Английская Википедия:Binet equation

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Шаблон:Short description Шаблон:Classical mechanics

The Binet equation, derived by Jacques Philippe Marie Binet, provides the form of a central force given the shape of the orbital motion in plane polar coordinates. The equation can also be used to derive the shape of the orbit for a given force law, but this usually involves the solution to a second order nonlinear ordinary differential equation. A unique solution is impossible in the case of circular motion about the center of force.

Equation

The shape of an orbit is often conveniently described in terms of relative distance <math>r</math> as a function of angle <math>\theta</math>. For the Binet equation, the orbital shape is instead more concisely described by the reciprocal <math>u = 1/r</math> as a function of <math>\theta</math>. Define the specific angular momentum as <math>h=L/m</math> where <math>L</math> is the angular momentum and <math>m</math> is the mass. The Binet equation, derived in the next section, gives the force in terms of the function <math> u(\theta) </math>: <math display="block">F(u^{-1}) = -m h^2 u^2 \left(\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^2}+u\right).</math>

Derivation

Newton's Second Law for a purely central force is <math display="block">F(r) = m \left(\ddot{r}-r\dot{\theta }^2\right).</math>

The conservation of angular momentum requires that <math display="block">r^{2}\dot{\theta } = h = \text{constant}.</math>

Derivatives of <math>r</math> with respect to time may be rewritten as derivatives of <math>u=1/r</math> with respect to angle: <math display="block">\begin{align} 

&\frac{\mathrm{d}u}{\mathrm{d}\theta } = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\theta }=-\frac{{\dot{r}}}{r^{2}\dot{\theta }}=-\frac{{\dot{r}}}{h} \\ 
& \frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}=-\frac{1}{h}\frac{\mathrm{d}\dot{r}}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\theta }=-\frac{\ddot{r}}{h\dot{\theta }} = -\frac{\ddot{r}}{h^2 u^2}

\end{align} </math>

Combining all of the above, we arrive at <math display="block">F = m\left(\ddot{r}-r\dot{\theta }^2\right) = -m\left(h^2 u^2 \frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^2} +h^{2}u^{3}\right)=-mh^{2}u^{2}\left(\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u\right)</math>

The general solution is [1] <math display="block">\theta = \int_{r_0}^r \frac{\mathrm dr}{r^2\sqrt{\frac{2m}{L^2} (E-V) - \frac{1}{r^2}}} + \theta_0</math> where <math>(r_0, \theta_0)</math> is the initial coordinate of the particle.

Examples

Kepler problem

Classical

The traditional Kepler problem of calculating the orbit of an inverse square law may be read off from the Binet equation as the solution to the differential equation <math display="block">-k u^2 = -m h^2 u^2 \left(\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u\right)</math> <math display="block">\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u = \frac{k}{mh^2} \equiv \text{constant}>0.</math>

If the angle <math>\theta</math> is measured from the periapsis, then the general solution for the orbit expressed in (reciprocal) polar coordinates is <math display="block">l u = 1 + \varepsilon \cos\theta.</math>

The above polar equation describes conic sections, with <math>l</math> the semi-latus rectum (equal to <math>h^2/\mu = h^2m/k</math>) and <math>\varepsilon</math> the orbital eccentricity.

Relativistic

The relativistic equation derived for Schwarzschild coordinates is[2] <math display="block">\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u=\frac{r_s c^2}{2 h^{2}}+\frac{3 r_s}{2}u^{2}</math> where <math>c</math> is the speed of light and <math>r_s</math> is the Schwarzschild radius. And for Reissner–Nordström metric we will obtain <math display="block">\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u=\frac{r_s c^2}{2 h^2}+\frac{3 r_s}{2} u^2-\frac{G Q^{2}}{4 \pi \varepsilon_0 c^{4}}\left(\frac{c^2}{h^2} u +2u^3\right)</math> where <math>Q</math> is the electric charge and <math>\varepsilon_0</math> is the vacuum permittivity.

