{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

396_PartUniversity Physics Solution

396_PartUniversity Physics Solution - Gravitation 12.55...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Gravitation 12-17 12.55. I DENTIFY and S ET U P : (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: E XECUTE : 3/2 E 2 r T Gm π = and 2 3 2 E 4 r T Gm π = 1 3 2 E 2 4 T Gm r π = = 1 3 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 π × × × = × 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 . r h R = + Thus 7 6 7 E 4.23 10 m 6.38 10 m 3.59 10 m. h r R = = × × = × E VALUATE : The orbital speed of the geosynchronous satellite is 2 / 3080 m/s. r T π = 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 6.38 10 m cos 4.23 10 m R r θ × = = × 81.3 θ = ° 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. I DENTIFY : 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. S ET U P : The radius of the orbit of the lander is 5 6 5.75 10 m 4.80 10 m × + × . E XECUTE : From Eq.(12.14), 2 3 2 P 4 r T GM π = and 2 3 2 5 6 3 24 P 2 11 2 2 3 2 4 4 (5.75 10 m 4.80 10 m) 2.731 10 kg (6.673 10 N m /kg )(5.8 10 s) r M GT π π × + × = = = × × × , 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 p a 2 2 6 p 6.673 10 N m /kg 2.731 10 kg 85.6 kg 677 N 4.80 10 m GM m w r × × = = = × . E VALUATE : At the surface of the earth the weight of the astronaut would be 839 N. 12.57. I DENTIFY : From Example 12.5, the escape speed is 2 GM v R = . Use / M V ρ = to write this expression in terms of ρ . S ET U P : For a sphere 3 4 3 V R π = . E XECUTE : In terms of the density , ρ the ratio ( ) 2 is 4 3 , M R R π ρ 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 π = × × = E VALUATE : This is much less than the escape speed for the earth, 11,200 m/s. 12.58. I DENTIFY : From Example 12.5, the escape speed is 2 GM v R = . Use / M V ρ = to write this expression in terms of ρ . On earth, the height h you can jump is related to your jump speed by 2 v gh = For part (b), apply Eq.(12.4) to Europa. S ET U P : For a sphere 3 4 3 V R π =
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
12-18 Chapter 12 E XECUTE : 3 4 3 /( ) M R ρ π = , so the escape speed can be written as 2 8 3 G R v π ρ = . Equating the two expressions for v and squaring gives 2 2 8 3 2 , or , 3 4 π gh gh GR R G ρ π 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 πρ = = .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern