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Physics 1 Problem Solutions 135

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

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Chapt. 14 Gravity 131 d) Kepler’s 3rd law P 2 = parenleftbigg 4 π 2 GM parenrightbigg r 3 , where P is orbital period. And also derive the form I think people ought to prefer: P = parenleftbigg 2 π GM parenrightbigg r 3 / 2 . e) Kepler’s 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) Kepler’s 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 Moon’s period in days from Kepler’s 3rd law given that Moon’s 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. HINT: The discrepancy is mostly not due
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