Universal Gravitation Chapter Problems File

6 Comments

Stephen McNally • 2 years, 10 months agologin to reply

On Universal Gravitation Chapter Problems, check out problem 34b. IT appears that the right answer is 1.69 x 10^7 m, not 2.04 x 10^7 m. Looks like a problem in adding in scientific notation. Could you check that out. BTW, that matches up with Wikipedia's report on the height for a Geosynchronous orbit around Mars. Could you comment and correct. Steve McNally, Glen Rock High School

Yuriy Zavorotniy • 2 years, 10 months agologin to reply

Steve, I got 5082.7 s and corrected this answer. Thank you. Yuriy

Alexander Henderson • 1 year, 10 months agologin to reply

36b is 26455s not 5082.7s 2*(pie)*(3*(6.4x10^6)) / 4560 = 26,455s

John Ennis • 1 year, 10 months agologin to reply

Alexander, you are correct, and a new version of the problems will be uploaded. I cut it to two significant figures and averaged the value you get when you use Kepler's Law and measurements of the earth's radius and mass to 4 sig figs. The answer now is 2.7 x 10^4 s. Thanks for the correction, John.

Melissa Axelsson • 1 year, 10 months agologin to reply

New version is now posted.

Stephen McNally • 9 months agologin to reply

Yurly, referring to question from one month ago, above. The Geostationary orbit for mars is 2.03 E7 m from the center of the planet, but the question was for the height above the surface. Doesn't that mean it should be 2.03 E7 m - 3.39 E6 m (mean radius of Mars) for a height of 1.7 E7 m. Could you check and update the file. Steve McNally, Glen Rock High School

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