IPhO2011_Th_S1

Theoretical Competition: Solution Question 1 Page 1 of 7 1 I. Solution 1.1 Let O be their centre of mass. Hence 0MR mr  ……………………… (1)     2 0 2 2 0 2 GMm m r R r GMm M R R r       ……………………… (2) From Eq. (2), or using reduced mass,     2 0 3 G M m R r     Hence, 2 0 3 2 2 ( ) ( ) ( ) ( ) G M m GM Gm R r r R r R R r . ……………………………… (3) O M m R r  1  r2 r1  2 2 1 Theoretical Competition: Solution Question 1 Page 2 of 7 2 1.2 Since  is infinitesimal, it has no gravitational influences on the motion of neither M nor m . For  to remain stationary relative to both M and m we must have:     2 1 2 0 32 2 1 2 cos cos G M mGM Gm r r R r             ……………………… (4) 1 22 2 1 2 sin sin GM Gm r r     ……………………… (5) Substituting 2 1 GM r from Eq. (5) into Eq. (4), and using the identity 1 2 1 2 1 2sin cos cos sin sin( )        , we get     1 2 132 2 sin( ) sin M m m r R r        ……………………… (6) The distances 2r and  , the angles 1 and 2 are related by two Sine Rule equations   1 1 1 21 2 sin sin sinsin R r R r          ……………………… (7) Substitute (7) into (6)     43 2 1 M mR r mR r    ……………………… (10) Since m R M m R r    ,Eq. (10) gives 2r R r  ……………………… (11) By substituting 2 2 Gm r from Eq. (5) into Eq. (4), and repeat a similar procedure, we get 1r R r  ……………………… (12) Alternatively,   1 1sinsin 180 r R    and 2 2sin sin r r    1 2 2 2 1 1 sin sin r rR m r r M r       Combining with Eq. (5) gives 1 2r r Theoretical Competition: Solution Question 1 Page 3 of 7 3 Hence, it is an equilateral triangle with 1 2 60 60       ……………………… (13) The distance  is calculated from the Cosine Rule. 2 2 2 2 2 ( ) 2 ( )cos60r R r r R r r rR R            ……………………… (14) Alternative Solution to 1.2 Since  is infinitesimal, it has no gravitational influences on the motion of neither M nor m .For  to remain stationary relative to both M and m we must have:     2 1 2 32 2 1 2 cos cos G M mGM Gm r r R r             ……………………… (4) 1 22 2 1 2 sin sin GM Gm r r     ……………………… (5) Note that   1 1sinsin 180 r R    2 2sin sin r r    (see figure) 1 2 2 2 1 1 sin sin r rR m r r M r       ……………………… (6) Equations (5) and (6): 1 2r r ……………………… (7) 1 2 sin sin m M    ……………………… (8) 1 2  ……………………… (9) The equation (4) then becomes:     2 1 2 13 cos cos M m M m r R r        ……………………… (10) Equations (8) and (10):     2 1 1 2 23 sin sin rM m M R r         ……………………… (11) Note that from figure, 2 2sin sin r    ……………………… (12) Theoretical Competition: Solution Question 1 Page 4 of 7 4 1.3 The energy of the mass is given by 2 2 21 2 1 2 (( ) ) GM Gm d E r r dt           ………………………..(15) Since the perturbation is in the radial direction, angular momentum is conserved ( 1 2r r and m M ), 4 2 2 0 01 2 2 2 ( ) GM d E dt                ………………………..(16) Since the energy is conserved, 0 dE dt  4 22 0 0 2 2 3 2 0 dE GM d d d d dt dt dt dt dt               ……………(17) d d d d dt d dt dt          …………….(18) 4 22 0 0 3 2 3 2 0 dE GM d d d d dt dt dt dt dt                …………….