Hw 7 Solutions - P1 The period T and orbit radius r are...

This preview shows page 1 - 3 out of 5 pages.

P1.- The period T and orbit radius r are related by the law of periods: T 2 = (4 π 2 / GM ) r 3 , where M is the mass of Mars. The period is 7 h 39 min, which is 2.754 × 10 4 s. We solve for M : ( ) 2 3 2 6 3 23 2 2 11 3 2 4 4 4 (9.4 10 m) 6.5 10 kg. (6.67 10 m /s kg) 2.754 10 s r M GT π π × = = = × × × P2.- Kepler’s law of periods, expressed as a ratio, is 3 2 2 3 (1.52) 1y M M M E E a T T a T = = where we have substituted the mean-distance (from Sun) ratio for the semimajor axis ratio. This yields T M = 1.87 y. The value in Appendix C (1.88 y) is quite close, and the small apparent discrepancy is not significant, since a more precise value for the semimajor axis ratio is a M / a E = 1.523 which does lead to T M = 1.88 y using Kepler’s law. A question can be raised regarding the use of a ratio of mean distances for the ratio of semimajor axes, but this requires a more lengthy discussion of what is meant by a mean distance” than is appropriate here. P3.- (a) If r is the radius of the orbit then the magnitude of the gravitational force acting on the satellite is given by GMm/r 2 , where M is the mass of Earth and m is the mass of the satellite. The magnitude of the acceleration of the satellite is given by v 2 /r , where v is its speed. Newton’s second law yields GMm/r 2 = mv 2 /r . Since the radius of Earth is 6.37 × 10 6 m the orbit radius is r = (6.37 × 10 6 m + 160 × 10 3 m) = 6.53 × 10 6 m. The solution for v is 11 3 2 24 3 6 (6.67 10 m /s kg)(5.98 10 kg) 7.82 10 m/s. 6.53 10 m GM v r × × = = = × × (b) Since the circumference of the circular orbit is 2 π r , the period is 6 3 3 2 2 (6.53 10 m) 5.25 10 s. 7.82 10 m/s r T v π π × = = = × × This is equivalent to 87.5 min
Image of page 1

Subscribe to view the full document.

P4.- . (a) The period is T = 27(3600) = 97200 s, and we are asked to assume that the orbit is circular (of radius r = 100000 m). Kepler’s law of periods provides us with an approximation to the asteroid’s mass: ( ) 2 3 2 16 4 (97200) 100000 6.3
Image of page 2
Image of page 3

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

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