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**Sample text**

4) The situation is illustrated in Fig. 4. We evaluate first the functional A[y] on the reference path y¯(s); this is A¯ = A[¯ y] = s1 G(¯ y , y¯′ ) ds. s0 We next evaluate the functional on a displaced path y(s) = y¯(s) + δy(s); this is s1 G(¯ y + δy, y¯′ + δy ′ ) ds, A[¯ y + δy] = s0 where δy ′ ≡ y ′ − y¯′ = d(y − y¯)/ds = d(δy)/ds. The change in the functional is δA = A[¯ y + δy] − A[¯ y] s1 = s0 G(¯ y + δy, y¯′ + δy ′ ) − G(¯ y , y¯′ ) ds, and we wish to find conditions on y¯(s) that will allow us to set δA = 0, up to corrections of second order in δy.

The eccentricity, on the other hand, can be related to the reduced energy ε; as we shall calculate in a following paragraph, ε=− GM 1 − e2 . 42) This equation is valid for e < 1, which means that ε < 0, and it is valid also for e ≥ 1, which means that ε ≥ 0. We have just observed that ε < 0 when e < 1. This is the case of bound motion, which takes place between two turning points at r = rmin = p/(1 + e) and r = rmax = p/(1 − e). As we see from Eq. 40), the motion proceeds from r = rmin (known as the orbit’s pericentre) when φ = 0, to r = rmax (known as the orbit’s apocentre) when φ = π, and then back to r = rmin when φ = 2π.

B) Find the angle βmax which maximizes the range. 2. A particle traveling in the positive x direction is subjected to a force F = kx3 . The particle started from an initial position x0 < 0. Draw an energy diagram for this situation and provide a qualitative description of the possible motions. 3. Two bodies of masses m1 and m2 are subjected to a mutual attractive force F12 = −km1 m2 r, where k is a constant and r = r1 −r2 is the relative position vector. (a) Show that the equation of motion for r(t) can be put in the form of an energy equation, 1 2 r˙ + ν(r) = ε, 2 and find an expression for ν(r), the effective potential.