
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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.


Tags 
equation, linear, problem, variation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Linear Equation Problem  willreyes93  Linear Algebra  2  September 18th, 2014 02:07 PM 
2nd order differential equation  variation of parameters  Akcope  Calculus  0  May 8th, 2014 12:26 PM 
Linear Variation Proportion  Xeon8  Algebra  4  April 27th, 2013 07:18 PM 
Linear equation problem  lumen8r  Linear Algebra  1  March 9th, 2010 04:06 PM 
Linear equation word problem  okcomputer76  Linear Algebra  0  December 31st, 1969 04:00 PM 