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Next: Stability of Lagrange points Up: Three-body problem Previous: Lagrange points

Zero-velocity surfaces

Consider the surface

$\displaystyle V(x,y,z) = C,$ (9.60)

where

$\displaystyle V(x,y,z) = - 2\,U = \frac{2\,\mu_1}{\rho_1} + \frac{2\,\mu_2}{\rho_2} +x^{\,2}+y^{\,2}.$ (9.61)

Note that $ V\geq 0$ . It follows, from Equation (9.35), that if the mass $ m_3$ has the Jacobi integral $ C$ and lies on the surface specified in Equation (9.60), then it must have zero velocity. Hence, such a surface is termed a zero-velocity surface. The zero-velocity surfaces are important because they form the boundary of regions from which the mass $ m_3$ is dynamically excluded; that is, regions where $ V< C$ . Generally speaking, the regions from which $ m_3$ is excluded grow in area as $ C$ increases, and vice versa.

Let $ C_i$ be the value of $ V$ at the $ L_i$ Lagrange point, for $ i=1,5$ . When $ \mu_2\ll 1$ , it is easily demonstrated that

    $\displaystyle C_1$ $\displaystyle \simeq 3 + 3^{4/3}\,\mu_2^{\,2/3}-\frac{10}{3}\,\mu_2,$ (9.62)
    $\displaystyle C_2$ $\displaystyle \simeq 3 + 3^{4/3}\,\mu_2^{\,2/3}-\frac{14}{3}\,\mu_2,$ (9.63)
    $\displaystyle C_3$ $\displaystyle \simeq 3 + \mu_2,$ (9.64)
    $\displaystyle C_4$ $\displaystyle \simeq 3 - \mu_2,$ (9.65)
and   $\displaystyle C_5$ $\displaystyle \simeq 3 - \mu_2.$ (9.66)

Note that $ C_1>C_2>C_3>C_4=C_5$ .

Figure: Zero-velocity surface $ V=C$ , where $ C> C_1$ , calculated for $ \mu _2=0.1$ . Mass $ m_3$ is excluded from the black region.
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Figure: Zero-velocity surface $ V=C$ , where $ C= C_1$ , calculated for $ \mu _2=0.1$ . Mass $ m_3$ is excluded from the black region.
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\epsfysize =3.25in
\centerline{\epsffile{Chapter08/fig8.08.eps}}
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Figure: Zero-velocity surface $ V=C$ , where $ C= C_2$ , calculated for $ \mu _2=0.1$ . Mass $ m_3$ is excluded from the black region.
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\epsfysize =3.25in
\centerline{\epsffile{Chapter08/fig8.09.eps}}
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Figure: Zero-velocity surface $ V=C$ , where $ C =C_3$ , calculated for $ \mu _2=0.1$ . Mass $ m_3$ is excluded from the back region.
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\epsfysize =3.25in
\centerline{\epsffile{Chapter08/fig8.10.eps}}
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Figure: Zero-velocity surface $ V=C$ , where $ C_4< C < C_3$ , calculated for $ \mu _2=0.1$ . Mass $ m_3$ is excluded from the black regions.
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\centerline{\epsffile{Chapter08/fig8.11.eps}}
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Figures 9.7 through 9.11 show the intersection of the zero-velocity surface $ V=C$ with the $ x$ -$ y$ plane for various different values of $ C$ , and illustrate how the region from which $ m_3$ is dynamically excluded--which we shall term the excluded region--evolves as the value of $ C$ is varied. Of course, any point not in the excluded region is in the so-called allowed region. For $ C> C_1$ , the allowed region consists of two separate oval regions centered on $ m_1$ and $ m_2$ , respectively, plus an outer region that lies beyond a large circle centered on the origin. All three allowed regions are separated from one another by an excluded region. (See Figure 9.7.) When $ C= C_1$ , the two inner allowed regions merge at the $ L_1$ point. (See Figure 9.8.) When $ C= C_2$ , the inner and outer allowed regions merge at the $ L_2$ point, forming a horseshoe-like excluded region. (See Figure 9.9.) When $ C =C_3$ , the excluded region splits in two at the $ L_3$ point. (See Figure 9.10.) For $ C_4< C < C_3$ , the two excluded regions are localized about the $ L_4$ and $ L_5$ points. (See Figure 9.11.) Finally, for $ C < C_4$ , there is no excluded region.

Figure: Zero-velocity surfaces and Lagrange points calculated for $ \mu _2=0.01$ .
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Figure 9.12 shows the zero-velocity surfaces and Lagrange points calculated for the case $ \mu _2=0.01$ . It can be seen that, at very small values of $ \mu _2$ , the $ L_1$ and $ L_2$ Lagrange points are almost equidistant from mass $ m_2$ . Furthermore, mass $ m_2$ , and the $ L_3$ , $ L_4$ , and $ L_5$ Lagrange points all lie approximately on a unit circle, centered on mass $ m_1$ . It follows that, when $ \mu _2$ is small, the Lagrange points $ L_3$ , $ L_4$ and $ L_5$ all share the orbit of mass $ m_2$ about $ m_1$ (in the inertial frame) with $ C_3$ being directly opposite $ m_2$ , $ L_4$ (by convention) $ 60^\circ$ ahead of $ m_2$ , and $ L_5$ $ 60^\circ$ behind.


next up previous
Next: Stability of Lagrange points Up: Three-body problem Previous: Lagrange points
Richard Fitzpatrick 2016-03-31