $P\_1\; \backslash$ and $P\_2\; \backslash$ that is furthest away from the point $F\_1\; \backslash$. The geometrical problem to solve is to find all ellipses that go through the points $P\_1\; \backslash$ and $P\_2\; \backslash$ and have a focus at the point $F\_1\; \backslash$ The points $F\_1\; \backslash$, $P\_1\; \backslash$ and $P\_2\; \backslash$ define a hyperbola going through the point $F\_1\; \backslash$ with foci at the points $P\_1\; \backslash$ and $P\_2\; \backslash$. The point $F\_1\; \backslash$ is either on the left or on the right branch of the hyperbola depending on the sign of $A\; \backslash$. The semi-major axis of this hyperbola is $|A|\; \backslash$ and the eccentricity $E\backslash$ is $\backslash frac\backslash$. This hyperbola is illustrated in figure 2. Relative the usual canonical coordinate system defined by the major and minor axis of the hyperbola its equation is :$\backslash frac\; -\; \backslash frac\; =\; 1\; \backslash quad\; (1)$ with :$B\; =\; |A|\; \backslash sqrt\; =\; \backslash sqrt\; \backslash quad\; (2)$ For any point on the same branch of the hyperbola as $F\_1\; \backslash$ the difference between the distances $r\_2\; \backslash$ to point $P\_2\; \backslash$ and $r\_1\; \backslash$ to point $P\_1\; \backslash$ is $r\_2\; -\; r\_1\; =\; 2A\; \backslash quad\; (3)$ For any point $F\_2\; \backslash$ on the other branch of the hyperbola corresponding relation is $s\_1\; -\; s\_2\; =\; 2A\; \backslash quad\; (4)$ i.e. :$r\_1\; +\; s\_1\; =\; r\_2\; +\; s\_2\; \backslash quad\; (5)$ But this means that the points $P\_1\; \backslash$ and $P\_2\; \backslash$ both are on the ellipse having the focal points $F\_1\; \backslash$ and $F\_2\; \backslash$ and the semi-major axis :$a\; =\; \backslash frac\; =\; \backslash frac\; \backslash quad\; (6)$ The ellipse corresponding to an arbitrary selected point $F\_2\; \backslash$ is displayed in figure 3. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Lambert's problem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.