Practice

**zorop** · July 19th 2009, 11:11 PM

A particle of mass m moves under the influence of a central force whose magnitude is given by f(r) = kr, where k > 0. (The particle is thus repelled by the force center.) At some instant $t_{0}$ , the particle is observed to be at ( $r=r_{0}$ $\theta=\theta_{0}$ ) and moving with a speed $v_{0}$ in a direction that makes an angle $\gamma$ with the polar axis (i.e., the positive x-axis). Assume $\gamma$ is not equal to { $0,\pi,2\pi$ } and $v_{0}^2$ is not equal to $\frac{k}{m}r_{0}^2$ .

(a) Show that the polar equation for the orbit of the particle has the form $\frac{\alpha}{r^2}=1+\varepsilon\sin[{2(\theta-\theta_{0})}]$
(b) Determine the values of the constants $\alpha$ and $\varepsilon$ in terms of the parameters m, k, E, and L.
(c) Determine the values of the constants $\alpha$ and $\varepsilon$ in terms of the parameters m, k, $r_{0},v_{0}$ , and $\gamma$ .
(d) Show that the orbit equation, in Cartesian coordinates, is a quadratic form A $x^2$ + Bxy + C $y^2$ + Dx + Fy + G = 0 and determine explicitly the constants A, B, C, D, F, and G in
terms of $\alpha,\varepsilon,\theta_{0}$
(e) The orbit of the particle is thus a conic section! Determine the type of conic section.
(f) Show that the conic section (whose precise nature you determined in part (e)) is one whose center is at the origin and whose axes are rotated by an angle $\beta$ (with respect to the polar axis)
where the value of $\beta$ is given by the equation $\tan(2\beta)=-\cot(\theta_{0})$
Thus, the orientation of the orbit (w.r.t. a set of fixed coordinate axes) depends only on $\theta_{0}$ !
(g) Prove that the shape of the orbit is independent of $\theta_{0}$ (i.e., $\theta_{0}$ only affects the orientation).
(h) Find the distance of closest approach to the center of force (i.e. $r_{min}$ ).
(i) Determine the eccentricity $e$ of the orbit.
(j) How do parts (a)-(i) change if the restriction ( $v_{0}^2$ is not equal to $\frac{k}{m}r_{0}^2$ ) is removed?

**zorop** · July 19th 2009, 11:15 PM

$\frac{\alpha}{r^2}=1+\theta*\sin2{2*(\theta-\theta_{0})}$

**zorop** · July 19th 2009, 11:16 PM

$\frac{\alpha}{r^2}=1+\epsilon\sin[{2(\theta-\theta_{0})}]$

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