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# The weight of an object varies inversely as the square of the distance

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Intern
Joined: 03 Jan 2018
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The weight of an object varies inversely as the square of the distance [#permalink]

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05 Jan 2018, 12:09
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The weight of an object varies inversely as the square of the distance from the center of the earth. At Sea Level (6,400 km from the center of the earth), an astronaut weighs 100 lbs. How far above the Earth must the astronaut be in order to weigh 64 lb?

[Reveal] Spoiler:
I am using the Variation Equation: yz = k as the object varies "inversely"

The correct answer is 1,600 km but I can't seem to get there.

Can someone shed some light on this one?

Thank you
Math Expert
Joined: 02 Sep 2009
Posts: 44636
Re: The weight of an object varies inversely as the square of the distance [#permalink]

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05 Jan 2018, 12:25
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Expert's post
sullivan88 wrote:
The weight of an object varies inversely as the square of the distance from the center of the earth. At Sea Level (6,400 km from the center of the earth), an astronaut weighs 100 lbs. How far above the Earth must the astronaut be in order to weigh 64 lb?

[Reveal] Spoiler:
I am using the Variation Equation: yz = k as the object varies "inversely"

The correct answer is 1,600 km but I can't seem to get there.

Can someone shed some light on this one?

Thank you

The weight of an object varies inversely as the square of the distance from the center of the earth:

$$weight = \frac{k}{distance^2}$$, where k is some constant.

$$100= \frac{k}{6,400^2}$$ --> $$k=6,400^2*100$$.

$$64=\frac{6,400^2*100}{x^2}$$ --> $$x = \sqrt{\frac{6,400^2*100}{64}}=\frac{6,400*10}{8}=8,000$$ from the center of the earth.

The distance from the Earth (sea level) = $$8,000 - 6,400 = 1,600$$.
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Math Expert
Joined: 02 Sep 2009
Posts: 44636
Re: The weight of an object varies inversely as the square of the distance [#permalink]

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05 Jan 2018, 12:27
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Expert's post
Bunuel wrote:
sullivan88 wrote:
The weight of an object varies inversely as the square of the distance from the center of the earth. At Sea Level (6,400 km from the center of the earth), an astronaut weighs 100 lbs. How far above the Earth must the astronaut be in order to weigh 64 lb?

[Reveal] Spoiler:
I am using the Variation Equation: yz = k as the object varies "inversely"

The correct answer is 1,600 km but I can't seem to get there.

Can someone shed some light on this one?

Thank you

The weight of an object varies inversely as the square of the distance from the center of the earth:

$$weight = \frac{k}{distance^2}$$, where k is some constant.

$$100= \frac{k}{6,400^2}$$ --> $$k=6,400^2*100$$.

$$64=\frac{6,400^2*100}{x^2}$$ --> $$x = \sqrt{\frac{6,400^2*100}{64}}=\frac{6,400*10}{8}=8,000$$ from the center of the earth.

The distance from the Earth (sea level) = $$8,000 - 6,400 = 1,600$$.

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Re: The weight of an object varies inversely as the square of the distance   [#permalink] 05 Jan 2018, 12:27
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# The weight of an object varies inversely as the square of the distance

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