SOLUTION: Can you please help me with this word problem? The gravitational attraction between two masses varies inversely as the square of the distance between them. The force of attracti

Algebra ->  Equations -> SOLUTION: Can you please help me with this word problem? The gravitational attraction between two masses varies inversely as the square of the distance between them. The force of attracti      Log On


   

Question 208695: Can you please help me with this word problem?
The gravitational attraction between two masses varies inversely as the square of the distance between them. The force of attraction is 4 lbs. when the masses are 3 ft apart, what is the attraction when the masses are 6 ft. apart?
Not sure how to even set up, algebra is hard for me!
Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
Let g = gravitational attraction force in terms of pounds between two masses.
Your equation is: g = k/d^2, where k is the constant of proportionality and d is our distance squared.
We know that g = 4.
We also know that 3 = d.
We first need to find k.
4 = k/(3)^2
4 = k/9
4*9 = k
36 = k
Now that we know k, we need to find g.
g = 36/(6)^2
g = 36/36
g = 1 pound.