Question

the weight w of an object varies inversely as the square of the distance d from the center of the earth. at sea - level (3978 mi from the center of the earth), an astronaut weighs 112 lb. find her weight when she is 350 mi above the surface of the earth and the spacecraft is not in motion.
her weight is □ lb.
(round to the nearest hundredth as needed.)

Explanation:

Step1: Establish inverse - square relationship formula

The weight $W$ varies inversely as the square of the distance $d$ from the center of the earth, so $W=\frac{k}{d^{2}}$. At sea - level, $d = 3978$ mi and $W = 112$ lb. We can find the constant $k$:
$112=\frac{k}{3978^{2}}$, then $k = 112\times3978^{2}$.

Step2: Calculate the new weight

The new distance $d'=3978 + 360=4338$ mi. Substitute $k = 112\times3978^{2}$ into $W=\frac{k}{d^{2}}$ with $d = 4338$:
$W=\frac{112\times3978^{2}}{4338^{2}}$.
First, calculate $3978^{2}=3978\times3978 = 15824484$ and $4338^{2}=4338\times4338 = 18818244$.
Then $112\times3978^{2}=112\times15824484 = 1772342208$.
$W=\frac{1772342208}{18818244}\approx94.18$.

Answer:

$94.18$