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\(r=\frac{l}{e\cos \theta+1}\), where $l$ is the semi-latus rectum, and \(e\) is the eccentricity. so \(\frac{1}{r}=\frac el\cos\theta+\frac 1l\). Since the most common orbits of heavenly bodies are circles and ellipses, we should be able to construct a linear differential equation that involves the derivatives of \(\frac 1r\) with respect to \(\theta\).Let‘s define \(u=\frac 1r\), and denote differentiation w.r.t. \(\theta\) by primes and that w.r.t. by a dot on top. Particularly, we should derive this equation from Newton‘s law of universal gravity, which is given by \(a=-\frac GMu^2e_r\). Since \(v=r‘\omega e_r+\omega re_{\theta}=-\frac u‘{u^2}\omega e_r+\frac {\omega}ue_{\theta}\), we have that \(a=v‘\omega=\omega(-\frac u‘{u^2}\omega e_r+\frac {\omega}ue_{\theta})‘=-GMu^2e_r\). To eliminate \(\omega\), notice that the angular momentum per unit mass \(h=\omega r^2\) is a constant w.r.t. \(\theta\). So since \(\omega=hu^2\), we get \(hu^2(-\frac {u‘}{u^2}hu^2e_r+hue_{\theta})‘=-GMu^2e_r\), \(h(-u‘he_r+hue_{\theta})‘=-GMe_r\), \(h(-u‘‘he_r-u‘he_{\theta}+hu‘e_{\theta}-hue_r)=-GMe_r\),\(u‘‘+u=\frac {GM}h^2\) where we used the fact that $(e_{\theta})‘=-e_r$ and $(e_r)‘=e_{\theta}.$ The solution of this equation is given by $u=C\cos(\theta-\theta_0)+\frac{GM}{h^2}$, where $C$ and $theta_0$ are real numbers. Given any initial condition, we can always modify our coordinate system so that when the object is about to move ahead, its initial angle $\theta$ vanishes. Therefore let $t=0$, and we only consider the case where $\theta=0$. Hence $\frac 1{r_0}=C\cos\theta_0+\frac{GM}{h^2}$, $C=(\frac 1{r_0}-\frac{GM}{h^2})\frac {1}{\cos\theta_0}$. Differentiate our solution, we get $u‘=-C\sin(\theta-\theta_0)$, so $u‘(0)=-C\sin\theta_0$. We have already replace $C$ by $r_0$ as one of our parameters for the initial state. Next we shall compute the dependency of $\theta_0$ upon $v^r(0)$,which is denoted by $v^r_0$. $r‘=-\frac {u‘}{u^2}$, $v^r=r‘\omega=-\frac{u‘}{u^2}\omega=-u‘h$, $u‘=\frac{-v^r}h$, $u‘(0)=\frac{-v^r_0}{h}$, $\frac{-v^r_0}{h}=-C\sin{\theta_0}$,$\frac{v^r_0}{h}=C\sin{\theta_0}$,$\tan(\theta_0)=\frac{v^r_0}{h^2}(\frac 1{r_0}-\frac{GM}{h^2})^{-1}$, $\theta_0=\arctan(\frac{v^r_0}{h^2}(\frac 1{r_0}-\frac{GM}{h^2})^{-1})$. One thing to keep in mind is that by convention we say that $G$ and $M$ are known and that $h$ is unknown. This makes perfect sense since the particle we are dealing with here is arbitrary and we already know about the fixed body. Therefore, since $h$ is conserved we can think of $h$ as part of the initial state parameters. Since $v^{\theta}(0)=v^{\theta}_0=\frac{h}r_0$, we see that $h$ depends upon $v^{\theta}_0$. Hence with the above formulae we‘ve derived, we can determine the trajectory of a particle given its initial radial, tangential speed and radial distance.
Let e=0.5,l=1,h=1, then
$\frac {du}{dt}=hu‘u^2=-0.5(1+0.5\cos\theta)^2\sin\theta$
$\frac {d^2u}{dt^2}=\frac {d}{d\theta}(\frac{du}{dt})hu^2=-\frac{0.5(1+0.5\cos\theta)^3(2\cos\theta+0.5(3\cos(2\theta)-1))}{2}$
$\frac{d^3u}{dt^3}=\frac{0.5(1+0.5\cos\theta)^4(2+3*0.5^2+20*0.5\cos\theta+15*0.5^2\cos2\theta)\sin\theta}{2}$
$\frac{d^4u}{dt^4}=\frac{0.5(1+0.5\cos\theta)^5((8-58*0.5^2)\cos\theta+0.5(-20(-5+3*0.5^2)\cos(2\theta)+3(-4+9*0.5^2+70*0.5\cos(3\theta)+35*0.5^2\cos(4\theta))))}{8}$标签：
原文地址：http://www.cnblogs.com/dieWeltderPhysik/p/4340939.html
