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7 we will describe this qualitatively by constructing an energy diagram, a plot of the effective potential ν(r) together with the constant value of the reduced energy ε. The energy diagram is shown in Fig. 11. Recall the two main features of such diagrams: (i) The difference between ε and ν(r) represents 21 r˙ 2 , the reduced kinetic energy, which must be positive; and (ii) points on the diagram for which ν(r) = ε represent turning points of the motion, at which r˙ changes sign, either from positive to negative, or from negative to positive.

R4 dφ2 + The equation that determines the shape of the orbit is therefore or 2h2 h2 d2 r − r4 dφ2 r5 dr dφ 2 1 d2 r 2 − 3 r2 dφ2 r dr dφ 2 or − 2 1 d2 r + 3 2 2 r dφ r + GM h2 − = 0, r2 r3 − 1 GM + 2 = 0, r h 2 dr dφ 1 GM = 2 . 38) This is a second-order, nonlinear differential equation for r(φ). The standard trick that is used to solve Eq. 38) is to adopt u = 1/r as the dependent variable. Then r = 1/u, 1 du dr =− 2 , dφ u dφ and d2 r 2 = 3 dφ2 u du dφ 2 − 1 d2 u . u2 dφ2 With these transformations Eq.

A) Show that the shape of the orbit is described by r= p , 1 + e cos(kφ) where p, e, and k are constants. Express p and k in terms of a, b, h2 , and µ. ) (b) Plot the orbit in the x-y plane. 99, and let φ range from 0 to 16π. What is happening to the major axis of the ellipse? 8 Additional problems 1. An inclined plane makes an angle α with the horizontal. A projectile is launched from point A at the bottom of the inclined plane. Its initial speed is v0 , and its initial velocity vector makes an angle β with the horizontal.

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