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# 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 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 2 2 2 2 1 2 1 g g U F x m x U F x m x α α = − = − = − = − (1) where ( ) ( ) ( ) ( ) 1 1 1 1 1 2 3 , , x x x = r and where α is the constant gravitational acceleration. Therefore, introducting the relative coordinate r and the center of mass coordinate R according to ( ) 1 2 1 1 2 2 1 2 m m m m = + = + r r r r r R (2) we can express r and r in terms of r and R by 1 2 2 1 1 2 1 2 1 2 m m m m m m = + + = − + + r r r r 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 : ( ) ( ) ( ) ( ) 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 1 2 2 1 1 2 1 2 1 2 1 1 2 2 1 1 2 2 g g L m m U U U m m m m U r m m m m m m m x X m x m m m m α α = + + + + + X = + + + + + + + r r r r R r R ± ± ± ± ± ± (4) where x and X are the components of r and R , respectively. Then, (4) becomes 1 x ( ) ( ) ( ) 2 2 2 2 2 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 m m L m m m m m m m m m m m m U r x m m X m m α α = + + + + + + + + r r ± ± ± 2 + R 2 2 (5) Hence, we can write the Lagrangian in the form ( ) ( ) ( ) 2 2 1 2 1 2 1 1 2 2 r m m m m L U X µ α = + + + + r R ± ± (6) where µ is the reduced mass: 1 2 1 2 m m m m µ = + (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 2 2 2 2 du E k u u θ µ µ = − + A A (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 2 2 1 2 sin const. 4 dx ax b a ax bx c b ac + = + + + (2) we have 2 1 2 2 2 2 2 const. sin 2 8 k r k E µ θ µ µ + + = + A A A (3)
CENTRAL-FORCE MOTION 235 or, equivalently, ( ) 2 2 2 2 2 2 sin const. 2 8 k r k E µ θ µ µ + + = + A A A (4) We can choose the point from which θ is measured so that the constant in (4) is 2 π . Then, 2 2 2 1 1 cos 2 1 k r E k µ θ µ = + A A (5) which is the desired expression. 8-3. When 2 k k , 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, 1 1 2 2 E T U U U 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, ( ) 1 2 = E U or ( ) 1 2 . 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
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CDIM08 - CHAPTER 8 Central-Force Motion 8-1 x3 m1 r1 r2 x2...

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