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**Special Applications: Central Force Motion**

Here we will explore the centripetal force that defines the orbit of one body around a second stationary body. Such as planetary orbits about the Sun, which is assumed to be stationary. What would our centripetal force be in this case? Gravity We will use forces written using polar coordinates to describe this motion πΉ π =π π βπ π 2 Where πΉ π =β πΊ π 0 π π 2 πΉ π =π π π +2 π π Where πΉ π =0 β πΊ π 0 π π 2 =π π βπ π 2 0=π π π +2 π π Let us start by looking in the q-direction: 0=π π π +2 π π β0= π π π π π +2 π π β0=π π 2 π +2π π π β0= π 2 π +2π π π Where π 2 π +2π π π = π ππ‘ π 2 π β π ππ‘ π 2 π =0 When the derivative of a quantity is zero that quantity must be constant. β π 2 π =h Where h is a constant

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**Note that the magnitude of the angular momentum for this situation is essentially:**

π» = π Γπ π£ βπ»=πππ£ sin π Assuming q = 90o and using π£=π π βπ»=πππ π =π π 2 π =πβ The angular momentum is constant. Note also that the net moment is zero, which also requires that the angular momentum is constant. Let us now consider the area swept out by the radius vector as the position of the orbiting mass changes. v Fg r dr Perigee (Perihelion) β Closest dA dA Apogee (Aphelion) β Furthest r dr Fg v π΄= 1 2 πβ βπ΄= 1 2 π ππ βππ΄= 1 2 π 2 ππ Where b = r and h = s = rq β ππ΄ ππ‘ = 1 2 π 2 ππ ππ‘ β π΄ = 1 2 π 2 π Looking at rate of change of the area Note that h= π 2 π is constant, therefore π΄ is also constant. This is Keplerβs Second Law. Keplerβs Second Law - The radius vector from the sun to the planet sweeps out equal areas in equal time intervals.

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**Let us now examine the radial direction:**

β πΊ π 0 π π 2 =π π βπ π 2 β πΊ π 0 π 2 = π βπ π 2 Let us define π= 1 π’ and use h= π 2 π π =β 1 π’ 2 π’ =ββ π’ π =ββ ππ’ ππ π = π ππ‘ ββ ππ’ ππ =βh π 2 π’ ππππ‘ ππ ππ =βh π 2 π’ ππ ππ ππ‘ =βh π 2 π’ ππ 2 π =β h 2 π’ 2 π 2 π’ ππ 2 ββπΊ π 0 π’ 2 =β h 2 π’ 2 π 2 π’ ππ 2 β 1 π’ β 2 π’ 4 β π 2 π’ ππ 2 +π’= πΊ π 0 β 2 non-homogeneous linear differential equation π’=πΆ cos π+πΏ + πΊ π 0 β 2 Where C and d are integration constants Solution to the differential equation: Let us rewrite for r and select our initial conditions such that r is a minimum when q = 0o, therefore d = 0. β 1 π =πΆ cos π + πΊ π 0 β 2 The form of this expression matches that for a conic section. r β distance from focus to particle on the path d - distance from focus to Directrix e - eccentricity β deviation from a circular path 1 π = 1 π cos π + 1 ππ

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Semiminor axis Conic Sections: Semimajor axis 2b 1 π = 1 π cos π + 1 ππ 2a Where: Foci c π= 1 πΆ ππ= β 2 πΊ π 0 Eccentricity: π= π dβrcos π d rcosq π= π π e = 0 Circle e < 1 ellipse e = 1 Parabola e > 1 Hyperbola

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**Elliptical path ( 0 < e < 1 ):**

q = 0 and q = p 1 π = 1 π cos π + 1 ππ β 1 π πππ = 1 π + 1 ππ β 1 π πππ = 1+π ππ β π πππ = ππ 1+π 1 π = 1 π cos π + 1 ππ β 1 π πππ₯ =β 1 π + 1 ππ β 1 π πππ₯ = 1βπ ππ β π πππ₯ = ππ 1βπ 2π= π πππ + π πππ₯ = ππ 1+π + ππ 1βπ = 2ππ 1β π 2 βπ= π 1β π 2 π 1 π = 1 π cos π + 1 ππ β 1 π = π cos π π 1β π π 1β π 2 β 1 π = 1+π cos π π 1β π 2 Keplerβs 1st Law β All planets move in elliptical orbits with the sun at one focus. Letβs now define the period of the elliptical orbit: Rewrite using ππ= β 2 πΊ π 0 , π= 1 πΆ , π= π 1β π 2 π , π=π 1β π 2 , π= πΊ π 0 π
2 and simplify π= π΄ π΄ = πππ π 2 π = 2πππ β βπ= 2π π
π π 3 2 β π 2 = 4 π 2 π
2 π π 3 Keplerβs 3rd Law

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Parabolic path ( e = 1 ): 1 π = 1 π cos π + 1 ππ β 1 π = 1 π cos π +1 As q goes from 0 to p, r goes to β. Hyperbolic path ( e > 1 ): 1 π = 1 π cos π + 1 ππ Beyond a certain angle q1, r goes to β. 1 β = 1 π cos π ππ β0= cos π π β cos π 1 =β 1 π Note that for a hyperbolic path there are always two solutions. 1 βπ = 1 π cos πβπ + 1 ππ Only for Repulsive forces β Does not make sense for this situation. 1 π = 1 π cos π β 1 ππ Attractive Forces

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