The Earth may not be quite spherical, but assuming a spherical Earth is a reasonable zerothorder assumption that is a reasonable starting point for understanding the planet. E$.$g$.,$ we can look at the properties $($like density and composition$)$ that make it distinctly different from planets that have different shapes like Jupiter. Jupiter has an angular size $($width$)$ of $43\mathrm{.}9$ between its north and south poles, a value that grows to $47\mathrm{.}2$ when this width is measured along the equator. For this problem, assume that the Earth is exactly between the Sun and Jupiter and that the Earth$-$Sun and Jupiter$-$Sun distances are exactly $1\mathrm{.}00$ and $5\mathrm{.}20$ AU$,$ respectively. 

Determine the polar and equatorial diameters of Jupiter units of both kilometers and Earth equatorial diameters. NB: the values you calculate here will not match the values quoted on the formula sheet; this exercise is meant for familiarity with the method.

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To determine the polar and equatorial diameters of Jupiter in both kilometers and Earth equatorial diameters we can use the concept of angular size an... View the full answer