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exercise9_sol - 251 College Physics I Exercise 9 Suggested...

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Unformatted text preview: 251 College Physics I Exercise 9 Suggested solutions 1. (a) Let the mass of the head and feet of the astronaut to be 771;. and m,, respec- tively. Since the astronaut of height h floats feet “down”, his head and feet are respectively at a distance of (r + h) and r from the center of Earth. For the head, the gravitational force is given by: GMEmh Fr. = mhah = (r + ’1)“ => 0 _ GIVE h — (r + h)2 where M3 is the mass of Earth. For the feet, the gravitational force is given by: 01W m F1=mlar= :5”; I GIVE => (1;: 1‘2 So the difference in the gravitational acceleration at his feet and at his head is: GME 'I'2 Hence. A“ _ (6.67 x 10-“N . mQ/kgxsss x 102‘kg) [1_ (6.77 x 10316)2 ] _ ( (6.77 x 100 In)? 6.77 x 106m +1.70 m)2 = 4.37 x 10-6 m/52 (b) If a black hole of mass M3 is around the astronaut instead of Earth, the difference in the gravitational acceleration becomes: GM” r2 A“ ,3 i“ <r+h>2i . x ' -m . x g . x m _ (6 67 10 11N 2 k )(199 10311: 1— (6 77 10° )2 _ (6.77 x 106 m)2 (6.77 x 105m +1.70m)2 = 14.5 2. To solve this problem, we need to use the shell theorems which state that: 1) A uniformly dense spherical shell attracts an external particle as if all the masses of the shell were concentrated at its center. 2) A uniform dense spherical shell exerts no gravitational force on a particle located anywhere inside it. (a) (b) (C) 3. (a) (b) The magnitude of the gravitational force on the point mass m which is at a distance 3a from the center of the shells is equal to: GmM1 , GmMg . 1‘ _ '7' _ Gm(A{1 + Ala) (30? (36)2 Fm = lfim.Mx +Fm.M5l = - —‘ 9a” The magnitude of the gravitational force on the point mass m which is at a distance 1.9a from the center of the shells is equal to: Glif‘ _ 0711M; (1.9a)2 — 3.6M2 Note that the shell M2 does not exert gravitational force on the mass m as it is located inside the shell. The magnitude of the gravitational force on the point mass m which is at a distance 0.9a from the center of the shells is equal to: Fm:0 Note that both the shells M1 and I": do not exert gravitational force on the mass m as it is located inside the shells. FM=IFMMII= - The kinetic energy of the bowling ball of mass m on Earth's surface is equal to: 2 KB = 1m1)2 = ~1-1n (2”RE) 2 2 T 1 Zn x 6.37 x10‘rn 2 where RE is the radius of Earth. The potential energy of the bowling ball of mass m on Earth’s surface is equal to: _ GmME PE — RE —11 _ 2 24 = _W = -451 x 1081 6.37 x 105m where M3 is the mass of Earth. The kinetic energy of the bowling ball is negligible compared to its potential energy. So its total mechanical energy E is given by: E a PE: —4.51x108J When the bowling ball moves in a circular orbit about Earth at an altitude h, its mechanical energy E' is equal to: GmME E" = —— 2(RE + h) _ _(6.67 x 10'uN . ma/kg2)(7.20kg)(5.98 x1024 kg) _ _214 x lOBJ — 2(6.37 x 106m+ 3.50 x 105 m) ‘ ‘ 2 (C) 4. (a) (b) 5. (a) The change in the hall’s mechanical energy is: AE = E" — E = —2.14 x 10% —(—4.51x103J)= 2.37 x 1033 Just as B is released. the gravitational potential energy of the two-sphere system is equal to: GmAmg _ _(6.67 x 10-11 N . m2/kg2)(20kg)(10 kg) _ _ -8 TAB (0.8m) — 1.67 x 10 J U:— where ”B is the distance between the spheres A and B. Let the initial and final positions of sphere B be 2:1 and 112 respectively. When sphere B is moving, there is no external force acting on the two-sphere system. By the conservation of mechanical energy, the mechanical energy of this system must be conserved. So we have: KE1=21 + PEzzzl = KEN: ‘l‘ PEzzrg 3 0+ (_GmAmB) = mm + (_G’m,4ma) I1 1'2 l 1 => KEM, = GMAMB 1—2 — z Note that sphere A is held at rest at the origin. Therefore, the kinetic energy of sphere B when it has moved 0.2 m toward A is equal to: KE,=,, = (6.67x 10-11 N - m2/kg2)(20kg)(10kg) (fi — fl) = 5.56x10-9J The gravitational force providw the necessary centripetal acceleration for the circular motion of the satellite. Therefore, if the satellite moves in a circular orbit at an altitude h, GmME _ mv2 (Rad-h)2 — RE-l-h => 11—” GME _ RE+h where ME and R3 are the mass and radius of Earth. . _ (6.67 x 10-!1 N - mfi/kgzxsss x 1024 kg) _ 3 "”‘V 6.37x106m+6.40x105m _7.54x10 m/s So the period of revolution is 21412,; + h) _ 21r(6.37 x 1on + 6.40 x 105 m) _ 3 v 7.54 X 103 m/s — 5'84 X 10 s T: (b) The initial total mechanical energ of the satellite is: GmMg —2(RE + h) (6.67 x 10-11 N - mfi/kg2x22o kg)(5.98 x 1024kg) ‘ 2(6.37 x 106 m + 6.40 x 105111) E: = —6.26 x 1093 After 1500 revolutions, the enery loss of the satellite is: |AE| = 1500 x 1.4 x 105.1 =2.1x 1031 So the total mechanical enery of the satellite at the end of its 1500th revolution is equal to: E": E+AE = —6.26 x 10°J+(——2.1x108J)= -6.47 x 109.] The corresponding satellite‘s altitude 12' can be obtained by: E, _ _ GmME - 2(RE + h’) , _ GmMg 5 h — - 2E’ - RE Putting in the data, we find h’=-[W _ 6 2(—6.47 x 10° J) 6.37 x 10 m =4.11x 105m When the satellite moves in a. circular orbit at an altitude h’, its kinetic energy is: lm‘va : GmME 2 2025 + 12’) So the speed of the satellite v’ is equal to: GME (6.67 x 10—11 N . mn/kg2)(5.98 x 1024155) 3 I _ _ —— _ " _VRE+h'_ 6.37x106m+4.11x105m 4'67“" “1/5 And the corresponding period of revolution is 27r(RE+ h’) _ 21r(6.37 x 10°m+4.11 x 105m) T = 2/ 7.67 x 103 m/s = 5.55 x 1035 (c) An external retarding force must do work on the satellite for it to lose mechanical energy. So there’s an external torque exerted on the satellite when it revolves around the Earth. Hence, the angular momentum of the satellite is NOT conserved during this process. 6. (it) Using Kepler’s third law to relate the orbits of the Earth and the comet around thfi Sun, we have 2/3 2/3 => remnantc‘m) =(1AU)(3010'¥) =208AU (b) The mean distance is the numeric average of the closest and farthest distances Therefore, 203AU=1AU# => rm = 415AU (c) Fig. 1 illustrates the Kepler’s second law for an object orbiting around the Sun. If the time for each shaded region is made much shorter. then the area of each region can be apprwcimated as a triangle. The area of each triangle is equal to half the base (speed of comet multiplied by the amount of time) times the height (distance from Sun). So we have the following: Areamln = Area“: => avatars. = gamer...“ ”"n "m =415 : —=— vim“ Twin ...
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