My Math Forum Variation Linear Equation Problem

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 June 4th, 2018, 10:27 AM #1 Newbie   Joined: May 2018 From: United States Posts: 6 Thanks: 0 Math Focus: Algebra Variation Linear Equation Problem Problem: "The weight of an object varies inversely as the square of the distance from the center of the earth. At sea level (6400 km from the center of the earth), an astronaut weighs 100 lb. How far above the earth must the astronaut be in order to weigh 64 lb?" I generally understand elementary Variation Linear Equations, but something about this question seems to be fundamentally throwing me off. Assuming that I understand the question correctly, I can successfully get the constant (4,096,000,000) , but something after that seems to have gone frightfully wrong as I consistently get an answer/distance of 80,000, which my book says is wrong. (w)(d²)=k (100)(6400²)=k (100)(40,960,000)=k 4,096,000,000=k 4,096,000,000=(64)(d²) 4,096,000,000/64=(64/64)(d²) 64,000,000=d² (Now I find the square root of both sides to remove d's exponent) 80,000=d Can anyone please tell me where I have gone wrong?
 June 5th, 2018, 07:43 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,307 Thanks: 1976 The square root of 64,000,000 is 8000, not 80,000. Subtract 6400 km from 8000 km to get the distance above the earth's surface. Thanks from Ebba Sen Pai

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