A body’s weight varies inversely with the square of its distance from the center of the earth; if a chicken weighs 20 pounds when it is 5000 miles from the earth’s center, how much will the chicken weigh when it is 10000 miles from the earth’s center?

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A body’s weight varies inversely with the square of its distance from the center of the earth; if a chicken weighs 20 pounds when it is 5000 miles from the earth’s center, how much will the chicken weigh when it is 10000 miles from the earth’s center?

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Answer 1 · 2016-03-26T02:41:25

5 pounds

Explanation:

The inverse relationship means that

$\frac{W_{1}}{W_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$ $W_1/W_2 = r_2^2/r_1^2$

or

$W_{1} r_{1}^{2} = W_{2} r_{2}^{2} = constant$ $W_1 r_1^2 = W_2 r_2^2 = "constant"$

where $W_{1}$ $W_1$ is the weight when the distance is $r_{1}$ $r_1$ , and $W_{2}$ $W_2$ is the weight when the distance is $r_{2}$ $r_2$ .

Now, we know that

$r_{1} = 5000 mile$ $r_1 = 5000 "mile"$
$W_{1} = 20 pound$ $W_1 = 20 "pound"$
$r_{2} = 10000 mile$ $r_2 = 10000 "mile"$

We want to find $W_{2}$ $W_2$ .

$\frac{W_{1}}{W_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$ $W_1/W_2 = r_2^2/r_1^2$

$\frac{20 pound}{W_{2}} = \frac{{(10000 mile)}^{2}}{{(5000 mile)}^{2}}$ $(20 "pound")/W_2 = (10000 "mile")^2/(5000 "mile")^2$

$= 4$ $= 4$

$W_{2} = \frac{20 pound}{4}$ $W_2 = (20 "pound")/4$

$= 5 pound$ $= 5 "pound"$

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A body’s weight varies inversely with the square of its distance from the center of the earth; if a chicken weighs 20 pounds when it is 5000 miles from the earth’s center, how much will the chicken weigh when it is 10000 miles from the earth’s center?

1 Answer

Explanation:

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