Problem_Set_9_Solutions

Problem_Set_9_Solutions - 14 Using Eq 13-1 we find FAB =...

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14. Using Eq. 13-1, we find F AB = 2 Gm A 2 d 2 j ^ and F AC = 4 Gm A 2 3 d 2 i ^ . Since the vector sum of all three forces must be zero, we find the third force (using magnitude-angle notation) is F AD = Gm A 2 d 2 (2.404 –56.3º) . This tells us immediately the direction of the vector r (pointing from the origin to particle D ), but to find its magnitude we must solve (with m D = 4 m A ) the following equation: 2.404 © ¨ § ¹ ¸ · Gm A 2 d 2 = Gm A m D r 2 . This yields r = 1.29 d . In magnitude-angle notation, then, r = (1.29 –56.3º) , with SI units understood. The “exact” answer without regard to significant figure considerations is r = ( 2 6 13 13 , –3 6 13 13 ) . (a) In ( x, y ) notation, the x coordinate is x =0.716 d . (b) Similarly, the y coordinate is y = 1.07 d .
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(d) This part refers specifically to the very large black hole treated in the previous part. With that mass for M in Eq. 13–16, and r = 2.002 GM / c 2 , we obtain ( ) ( ) ( ) 6 3 3 2 2 2 = 2 = 2.002 2.002 / g GM c da dr dr GM GM c where dr 1.70 m as in Sample Problem 13-3. This yields (in absolute value) an acceleration difference of 7.30 × 10 15 m/s 2 . (e) The miniscule result of the previous part implies that, in this case, any effects due to the differences of gravitational forces on the body are negligible. 22. (a) Plugging R h = 2 GM h / c 2 into the indicated expression, we find ( ) ( ) ( ) ( ) 4 2 2 2 2 2 1 = = = 1.001 2.002 1.001 2 / h h g h h h GM GM c a M R G GM c which yields a g = (3.02 × 10 43 kg·m/s 2 ) / M h . (b) Since M h is in the denominator of the above result, a g decreases as M h increases.
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