A particle of mass \(m\) is projected outward radially from the surface \((q=R)\) of a spherical planet. Show that the Hamiltonian is given by \(H=p^{2} / 2 m-|k| / q\) (with \(k\) constant), so that \(H\) is a constant of the motion (\equiv energy \(E)\). Sketch the phase portrait in the \((q, p)\) phase plane for \(q \geq R\), distinguishing between trajectories which correspond to the particle returning and not returning to the planet's surface. When the particle does return show that the time taken to do this is $$ t_{0}=\sqrt{2 m} \int_{R}^{h} \frac{\mathrm{d} q}{\sqrt{E+|k| / q}}, \quad \text { where } \quad h=\frac{|k|}{|E|} $$ Evaluate this integral to find \(t_{0}\) in terms of \(h, R,|k| / m\). (Hint: the substitution \(q=h \sin ^{2} \theta\) is helpful!) By considering the limit \(R / h \rightarrow 0\) show that the result is in accord with Kepler's third law (4.32).

Step by step solution

01

1. Find the Hamiltonian

To find the Hamiltonian, we start with the general expression \(H = T + U\), where \(T\) represents the kinetic energy of the system, and \(U\) represents the potential energy. For the particle system, the kinetic energy is given by \(T = \frac{p^2}{2m}\), and the potential energy due to gravity is \(U = -\frac{|k|}{q}\). Therefore, the Hamiltonian becomes: $$ H = \frac{p^2}{2m} - \frac{|k|}{q}. $$

02

2. Sketch phase portrait

In the \((q, p)\) phase plane, the phase portrait can be sketched by dividing the plane into two regions: where \(H > 0\) (particle does not return) and \(H < 0\) (particle returns). In the region where \(H > 0\), the particle's trajectory will be outward, and its velocity will increase as the distance from the planet increases. In the region where \(H < 0\), the particle's trajectory will be inward, and its velocity will decrease as the distance from the planet decreases. The region where \(H = 0\) represents the transition between the two regions and can be solved by \(p^2 = 2m |k| / q\). The phase portrait can be sketched following these conditions.

03

3. Integrate to find time taken to return

When the particle returns to the surface, the time taken can be found using: $$ t_0 = \sqrt{2m} \int_R^h \frac{dq}{\sqrt{E+|k|/q}}, $$ where \(h = \frac{|k|}{|E|}\). We are given a hint to use the substitution \(q = h\sin^2 \theta\). It implies that \(dq = 2h\sin\theta\cos\theta d\theta\). Substitute \(q\) and \(dq\) into the integral and solve: \begin{align*} t_0 &= \sqrt{2m} \int_{\arcsin\sqrt{\frac{R}{h}}}^{\frac{\pi}{2}} \frac{2h\sin\theta\cos\theta d\theta}{\sqrt{E+|k|/h\sin^2\theta}} \\ &= \sqrt{\frac{2m}{|E|}} \int_{\arcsin\sqrt{\frac{R}{h}}}^{\frac{\pi}{2}} \frac{\sin\theta\cos\theta d\theta}{\sqrt{1-\sin^2\theta}} = \sqrt{\frac{2m}{|E|}} \int_{\arcsin\sqrt{\frac{R}{h}}}^{\frac{\pi}{2}} d\theta \\ &= \sqrt{\frac{2m}{|E|}} \left[\frac{\pi}{2} - \arcsin\sqrt{\frac{R}{h}}\right]. \end{align*} Now, we have found \(t_0\) in terms of \(h, R, |k|/m\).

04

4. Verify Kepler's Third Law

To show that the result is in accord with Kepler's third law, we consider the limit \(R/h \rightarrow 0\). When \(R/h \rightarrow 0\), the expression inside the arcsine function will go to zero, and \(\arcsin(0) = 0\). Therefore, the time taken in the limit becomes: $$ t_0 = \sqrt{\frac{2m}{|E|}} \frac{\pi}{2}, $$ which is in accordance with Kepler's third law (4.32).

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