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Next: Lagrange brackets Up: Derivation of Lagrange planetary Previous: Introduction

Preliminary analysis

According to Equation (G.6), we have

$\displaystyle \frac{dX}{dt} = \frac{\partial f_1}{\partial t} + \sum_{k=1,6}\frac{\partial f_1}{\partial c_k}\frac{d c_k}{dt}.$ (G.13)

If this expression, and the analogous expressions for $ dY/dt$ and $ dZ/dt$ , were differentiated with respect to time, and the results substituted into Equations (G.3)-(G.5), then we would obtain three time evolution equations for the six variables $ c_1,\cdots,c_6$ . In order to make the problem definite, three additional conditions must be introduced into the problem. It is convenient to choose

$\displaystyle \sum_{k=1,6}\frac{\partial f_l}{\partial c_k}\frac{dc_k}{dt} = 0,$ (G.14)

for $ l=1,2,3$ . Hence, it follows from Equations (G.12) and (G.13) that

    $\displaystyle \frac{dX}{dt}$ $\displaystyle = \frac{\partial f_1}{\partial t} = g_1,$ (G.15)
    $\displaystyle \frac{dY}{dt}$ $\displaystyle = \frac{\partial f_2}{\partial t} = g_2,$ (G.16)
and   $\displaystyle \frac{dZ}{dt}$ $\displaystyle = \frac{\partial f_3}{\partial t} = g_3.$     (G.17)

Differentiation of these equations with respect to time yields

    $\displaystyle \frac{d^{\,2}X}{dt^{\,2}}$ $\displaystyle = \frac{\partial^{\,2} f_1}{\partial t^{\,2}} +\sum_{k=1,6} \frac{\partial g_1}{\partial c_k}\,\frac{d c_k}{dt},$ (G.18)
    $\displaystyle \frac{d^{\,2}Y}{dt^{\,2}}$ $\displaystyle = \frac{\partial^{\,2} f_2}{\partial t^{\,2}}+\sum_{k=1,6} \frac{\partial g_2}{\partial c_k}\,\frac{d c_k}{dt},$ (G.19)
and   $\displaystyle \frac{d^{\,2}Z}{dt^{\,2}}$ $\displaystyle = \frac{\partial^{\,2} f_3}{\partial t^{\,2}}+\sum_{k=1,6} \frac{\partial g_3}{\partial c_k}\,\frac{d c_k}{dt}.$     (G.20)

Substitution into Equations (G.3)-(G.5) gives

    $\displaystyle \frac{\partial^{\,2} f_1}{\partial t^{\,2}} +\mu\,\frac{f_1}{r^{\,3}} +\sum_{k=1,6} \frac{\partial g_1}{\partial c_k}\,\frac{d c_k}{dt}$ $\displaystyle = \frac{\partial {\cal R}}{\partial X},$ (G.21)
    $\displaystyle \frac{\partial^{\,2} f_2}{\partial t^{\,2}}+\mu\,\frac{f_2}{r^{\,3}}+\sum_{k=1,6} \frac{\partial g_2}{\partial c_k}\,\frac{d c_k}{dt}$ $\displaystyle = \frac{\partial {\cal R}}{\partial Y},$ (G.22)
and   $\displaystyle \frac{\partial^{\,2} f_3}{\partial t^{\,2}}+\mu\,\frac{f_3}{r^{\,3}}+\sum_{k=1,6} \frac{\partial g_3}{\partial c_k}\,\frac{d c_k}{dt}$ $\displaystyle = \frac{\partial {\cal R}}{\partial Z},$ (G.23)

where $ r=(f_1^{\,2}+f_2^{\,2}+f_3^{\,2})^{1/2}$ . Because $ f_1$ , $ f_2$ , and $ f_3$ are the respective solutions to Equation (G.3)-(G.5) when the right-hand sides are zero, and the orbital elements are thus constants, it follows that the first two terms in each of the preceding three equations cancel one another. Hence, writing $ f_1$ as $ X$ , and $ g_1$ as $ \dot{X}$ , and so on, Equations (G.14) and (G.21)-(G.23) yield

    $\displaystyle \sum_{k=1,6}\frac{\partial X}{\partial c_k} \,\frac{dc_k}{dt}$ $\displaystyle = 0,$ (G.24)
    $\displaystyle \sum_{k=1,6}\frac{\partial Y}{\partial c_k} \,\frac{dc_k}{dt}$ $\displaystyle = 0,$ (G.25)
    $\displaystyle \sum_{k=1,6}\frac{\partial Z}{\partial c_k} \,\frac{dc_k}{dt}$ $\displaystyle = 0,$ (G.26)
    $\displaystyle \sum_{k=1,6}\frac{\partial \dot{X}}{\partial c_k} \,\frac{dc_k}{dt}$ $\displaystyle = \frac{\partial {\cal R}}{\partial X},$ (G.27)
    $\displaystyle \sum_{k=1,6}\frac{\partial \dot{Y}}{\partial c_k} \,\frac{dc_k}{dt}$ $\displaystyle = \frac{\partial {\cal R}}{\partial Y},$ (G.28)
and   $\displaystyle \sum_{k=1,6}\frac{\partial \dot{Z}}{\partial c_k} \,\frac{dc_k}{dt}$ $\displaystyle = \frac{\partial {\cal R}}{\partial Z}.$ (G.29)

These six equations are equivalent to the three original equations of motion [(G.3)-(G.5)].


next up previous
Next: Lagrange brackets Up: Derivation of Lagrange planetary Previous: Introduction
Richard Fitzpatrick 2016-03-31