(19) Equations (11) and (12):     2 1 1 2 23 sin sin r rM m M R r        ……………………… (13) Also from figure,       2 2 2 2 2 1 2 1 2 1 1 1 22 cos 2 1 cosR r r rr r r             ……………… (14) Equations (13) and (14):     2 1 2 1 2 sin sin 2 1 cos            ……………………… (15) 1 2 1 2 2180 180 2          (see figure) 2 2 1 1 cos , 60 , 60 2       Hence M and m from an equilateral triangle of sides  R r Distance  to M is R r Distance  to m is R r Distance  to O is   22 2 23 2 2 R r R R r R Rr r                     R R  60 o O Theoretical Competition: Solution Question 1 Page 5 of 7 5 Since 0 d dt   , we have 4 22 0 0 3 2 3 2 0 GM d dt         or 4 22 0 0 2 3 3 2d GM dt         . …………………………(20) The perturbation from 0 and 0 gives 0 0 1         and 0 0 1            . Then 4 22 2 0 0 0 03 32 2 03 3 0 0 0 0 2 ( ) 1 1 1 d d GM dt dt                                    ………………(21) Using binomial expansion (1 ) 1n n    , 2 2 0 0 02 3 0 0 0 0 2 3 3 1 1 1 d GM dt                                 . ……………….(22) Using       , 2 20 0 0 02 3 2 0 0 0 0 32 3 1 1 d GM dt                              . ……………….(23) Since 2 0 3 0 2GM    , 2 2 20 0 0 0 02 2 0 0 0 3 3 1 1 d dt                              ……………….(24) 2 2 0 0 02 2 0 0 34d dt                ……………….(25) 22 2 0 02 2 0 3 4 d dt             ……………….(26) From the figure, 0 0 cos30  or 2 0 2 0 3 4    , 2 2 2 0 02 9 7 4 4 4 d dt                   . …………….…(27) Theoretical Competition: Solution Question 1 Page 6 of 7 6 Angular frequency of oscillation is 0 7 2  . Alternative solution: M m gives R r and 20 3 3 ( ) ( ) 4 G M M GM R R R      . The unperturbed radial distance of  is 3R , so the perturbed radial distance can be represented by 3R  where 3R  as shown in the following figure. Using Newton’s 2nd law, 2 2 22 2 3/2 2 ( 3 ) ( 3 ) ( 3 ) { ( 3 ) } GM d R R R dtR R                . (1) The conservation of angular momentum gives 2 2 0( 3 ) ( 3 )R R    . (2) Manipulate (1) and (2) algebraically, applying 2 0  and binomial approximation. 22 0 22 2 3/2 3 32 ( 3 ) { ( 3 ) } (1 / 3 ) RGM d R dtR R R            22 0 22 3/2 3 32 ( 3 ) {4 2 3 } (1 / 3 ) RGM d R dtR R R           22 0 3 23/2 3 3(1 / 3 ) 3 4 (1 3 / 2 ) (1 / 3 ) RGM R d R R dtR R           2 2 2 0 02 3 3 3 3 1 1 3 1 4 3 3 d R R R dtR R                           2 2 02 7 4 d dt           1.4 Relative velocity Let v = speed of each spacecraft as it moves in circle around the centre O. The relative velocities are denoted by the subscripts A, B and C. For example, BAv is the velocity of B as observed by A. The period of circular motion is 1 year 365 24 60 60T     s. ………… (28) The angular frequency 2 T    The speed 575 m/s 2cos30 L v    ………… (29) Theoretical Competition: Solution Question 1 Page 7 of 7 7 The speed is much less than the speed light  Galilean transformation. In Cartesian coordinates, the velocities of B and C (as observed by O) are For B, ˆ ˆcos60 sin 60Bv v v   i j For C, ˆ ˆcos60 sin 60Cv v v   i j Hence BC ˆ ˆ2 sin 60 3v v v    j j The speed of B as observed by C is 3 996 m/sv  ………… (30) Notice that the relative velocities for each pair are anti-parallel. Alternative solution for 1.4 One can obtain BCv by considering the rotation about the axis at one of the spacecrafts. 6 BC 2 (5 10 km) 996 m/s 365 24 60 60 s v L         C B A v v v O BCv BAv ACv CAv CBv ABv L L L jˆ iˆ