# Math Help - Inverse Variation

1. ## Inverse Variation

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?

My Work:

Let g = gravitational attraction force in terms of pounds between two masses.

My equation is: g = k/d^2, where k is the constant of proportionality and d is the distance squared.

I know that g = 4.
I also know that 3 = d.
I first need to find k.
4 = k/(3)^2
4 = k/9
4*9 = k
36 = k
Now that I know k, I need to find g.
g = 36/(6)^2
g = 36/36
g = 1 pound.

Is this correct?

2. Originally Posted by sharkman
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?

My Work:

Let g = gravitational attraction force in terms of pounds between two masses.

My equation is: g = k/d^2, where k is the constant of proportionality and d is the distance squared.

I know that g = 4.
I also know that 3 = d.
I first need to find k.
4 = k/(3)^2
4 = k/9
4*9 = k
36 = k
Now that I know k, I need to find g.
g = 36/(6)^2
g = 36/36
g = 1 pound.

Is this correct?

correct