Physics 1 Problem Solutions 135

Physics 1 Problem Solutions 135 - Chapt. 14 Gravity 131 d)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapt. 14 Gravity 131 d) Keplers 3rd law P 2 = parenleftbigg 4 2 GM parenrightbigg r 3 , where P is orbital period. And also derive the form I think people ought to prefer: P = parenleftbigg 2 GM parenrightbigg r 3 / 2 . e) Keplers 3rd law specialized for the case of planets orbiting the Sun with period measured in years and distance measured in astronomical units: i.e., P 2 yr = Xr 3 AU or P yr = Xr 3 / 2 AU , where X and X are fiducial constants you must solve for. f) Keplers 3rd law specialized for the case of satellites orbiting the Earth with period measured in days and distance measured in mean Earth radii: i.e., P 2 d = Xr 3 Earth- unit or P d = Xr 3 / 2 Earth- unit , where X and X are fiducial constants you must solve for. g) The Moons period in days from Keplers 3rd law given that Moons mean distance from the Earth is 60 . 33 in mean Earth radii. Why is the value obtained significantly different from the mean lunar month of 29.531 days.from the mean lunar month of 29....
View Full Document

This note was uploaded on 11/16/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

Ask a homework question - tutors are online