Question

The weight of a body varies inversely with the square of its distance from the center of the earth. assuming that the radius of earth is 3960 miles, how much would a man weigh at an altitude of 1 mile above earths surface if he weighs 200 pounds on earth's surface?

Answer 1

199.9 lb

Explanation:

Let w = weight and d = distance from the center of the earth.

If y varies inversely with x, the standard form of the equation is

y = k/x

Here, the weight varies inversely with the square of the distance, so the form of the equation is

w = k/d^2

We can find k, the constant of proportionality, by using the given information. At the earth's surface, the distance from the center of the earth is 3960 miles, and the weight of the man is 200 lb.

w = 200; d = 3960

w = k/d^2

200 = k/(3960^2)

k = 200 * 3960^2

k = 3.136 * 10^9

The equation is

w = (3.13632 * 10^9)/d^2

When the man is 1 mile above the surface of the earth, the distance to the center of the earth is 3960 miles + 1 mile = 3961 miles

w = (3.13632 * 10^9)/(3961^2)

w = 199.9

Answer: 199.9 lb

Answer 2

The weight at an altitude of 1 mile above earths surface if he weighs 200 pounds on earth's surface is 200.10 pounds.

The relationship between weight (W), distance from the center of the Earth (d), and the radius of the Earth (r) can be expressed using the inverse square law:

`\[ W \propto (1)/(d^2) \]`

Given that the radius of the Earth (r) is 3960 miles and the man weighs 200 pounds on the Earth's surface, we can set up a proportion to find the weight at an altitude of 1 mile above the Earth's surface.

Let
`\( W_1 \)` be the weight at 1 mile above the Earth's surface, and
`\( d_1 \)` be the distance from the center of the Earth at that altitude.

`\[ (W)/(W_1) = \left((d)/(d_1)\right)^2 \]`

Given that
`\( W = 200 \)` pounds and
`\( r = 3960 \)` miles (radius of the Earth), and the altitude is 1 mile, we can substitute these values into the equation:

`\[ (200)/(W_1) = \left((3960)/(3961)\right)^2 \]`

Now, solve for
`\( W_1 \)`:

`\[ W_1 = (200)/(\left((3960)/(3961)\right)^2) \]`

Calculating this expression will give you the weight of the man at an altitude of 1 mile above the Earth's surface.

`\[ W_1 = 200.10`