Inverse Kepler problem

Consider the inverse Kepler problem. What kind of force law produces a noncircular elliptical orbit (or more generally a noncircular conic section) around a focus of the ellipse?

Differentiating twice the above polar equation for an ellipse gives <math display="block">l \, \frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^2} = - \varepsilon \cos \theta.</math>

The force law is therefore <math display="block">F = -mh^{2}u^{2} \left(\frac{- \varepsilon \cos \theta}{l}+\frac{1 + \varepsilon \cos \theta}{l}\right)=-\frac{m h^2 u^2}{l}=-\frac{m h^2}{l r^2},</math> which is the anticipated inverse square law. Matching the orbital <math>h^2/l = \mu</math> to physical values like <math>GM</math> or <math>k_e q_1 q_2/m</math> reproduces Newton's law of universal gravitation or Coulomb's law, respectively.

The effective force for Schwarzschild coordinates is[3] <math display="block">F = -GMmu^2 \left(1+3\left(\frac{hu}{c}\right)^2\right)= - \frac{GMm}{r^2} \left(1+3\left(\frac{h}{rc}\right)^2\right).</math> where the second term is an inverse-quartic force corresponding to quadrupole effects such as the angular shift of periapsis (It can be also obtained via retarded potentials[4]).

In the parameterized post-Newtonian formalism we will obtain <math display="block">F = -\frac{GMm}{r^2} \left(1+(2+2\gamma-\beta)\left(\frac{h}{rc}\right)^2\right).</math> where <math>\gamma = \beta = 1</math> for the general relativity and <math>\gamma = \beta = 0</math> in the classical case.

Cotes spirals

An inverse cube force law has the form <math display="block">F(r) = -\frac{k}{r^3}.</math>

The shapes of the orbits of an inverse cube law are known as Cotes spirals. The Binet equation shows that the orbits must be solutions to the equation <math display="block">\frac{\mathrm{d}^2 u}{\mathrm{d}\theta^2}+u=\frac{k u}{m h^2} = C u.</math>

The differential equation has three kinds of solutions, in analogy to the different conic sections of the Kepler problem. When <math>C < 1</math>, the solution is the epispiral, including the pathological case of a straight line when <math>C = 0</math>. When <math>C = 1</math>, the solution is the hyperbolic spiral. When <math>C > 1</math> the solution is Poinsot's spiral.

Off-axis circular motion

Although the Binet equation fails to give a unique force law for circular motion about the center of force, the equation can provide a force law when the circle's center and the center of force do not coincide. Consider for example a circular orbit that passes directly through the center of force. A (reciprocal) polar equation for such a circular orbit of diameter <math>D</math> is <math display="block">D \, u(\theta)= \sec \theta.</math>

Differentiating <math>u</math> twice and making use of the Pythagorean identity gives <math display="block">D \, \frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^2} = \sec \theta \tan^2 \theta + \sec^3 \theta = \sec \theta (\sec^2 \theta - 1) + \sec^3 \theta = 2 D^3 u^3-D \, u.</math>

The force law is thus <math display="block">F = -mh^2u^2 \left( 2 D^2 u^3- u + u\right) = -2mh^2D^2u^5 = -\frac{2mh^2D^2}{r^5}.</math>

Note that solving the general inverse problem, i.e. constructing the orbits of an attractive <math>1/r^5</math> force law, is a considerably more difficult problem because it is equivalent to solving <math display="block">\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u=Cu^3</math>

which is a second order nonlinear differential equation.

See also

Шаблон:Portal

References

Шаблон:Reflist

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  1. ↑ Шаблон:Cite book
  2. ↑ Шаблон:Cite web
  3. ↑ http://chaos.swarthmore.edu/courses/PDG07/AJP/AJP000352.pdf - The first-order orbital equation
  4. ↑ Шаблон:Cite arXiv
Источник — http://wikihandbk.com/ruwiki/index.php?title=Английская_Википедия:Binet_equation&oldid=13528521