Chapt. 14Gravity131d) Kepler’s 3rd lawP2=parenleftbigg4π2GMparenrightbiggr3,wherePis orbital period. And also derive the form I think people ought to prefer:P=parenleftbigg2π√GMparenrightbiggr3/2.e) Kepler’s 3rd law specialized for the case of planets orbiting the Sun with period measuredin years and distance measured in astronomical units: i.e.,P2yr=Xr3AUorPyr=√Xr3/2AU,whereXand√Xare fiducial constants you must solve for.f) Kepler’s 3rd law specialized for the case of satellites orbiting the Earth with periodmeasured in days and distance measured in mean Earth radii: i.e.,P2d=Xr3Earth-unitorPd=√Xr3/2Earth-unit,whereXand√Xare fiducial constants you must solve for.g) The Moon’s period in days from Kepler’s 3rd law given that Moon’s mean distance fromthe Earth is 60.33 in mean Earth radii. Why is the value obtained significantly differentfrom the mean lunar month of 29.531 days.HINT:The discrepancy is mostly not due
This is the end of the preview.
access the rest of the document.