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 123 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, 3.2 miles above sea​ level?

Answer 1

Answer:

Her weight is approximately 122.8lb

Step-by-step explanation:

Given

Inverse proportion;

Weight = 123lb when distance = 3960 miles from center of earth

Required

Calculate the weight when distance is 3.2 miles above sea level

Let weight be represented by W and distance by D

From the question, we understand that;

Weight is inversely proportional to square of distance;

Mathematically; this is

$W \alpha \ \frac{1}{D^2}$

Convert proportion to equation

$W = \frac{k}{D^2}$

Where k is the constant of  proportionality

When W = 123; D = 3960.

This implies that

$123 = \frac{k}{3960^2}$

Make k the subject of formula

$k = 123 * 3960^2$

$k = 1928836800$

Calculating her weight when she's at the top of mountain, 3.2 miles above sea level

First, her distance from center of earth has to be calculated

Distance = Previous distance + 3.2

Distance = 3960 + 3.2

Distance = 3963.2

Now, her weight can be calculated using $W = \frac{k}{D^2}$

Substitute for k and D

$W = \frac{1928836800}{3963.2^2}$

$W = \frac{1928836800}{15706954.24}$

$W = 122.801452817$

$W = 122.8\ (Approximated)$