the weight of a person on or above the surface of the earth varies inversely as the square of the distance…

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 179 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.\na. $w = \\frac{2,722,590,000}{d}$\nb. $w = \\frac{698,100}{d}$\nc. $w = \\frac{2,722,590,000}{d^{2}}$\nd. $w = \\frac{698,100}{d^{2}}$

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 179 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.\na. $w = \\frac{2,722,590,000}{d}$\nb. $w = \\frac{698,100}{d}$\nc. $w = \\frac{2,722,590,000}{d^{2}}$\nd. $w = \\frac{698,100}{d^{2}}$

Answer

Explanation:

Step1: Recall inverse - square formula

The general formula for inverse - square variation is $W=\frac{k}{d^{2}}$, where $W$ is the weight, $d$ is the distance from the center of the earth, and $k$ is a constant.

Step2: Find the value of the constant $k$

When the person is on the surface of the earth, $d = 3900$ miles and $W=179$ pounds. Substitute these values into the formula $W=\frac{k}{d^{2}}$: $179=\frac{k}{3900^{2}}$. Solve for $k$: $k = 179\times3900^{2}=179\times15210000 = 2722590000$.

Step3: Write the equation

The equation that relates weight $W$ to the distance $d$ from the center of the earth is $W=\frac{2722590000}{d^{2}}$.

Answer:

C. $W=\frac{2722590000}{d^{2}}$