Presentation on theme: "Special Applications: Central Force Motion"— Presentation transcript:
1 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?GravityWe will use forces written using polar coordinates to describe this motion𝐹 𝑟 =𝑚 𝑟 −𝑟 𝜃 2Where 𝐹 𝑟 =− 𝐺 𝑚 0 𝑚 𝑟 2𝐹 𝜃 =𝑚 𝑟 𝜃 +2 𝑟 𝜃Where 𝐹 𝜃 =0− 𝐺 𝑚 0 𝑚 𝑟 2 =𝑚 𝑟 −𝑟 𝜃 20=𝑚 𝑟 𝜃 +2 𝑟 𝜃Let 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 𝜃Where 𝑟 2 𝜃 +2𝑟 𝑟 𝜃 = 𝑑 𝑑𝑡 𝑟 2 𝜃→ 𝑑 𝑑𝑡 𝑟 2 𝜃 =0When the derivative of a quantity is zero that quantity must be constant.→ 𝑟 2 𝜃 =hWhere h is a constant
2 Note that the magnitude of the angular momentum for this situation is essentially: 𝐻 = 𝑟 ×𝑚 𝑣 →𝐻=𝑟𝑚𝑣 sin 𝜃Assuming q = 90o and using 𝑣=𝑟 𝜃→𝐻=𝑟𝑚𝑟 𝜃 =𝑚 𝑟 2 𝜃 =𝑚ℎThe 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.vFgrdrPerigee (Perihelion) – ClosestdAdAApogee (Aphelion) – FurthestrdrFgv𝐴= 1 2 𝑏ℎ→𝐴= 1 2 𝑟 𝑟𝜃→𝑑𝐴= 1 2 𝑟 2 𝑑𝜃Where b = r and h = s = rq→ 𝑑𝐴 𝑑𝑡 = 1 2 𝑟 2 𝑑𝜃 𝑑𝑡→ 𝐴 = 1 2 𝑟 2 𝜃Looking at rate of change of the areaNote 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.
3 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 2Let us define 𝑟= 1 𝑢 and use h= 𝑟 2 𝜃𝑟 =− 1 𝑢 2 𝑢 =−ℎ 𝑢 𝜃 =−ℎ 𝑑𝑢 𝑑𝜃𝑟 = 𝑑 𝑑𝑡 −ℎ 𝑑𝑢 𝑑𝜃 =−h 𝑑 2 𝑢 𝑑𝜃𝑑𝑡 𝑑𝜃 𝑑𝜃 =−h 𝑑 2 𝑢 𝑑𝜃 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 2non-homogeneous linear differential equation𝑢=𝐶 cos 𝜃+𝛿 + 𝐺 𝑚 0 ℎ 2Where C and d are integration constantsSolution 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 2The form of this expression matches that for a conic section.r – distance from focus to particle on the pathd - distance from focus to Directrixe - eccentricity – deviation from a circular path1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑
5 Elliptical path ( 0 < e < 1 ): q = 0 and q = p1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑→ 1 𝑟 𝑚𝑖𝑛 = 1 𝑑 + 1 𝑒𝑑→ 1 𝑟 𝑚𝑖𝑛 = 1+𝑒 𝑒𝑑→ 𝑟 𝑚𝑖𝑛 = 𝑒𝑑 1+𝑒1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑→ 1 𝑟 𝑚𝑎𝑥 =− 1 𝑑 + 1 𝑒𝑑→ 1 𝑟 𝑚𝑎𝑥 = 1−𝑒 𝑒𝑑→ 𝑟 𝑚𝑎𝑥 = 𝑒𝑑 1−𝑒2𝑎= 𝑟 𝑚𝑖𝑛 + 𝑟 𝑚𝑎𝑥 = 𝑒𝑑 1+𝑒 + 𝑒𝑑 1−𝑒 = 2𝑒𝑑 1− 𝑒 2→𝑑= 𝑎 1− 𝑒 2 𝑒1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑→ 1 𝑟 = 𝑒 cos 𝜃 𝑎 1− 𝑒 𝑎 1− 𝑒 2→ 1 𝑟 = 1+𝑒 cos 𝜃 𝑎 1− 𝑒 2Kepler’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 simplify𝜏= 𝐴 𝐴 = 𝜋𝑎𝑏 𝑟 2 𝜃 = 2𝜋𝑎𝑏 ℎ→𝜏= 2𝜋 𝑅 𝑔 𝑎 3 2→ 𝜏 2 = 4 𝜋 2 𝑅 2 𝑔 𝑎 3Kepler’s 3rd Law
6 Parabolic path ( e = 1 ):1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑→ 1 𝑟 = 1 𝑑 cos 𝜃 +1As q goes from 0 to p, r goes to ∞.Hyperbolic path ( e > 1 ):1 𝑟 = 1 𝑑 cos 𝜃 + 1 𝑒𝑑Beyond a certain angle q1, r goes to ∞.1 ∞ = 1 𝑑 cos 𝜃 𝑒𝑑→0= cos 𝜃 𝑒→ cos 𝜃 1 =− 1 𝑒Note that for a hyperbolic path there are always two solutions.1 −𝑟 = 1 𝑑 cos 𝜃−𝜋 + 1 𝑒𝑑Only for Repulsive forces – Does not make sense for this situation.1 𝑟 = 1 𝑑 cos 𝜃 − 1 𝑒𝑑Attractive Forces