CDIM08 - CHAPTER 8 Central-Force Motion 8-1. x3 m1 r1 r2 x2...

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CHAPTER 8 Central-Force Motion 8-1. x 3 m 2 m 1 x 2 r 1 r 2 x 1 In a uniform gravitational field, the gravitational acceleration is everywhere constant. Suppose the gravitational field vector is in the x direction; then the masses and have the gravitational potential energies: 1 1 m 2 m () 11 1 1 1 22 2 2 12 1 g g UF xm x x α =− (1) where ( ) 111 2 3 ,, xxx = r and where is the constant gravitational acceleration. Therefore, introducting the relative coordinate r and the center of mass coordinate R according to 1 2 mm mm +=+ rr r rr R (2) we can express r and r in terms of r and R by 1 2 2 1 1 2 m mm m =+ + + + R R (3) 233
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234 CHAPTER 8 which differ from Eq. (8.3) in the text by R . The Lagrangian of the two-particle system can now be expressed in terms of r and R : () 22 12 11 21 gg Lm m U UU mm mmU r mx X m x αα =+− + + ++ X =+  + +   rr r rR ±± (4) where x and X are the components of r and R , respectively. Then, (4) becomes 1 x 2 2 121 111 2 m Ur x m m X  =++   −+ + ± 2 + R (5) Hence, we can write the Lagrangian in the form 2 2 r m m m m LU X µ α =− ++ + + ± ± (6) where is the reduced mass: = + (7) Therefore, this case is reducible to an equivalent one-body problem. 8-2. Setting 1 u = r , Eq. (8.38) can be rewritten as 2 du Ek uu θ µµ +− AA (1) where we have used the relation ( ) 2 1 du r dr . Using the standard form of the integral [see Eq. (E.8c), Appendix E]: 1 sin const. 4 dx ax b a ax bx c b ac = + (2) we have 2 1 2 const. sin 2 8 k r kE += + A (3)
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CENTRAL-FORCE MOTION 235 or, equivalently, () 2 2 22 sin const. 2 8 k r kE µ θ −+ +=  +   A AA (4) We can choose the point from which is measured so that the constant in (4) is 2 π . Then, 2 2 2 1 1 cos 2 1 kr E k = + A A (5) which is the desired expression. 8-3. When 2 kk , the potential energy will decrease to half its former value; but the kinetic energy will remain the same. Since the original orbit is circular, the instantaneous values of T and U are equal to the average values, T and U . For a 2 r 1 force, the virial theorem states 1 2 T =− U (1) Hence, 11 ETU UU U =+= (2) Now, consider the energy diagram E D C A r k / r B where CB E = original total energy CA U = original potential energy C CD U = original centrifugal energy The point B is obtained from CB CA CD . According to the virial theorem, ( ) 12 = EU or ( ) . CB CA = Therefore, CD CB BA ==
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236 CHAPTER 8 Hence, if U suddenly is halved, the total energy is raised from B by an amount equal to ( ) 12 CA or by CB . Thus, the total energy point is raised from B to C ; i.e., E (final) = 0 and the orbit is parabolic . 8-4. Since the particle moves in a central, inverse-square law force field, the potential energy is k U r = − (1) so that the time average is 0 1 k U r τ =− d t (2) Since this motion is a central motion, the angular momentum is a constant of motion. Then, (3) 2 .
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This note was uploaded on 11/03/2008 for the course PHYS 3210 taught by Professor Ford during the Fall '08 term at Colorado.

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CDIM08 - CHAPTER 8 Central-Force Motion 8-1. x3 m1 r1 r2 x2...

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