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 𝐹 𝑟 =𝑚 𝑟 −𝑟 𝜃 2 Where 𝐹 𝑟 =− 𝐺 𝑚 0 𝑚 𝑟 2 𝐹 𝜃 =𝑚 𝑟 𝜃 +2 𝑟 𝜃 Where 𝐹 𝜃 =0 − 𝐺 𝑚 0 𝑚 𝑟 2 =𝑚 𝑟 −𝑟 𝜃 2 0=𝑚 𝑟 𝜃 +2 𝑟 𝜃 Let us start by looking in the q-direction: 0=𝑚 𝑟 𝜃 +2 𝑟 𝜃 →0= 𝑟 𝑟 𝑚 𝑟 𝜃 +2 𝑟 𝜃 →0=𝑚 𝑟 2 𝜃 +2𝑟 𝑟 𝜃 →0= 𝑟 2 𝜃 +2𝑟 𝑟 𝜃 Where 𝑟 2 𝜃 +2𝑟 𝑟 𝜃 = 𝑑 𝑑𝑡 𝑟 2 𝜃 → 𝑑 𝑑𝑡 𝑟 2 𝜃 =0 When the derivative of a quantity is zero that quantity must be constant. → 𝑟 2 𝜃 =h Where h is a constant

Note that the magnitude of the angular momentum for this situation is essentially: 𝐻 = 𝑟 ×𝑚 𝑣 →𝐻=𝑟𝑚𝑣 sin 𝜃 Assuming q = 90o and using 𝑣=𝑟 𝜃 →𝐻=𝑟𝑚𝑟 𝜃 =𝑚 𝑟 2 𝜃 =𝑚ℎ 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) – Closest dA dA Apogee (Aphelion) – Furthest r dr Fg v 𝐴= 1 2 𝑏ℎ →𝐴= 1 2 𝑟 𝑟𝜃 →𝑑𝐴= 1 2 𝑟 2 𝑑𝜃 Where b = r and h = s = rq → 𝑑𝐴 𝑑𝑡 = 1 2 𝑟 2 𝑑𝜃 𝑑𝑡 → 𝐴 = 1 2 𝑟 2 𝜃 Looking at rate of change of the area Note that h= 𝑟 2 𝜃 is constant, therefore 𝐴 is also constant. This is Kepler’s Second Law. Kepler’s Second Law - The radius vector from the sun to the planet sweeps out equal areas in equal time intervals.

Let us now examine the radial direction: − 𝐺 𝑚 0 𝑚 𝑟 2 =𝑚 𝑟 −𝑟 𝜃 2 − 𝐺 𝑚 0 𝑟 2 = 𝑟 −𝑟 𝜃 2 Let us define 𝑟= 1 𝑢 and use h= 𝑟 2 𝜃 𝑟 =− 1 𝑢 2 𝑢 =−ℎ 𝑢 𝜃 =−ℎ 𝑑𝑢 𝑑𝜃 𝑟 = 𝑑 𝑑𝑡 −ℎ 𝑑𝑢 𝑑𝜃 =−h 𝑑 2 𝑢 𝑑𝜃𝑑𝑡 𝑑𝜃 𝑑𝜃 =−h 𝑑 2 𝑢 𝑑𝜃 2 𝑑𝜃 𝑑𝑡 =−h 𝑑 2 𝑢 𝑑𝜃 2 𝜃 =− h 2 𝑢 2 𝑑 2 𝑢 𝑑𝜃 2 →−𝐺 𝑚 0 𝑢 2 =− h 2 𝑢 2 𝑑 2 𝑢 𝑑𝜃 2 − 1 𝑢 ℎ 2 𝑢 4 → 𝑑 2 𝑢 𝑑𝜃 2 +𝑢= 𝐺 𝑚 0 ℎ 2 non-homogeneous linear differential equation 𝑢=𝐶 cos 𝜃+𝛿 + 𝐺 𝑚 0 ℎ 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. → 1 𝑟 =𝐶 cos 𝜃 + 𝐺 𝑚 0 ℎ 2 The form of this expression matches that for a conic section. r – distance from focus to particle on the path d - distance from focus to Directrix e - eccentricity – deviation from a circular path 1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑

Semiminor axis Conic Sections: Semimajor axis 2b 1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑 2a Where: Foci c 𝑑= 1 𝐶 𝑒𝑑= ℎ 2 𝐺 𝑚 0 Eccentricity: 𝑒= 𝑟 d−rcos 𝜃 d rcosq 𝑒= 𝑐 𝑎 e = 0 Circle e < 1 ellipse e = 1 Parabola e > 1 Hyperbola

Elliptical path ( 0 < e < 1 ): rmin @ q = 0 and rmax @ q = p 1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑 → 1 𝑟 𝑚𝑖𝑛 = 1 𝑑 + 1 𝑒𝑑 → 1 𝑟 𝑚𝑖𝑛 = 1+𝑒 𝑒𝑑 → 𝑟 𝑚𝑖𝑛 = 𝑒𝑑 1+𝑒 1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑 → 1 𝑟 𝑚𝑎𝑥 =− 1 𝑑 + 1 𝑒𝑑 → 1 𝑟 𝑚𝑎𝑥 = 1−𝑒 𝑒𝑑 → 𝑟 𝑚𝑎𝑥 = 𝑒𝑑 1−𝑒 2𝑎= 𝑟 𝑚𝑖𝑛 + 𝑟 𝑚𝑎𝑥 = 𝑒𝑑 1+𝑒 + 𝑒𝑑 1−𝑒 = 2𝑒𝑑 1− 𝑒 2 →𝑑= 𝑎 1− 𝑒 2 𝑒 1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑 → 1 𝑟 = 𝑒 cos 𝜃 𝑎 1− 𝑒 2 + 1 𝑎 1− 𝑒 2 → 1 𝑟 = 1+𝑒 cos 𝜃 𝑎 1− 𝑒 2 Kepler’s 1st Law – All planets move in elliptical orbits with the sun at one focus. Let’s now define the period of the elliptical orbit: Rewrite using 𝑒𝑑= ℎ 2 𝐺 𝑚 0 , 𝑑= 1 𝐶 , 𝑑= 𝑎 1− 𝑒 2 𝑒 , 𝑏=𝑎 1− 𝑒 2 , 𝑔= 𝐺 𝑚 0 𝑅 2 and simplify 𝜏= 𝐴 𝐴 = 𝜋𝑎𝑏 1 2 𝑟 2 𝜃 = 2𝜋𝑎𝑏 ℎ →𝜏= 2𝜋 𝑅 𝑔 𝑎 3 2 → 𝜏 2 = 4 𝜋 2 𝑅 2 𝑔 𝑎 3 Kepler’s 3rd Law

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