F(r) = r f (r) 4.8. Central forces The most interesting problems in classical mechanics are about central forces.

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1 4.8. Cental foces The most inteesting poblems in classical mechanics ae about cental foces. Definition of a cental foce: (i) the diection of the foce F() is paallel o antipaallel to ; in othe wods, fo any position of the object on which F is acting, the diection of F points towad o away fom the oigin. {attaction to O o epulsion fom O} (ii) the magnitude of the foce F() depends only of the distance fom the oigin. {spheically symmetic} ^ F() = f () The simplest way to analyze the motion of the object is to use spheical pola coodinates. Spheical Pola Coodinates {,θ,φ} Figue 4.16 {, θ, φ } {, π/2, φ } 1

2 Spheical Pola Coodinates The diection vectos fo spheical pola ^ ^ ^ coodinates ae θ φ A cental and spheically symmetic foce, using spheical pola coodinates Figue 4.17 z z m F = f e θ x φ y y x "The oigin is the cente of foce." 2

3 Consevative cental foces O, handwitten, F() = U() You should know the gadient opeato in spheical pola coodinates, In wods, the potential enegy function U (which is a scala) depends only on the distance fom the cente of foce,. U(x,y,z) = U() whee = x 2 +y 2 +z 2 U() is called a spheically symmetic potential. ( study the fomulas inside the back cove of the book ) So, fo a spheically symmetic potential, U() F () = du/d 3

4 The Coulomb foce and the coesponding potential enegy q F C enegy U c () Q, fixed at O 4

5 NEWTON'S THEORY OF UNIVERSAL GRAVITATION Imagine a lage mass M (like the sun; o the Eath) and a smalle mass m (like a planet; o a satellite). The foce on mass m is F = GMm 2 e Moe complete Both masses ae attacted towad each othe Newton's thid law. Both masses evolve aound the CoM, and the CoM is fixed. m, a small mass; e.g., a planet F M, the gavitational souce; e.g., the sun enegy GMm / The potential enegy function is U() = GMm / whee G = 6.67 x m 3 s --2 kg --1. We have a spheically symmetic potential assuming the bodies ae themselves spheically symmetic. The Eath is slightly oblate, so the potential enegy of the Eath-Moon system is not tuly cental. Chapte 8 5

6 Futhemoe Thee ae tides. Synchonicity of the Moon τ evolutions = τ otations Moon ; eason because of ancient tides in the Moon; "tidal evolution". Today tides in the Eath ae dissipating enegy; the obit adius is inceasing and the peiod of otation of the Eath is inceasing. Angula momentum is conseved, L ote + L eve = 0. Tidal evolution of the Eath moon system today = 38 mm /y and τ ote = 23 μs /y 6

7 Radial motion in a spheically symmetic potential the oigin, O This is an example of "1-dimensional motion" in thee dimensions. If the angula momentum is 0 then the mass m just moves on a adial line though the oigin. m adial motion Example An asteoid falls adially towad the Eath, with zeo angula momentum elative the Eath. Assume these paametes: When the asteoid is at distance 0 fom the Eath ( = distance fom the cente of the Eath) the velocity is 0. Pick a numbe : let's ty 0 = eath-moon distance = 384,000 km. Neve mind how this unusual initial condition might be ceated; just take it as given. The asteoid mass; suppose R a = 1 km ; then the mass is m = 4/3 π R a 3. ( kg/m 3 ) = kg 7

8 Fist question Calculate the kinetic enegy of the asteoid when it hits the Eath. Solution Fo this we only need an algebaic calculation using consevation of enegy E = ½ m v 2 -- GMm / T hit GMm /R = GMm / 0 T hit = GMm ( 1/R 1/ 0 ) Second question... Calculate the time it will take fo the asteoid to fall to the suface of the Eath, fom the initial distance 0. Solution This is a time calculation, so we need to solve a diffeential equation. GM /R 2 = g and 1/ 0 0, so T hit = m g R = 2E E6 J T hit = J = petajoules Compae lagest eve H bomb = "Tsa Bomba" = 50 megaton TNT = 210 PJ 8

9 Inteesting compaisons: Moon to Eath tavel time (fom TEI SPS ignition to splashdown): { Tans Eath Injection ; Sevice Populsion System } Apollo 11: 2 days 11 hous 55 minutes Apollo 12: 3 days 9 minutes Apollo 14: 2 days 19 hous 26 minutes Apollo 15: 2 days 23 hous 23 minutes Apollo 16: 2 days 17 hous 30 minutes Apollo 17: 2 days 19 hous 49 minutes Homewok Assignment #8 due in class Wednesday, Octobe 25 [37] Poblem 4.26 * [38] Poblem 4.28 ** and Poblem 4.29 ** [Compute] [39] Poblem 4.33 ** [Compute] [40] Poblem 4.34 ** [41] Poblem 4.37 *** [Compute] [42] Poblem 4.38 *** [Compute] Use the cove page. This is a long assignment, so stat woking on it now. 9

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