(Precession Of Perihelion.) When A Small Mass Moves In A Bound Orbit Around A Mass M, We Found That The Angle AQ Between Successive Inner (Or Outer) Turning Points Is ?? = 14(Ec-(1–26M) (62 +9) Dr -1/2 Dr (I Have Restored The Factors Of G And C.) We Also Mentioned That Care Must Be Taken When Evaluating The Integral, Because The Integrand Blows Up At The

 

(Precession of perihelion.) When a small mass moves in a bound orbit around a mass M, we found that the angle A between successive inner (or outer) turning points is = - 2GM cr c + dr (I have restored the factors of G and c.) We also mentioned that care must be taken when evaluating the integral, because the integrand blows up at the inner and outer turning points, that is, at r_ and r+, respectively. To do the integral, first factor out (1 - 2GM) and write it as A = 2GM)=1/ [c (1 2GM)~ 121-1/2 dr (a) Now assume 2GM cr is small and expand the factors of (1 - 2GM) in the above expression in powers of 1/c, keeping only terms to order 1/c. Then define the new variable of integration u = 1/r and show that, to order 1/c, ~ 1 + = [1. 2G2M21 c2f2 2 . du 2GM udu - - u+ [(u_ u)(u u+)]1/2 c2 + - u+ [(u_ u)(u u+)]1/2 - (b) Show that the first integral is equal to . Also show that the second integral equals (/2)(u+u+), and that this is equal to GM/ to lowest order in 1/c. (c) From your results in part (b), show that, to order 1/c, 2 GM = - 2 ~ 6 cl