396_PartUniversity Physics Solution

# 396_PartUniversity Physics Solution - Gravitation 12.55....

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Gravitation 12-17 12.55. IDENTIFY and SET UP: (a) To stay above the same point on the surface of the earth the orbital period of the satellite must equal the orbital period of the earth: 4 1 d(24 h/1 d)(3600 s/1 h) 8.64 10 s T == × Eq.(12.14) gives the relation between the orbit radius and the period: EXECUTE: 3/2 E 2 r T Gm π = and 23 2 E 4 r T Gm = 13 2 E 2 4 TGm r ⎛⎞ ⎜⎟ ⎝⎠ 4 2 11 2 2 24 7 2 (8.64 10 s) (6.673 10 N m /kg )(5.97 10 kg) 4.23 10 m 4 ×× ⋅× This is the radius of the orbit; it is related to the height h above the earth&s surface and the radius E R of the earth by E . rhR =+ Thus 767 E 6.38 10 m 3.59 10 m. hrR =− = × × = × EVALUATE: The orbital speed of the geosynchronous satellite is 2 / 3080 m/s. rT = The altitude is much larger and the speed is much less than for the satellite in Example 12.6. (b) Consider Figure 12.55. 6 E 7 cos R r θ × × 81.3 = ° Figure 12.55 A line from the satellite is tangent to a point on the earth that is at an angle of 81.3 ° above the equator. The sketch shows that points at higher latitudes are blocked by the earth from viewing the satellite. 12.56. IDENTIFY: Apply Eq.(12.12) to relate the orbital period T and P M , the planet’s mass, and then use Eq.(12.2) applied to the planet to calculate the astronaut’s weight. SET UP: The radius of the orbit of the lander is 56 5.75 10 m 4.80 10 m ×+ × . EXECUTE: From Eq.(12.14), 2 P 4 r T GM = and 2 5 6 3 24 P 21 1 2 2 3 2 44 ( 5 . 7 5 1 0 m 4 . 8 0 1 0 m ) 2.731 10 kg (6.673 10 N m /kg )(5.8 10 s) r M GT ππ × = × ×⋅ × , or about half the earth’s mass. Now we can find the astronaut&s weight on the surface from Eq.(12.2). (The landing on the north pole removes any need to account for centripetal acceleration.) ( )( ) () 11 2 2 24 pa 2 2 6 p 6.673 10 N m /kg 2.731 10 kg 85.6 kg 677 N GM m w r × = × . EVALUATE: At the surface of the earth the weight of the astronaut would be 839 N. 12.57. IDENTIFY: From Example 12.5, the escape speed is 2 GM v R = . Use / M V ρ = to write this expression in terms of . SET UP: For a sphere 3 4 3 VR = . EXECUTE: In terms of the density , the ratio ( ) 2 is 4 3 , M RR πρ and so the escape speed is ( ) ( ) 2 11 2 2 3 3 8 /3 6.673 10 N m /kg 2500 kg/m 150 10 m 177 m/s. v × = EVALUATE: This is much less than the escape speed for the earth, 11,200 m/s. 12.58. IDENTIFY: From Example 12.5, the escape speed is 2 GM v R = . Use / M V = to write this expression in terms of . On earth, the height h you can jump is related to your jump speed by 2 vg h = For part (b), apply Eq.(12.4) to Europa. SET UP: For a sphere 3 4 3 =

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12-18 Chapter 12 EXECUTE: 3 4 3 /( ) M R ρπ = , so the escape speed can be written as 2 8 3 GR v πρ = . Equating the two expressions for v and squaring gives 22 83 2, o r , 34 π gh gh GR R G ρ == where 2 9.80 m/s g = is for the surface of the earth, not the asteroid. Estimate 1 m h = (variable for different people, of course), 3.7 km. R = For Europa, 2 4 3 GM RG g R . 2 33 61 1 2 2 ( 1 . 3 3 m / s ) 3.03 10 kg/m 4 4 (1.57 10 m)(6.673 10 N m /kg ) g RG ππ = × ×× . EVALUATE: The earth has average density 3 5500 kg/m . The average density of Europa is about half that of the earth but a little larger than the average density of most asteroids.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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396_PartUniversity Physics Solution - Gravitation 12.55....

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