401_PartUniversity Physics Solution

# 401_PartUniversity Physics Solution - 12-22 Chapter 12...

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12-22 Chapter 12 E XECUTE : (a) 2 E g 2 E (10.0 kg)(9.80 m/s ) 98.0 N m m F G mg R = = = = . (b) The mass of the inner core is 3 3 4 3 6 3 22 4 4 inner inner 2 1 3 3 ( ) (1.3 10 kg/m ) (1.2 10 m) 9.4 10 kg m R R ρ π π = = × × = × . The mass of the outer core is 4 3 6 3 6 3 24 4 outer 3 (1.1 10 kg/m ) ([3.6 10 m] [1.2 10 m] ) 2.1 10 kg m π = × × × = × . Only the inner and outer cores contribute to the force. 22 24 11 2 2 g 6 2 (9.4 10 kg 2.1 10 kg)(10.0 kg) (6.67 10 N m /kg ) 110 N (3.6 10 m) F × + × = × = × . (c) Only the inner core contributes to the force and 22 11 2 2 g 6 2 (9.4 10 kg)(10.0 kg) (6.67 10 N m /kg ) 44 N (1.2 10 m) F × = × = × . (d) At 0 r = , g 0 F = . E VALUATE : In this model the earth is spherically symmetric but not uniform, so the result of Example 12.10 doesn’t apply. In particular, the force at the surface of the outer core is greater than the force at the surface of the earth. 12.71. I DENTIFY : Eq.(12.12) relates orbital period and orbital radius for a circular orbit. S ET U P : The mass of the sun is 30 1.99 10 kg M = × . E XECUTE : (a) The period of the asteroid is 3 2 11 2 . Inserting 3 10 m a T GM π = × for a gives 11 2.84 y and 5 10 m × gives a period of 6.11 y. (b) If the period is 11 5.93 y, then 4.90 10 m. a = × (c) This happens because 0.4 2 5, = another ratio of integers. So once every 5 orbits of the asteroid and 2 orbits of Jupiter, the asteroid is at its perijove distance. Solving when 11 4.74 , 4.22 10 m. T y a = = × E VALUATE : The orbit radius for Jupiter is 11 7.78 10 m × and for Mars it is 11 2.21 10 m × . The asteroid belt lies between Mars and Jupiter. The mass of Jupiter is about 3000 times that of Mars, so the effect of Jupiter on the asteroids is much larger. 12.72. I DENTIFY : Apply the work-energy relation in the form W E = Δ , where E K U = + . The speed v is related to the orbit radius by Eq.(12.10). S ET U P : 24 E 5.97 10 kg m = × E XECUTE : (a) In moving to a lower orbit by whatever means, gravity does positive work, and so the speed does increase. (b) ( ) 1/ 2 1/ 2 E v Gm r = , so ( ) 1/ 2 3/ 2 E E 3 2 2 r r Gm v Gm r r −Δ Δ Δ = = Note that a positive r Δ is given as a decrease in radius. Similarly, the kinetic energy is ( ) ( ) 2 E 1/2 1/2 / K mv Gm m r = = , and so ( ) ( ) 2 E 1/2 / K Gm m r r Δ = Δ and ( ) 2 E / U Gm m r r Δ = − Δ . ( ) 2 E /2 W U K Gm m r r = Δ + Δ = − Δ (c) 3 E / 7.72 10 m/s, v Gm r = = × \ ( ) 3 E /2 / 28.9 m/s, v r Gm r Δ = Δ = 10 E /2 8.95 10 J E Gm m r = − = − × (from Eq.(12.15)), ( ) ( ) 2 8 E /2 6.70 10 J, K Gm m r r Δ = Δ = × 9 2 1.34 10 J U K Δ = − Δ = − × , and 8 6.70 10 J. W K = −Δ = − × (d) As the term °burns up± suggests, the energy is converted to heat or is dissipated in the collisions of the debris with the ground. E VALUATE : When r decreases, K increases and U decreases (becomes more negative). 12.73. I DENTIFY : Use Eq.(12.2) to calculate g . F Apply Newton²s 2nd law to circular motion of each star to find the orbital speed and period. Apply the conservation of energy expression, Eq.(7.13), to calculate the energy input (work) required to separate the two stars to infinity.

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