Question

The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth. A particular person weighs 192 pounds on the surface of the earth and the radius of the earth is 3900 miles. Determine the equation that relates weight, W, to the distance from the center of the earth, d, for this person.

Answer 1

$\bf \qquad \textit{ inverse proportional variation}\\\\\\ \begin{array}{llllll} \textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\ \textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\ y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x} &&y=\cfrac{{{ k}}}{x} \end{array}$

"The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth"

namely "w" varies inversely to $\bf d^2$   or $\bf w=\cfrac{k}{d^2}$

then
"A particular person weighs 192 pounds on the surface of the earth and the radius of the earth is 3900 miles. "
namely when w = 192, d = 3900

so   $\bf w=\cfrac{k}{d^2}\qquad \begin{cases} w=192\\ d=3900 \end{cases}\implies 192=\cfrac{k}{3900^2} \\\\\\ \textit{solve for "k", to find the}\\ \textit{"constant of variation"}\\ \textit{once you've found it, plug it back in at }w=\cfrac{k}{d^2}$