... solve for m, 1074 8.1,6aathe semi –major axis of the new elliptical orbit. The old distance of m 1079 .16 is now the apolune distance, and the perilune can be found from m. 1070 6.1,62prrrapaObviously ... KUrrmGmKrmGmErGmrvrGmvrrmGmKUWrrmGmUrrmGmKrmGmmvK2,J 1070 .62/ (12.15)),Eq.(fromJ1095.82/m/s,9.282,m/s 1072 .7c) (a).part with agreement is,2/and/,/21soand,/212182E10E3E3E2E2E2EE2J. 1070 .6andJ1034.189 ... m.104.55m)1050.4)(010.01(1212 T12.36: a) m.1 007. 71021FmGmrb) From Eq. (12.19), using the result of part (a),days.121s1005.1kg)1090.1)(kgmN10673.6(m)1 007. 7(273022112310T...