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the weight of a body above sea level varies inversely with the square o…

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 116 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, 2.7 miles above sea level?

Explanation:

Step1: Set up the inverse - square relationship formula

Let $W$ be the weight and $d$ be the distance from the center of the Earth. The inverse - square relationship is $W=\frac{k}{d^{2}}$. When $W = 116$ pounds and $d = 3960$ miles, we can find the constant $k$. Substitute into the formula: $116=\frac{k}{3960^{2}}$, so $k = 116\times3960^{2}$.

Step2: Calculate the new distance

The woman is 2.7 miles above sea - level. The new distance $d_{new}=3960 + 2.7=3962.7$ miles.

Step3: Find the new weight

Since $k = 116\times3960^{2}$ and $W_{new}=\frac{k}{d_{new}^{2}}$, substitute $k$ and $d_{new}$ into the formula: $W_{new}=\frac{116\times3960^{2}}{3962.7^{2}}$.
First, calculate $3960^{2}=3960\times3960 = 15681600$ and $3962.7^{2}=3962.7\times3962.7=15702991.29$.
Then, $116\times15681600 = 1819065600$.
Finally, $W_{new}=\frac{1819065600}{15702991.29}\approx115.84$ pounds.

Answer:

115.84 pounds