Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page of 8 I. Solution M O m. Let O be thei cente of mass. Hence M m () m M GMm GMm () G M m Fom Eq. (), o using educed mass, Hence, GM ( m) GM Gm ( ) ( ) ( ). ()
Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page of 8. Since is infinitesimal, it has no gaitational influences on the motion of neithe M no m. Fo to emain stationay elatie to both M and m we must hae: GM Gm G M m cos cos GM Substituting (4) GM Gm sin sin (5) fom Eq. (5) into Eq. (4), and using the identity sin cos cos sin sin( ), we get sin( ) M m m sin (6) The distances and, the angles and ae elated by two Sineule equations Substitute (7) into (6) sin sin sin sin (7) M m 4 m () m Since,Eq. () gies M m () Gm By substituting fom Eq. (5) into Eq. (4), and epeat a simila pocedue, we get () Altenatiely, sin 8 sin and sin m sin M sin sin
Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page of 8 Combining with Eq. (5) gies Hence, it is an equilateal tiangle with 6 6 The distance is calculated fom the Cosine ule. ( ) ( )cos6 Altenatie Solution to. () (4) Since is infinitesimal, it has no gaitational influences on the motion of neithe M no m.fo to emain stationay elatie to both M and m we must hae: Note that GM Gm G M m cos cos (4) GM Gm sin sin (5) sin 8 sin (see figue) sin sin sin sin m (6) M Equations (5) and (6): (7) sin m (8) sin M (9) The equation (4) then becomes: M m M cosmcos ()
Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page 4 of 8 Equations (8) and (): sin Note that fom figue, M m sin M sin Equations () and (): sin Also fom figue, () () sin M m M sin cos cos () (4) sin sin (5) cos 8 8 (see figue) cos, 6, 6 Equations () and (4): Hence M and m fom an equilateal tiangle of sides Distance to M is Distance to m is Distance to O is. The enegy of the mass is gien by E GM Gm d (( ) )..(5) Since the petubation is in the adial diection, angula momentum is conseed and m M ), ( 4 GM d E ( )..(6) Since the enegy is conseed, de 4
Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page 5 of 8 4 de GM d d d d (7) d d d d d.(8) 4 de GM d d d d.(9) d Since, we hae 4 GM d o d GM 4 The petubation fom and gies. () and. O Then 4 d d GM ( ) () n Using binomial expansion ( ) n, d GM..() Using, d GM..() GM Since, d d 4.(4).(5) 5
Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page 6 of 8 d 4 Fom the figue, coso, 4.(6) d 9 7 4.. (7) 4 4 7 Angula fequency of oscillationis. Altenatie solution: M mgies and GM ( M) GM. The unpetubed adial distance of is, ( ) 4 so the petubed adial distance can be epesented by whee as shown in the following figue. Using Newton s nd GM d law, ( ) ( ) ( ). / { ( ) } () The conseation of angula momentum gies ( ) ( ). () Manipulate () and () algebaically, applying and binomial appoximation. GM d { ( ) } ( / ) ( ) / GM d {4 } ( / ) ( ) / GM ( / ) d 4 ( / ) ( / ) / d 4 d 7 4.4 elatie elocity 6
Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page 7 of 8 Let = speed of each spacecaft as it moes in cicle aound the cente O. The elatie elocities ae denoted by the subscipts A, B and C. Fo example, BA is the elocity of B as obseed by A. The peiod of cicula motion is yea T 6546 6 s. (8) The angula fequency T L The speed 575 m/s (9) cos The speed is much less than the speed light Galilean tansfomation. In Catesian coodinates, the elocities of B and C (as obseed by O) ae CB ĵ î C L CA O L L B BC BA AB A AC Fo B, cos6ˆisin 6ˆj B 7
Q_THEOY_SOLUTION_7_SENT_TO_LEADE.DOCX Theoetical Competition:Solution Question Page 8 of 8 Fo C, cos6ˆisin 6ˆj C Hence ˆ ˆ BC sin6j j The speed of B as obseed by C is 996 m/s () Notice that the elatie elocities fo each pai ae anti-paallel. Altenatie solution fo.4 One can obtain BC by consideing the otation about the axis at one of the spacecafts. BC L 65466 s 6 (5 km) 996 m/s 8