Question

the weight of a body above sea level varies inversely with the square of the distance from the center of earth. if a woman weighs 140 pounds when she is at sea - level, 3960 miles from the center of earth, how much will she weigh when she is at the top of a mountain, 3 miles above sea level? the woman will weigh approximately (type an integer or decimal rounded to two decimal places as needed.)

Explanation:

Step1: Define the inverse - square relationship

Let $w$ be the weight and $d$ be the distance from the center of the Earth. The relationship is $w=\frac{k}{d^{2}}$, where $k$ is a constant. When $w = 140$ pounds and $d = 3960$ miles (sea - level distance from the center of the Earth), we can find $k$. Substitute into the formula: $140=\frac{k}{3960^{2}}$, so $k = 140\times3960^{2}$.

Step2: Calculate the new distance

The woman is at the top of a mountain 3 miles above sea - level. So the new distance $d_{new}=3960 + 3=3963$ miles.

Step3: Calculate the new weight

Substitute $k = 140\times3960^{2}$ and $d_{new}=3963$ into the formula $w=\frac{k}{d^{2}}$. We have $w=\frac{140\times3960^{2}}{3963^{2}}$.
\[

$$\begin{align*} w&=140\times(\frac{3960}{3963})^{2}\\ &=140\times(\frac{3960}{3963})\times(\frac{3960}{3963})\\ &\approx140\times0.9985\times0.9985\\ &\approx140\times0.9970\\ & = 139.58 \end{align*}$$

\]

Answer:

139.58