Question 1105280
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Think "reciprocal" when dealing with inverse variation.  The gravitational force varies inversely as the square of the distance, so if the distance is reduced to 2/3 of its original value, the factor by which the force is increased is the reciprocal of the square of 2/3:
{{{1/(2/3)^2 = 1/(4/9) = 9/4}}}<br>
The gravitational force is increased by a factor of 9/4 when the distance is decreased to 2/3 of its original distance.<br>
I think it is awkward to try to see this if you use the whole formula; here is what it might look like.<br>
Let the original force be F1: {{{F1 = (m1*m2)/d^2}}}<br>
The new force F2, with the distance cut to 2/3 of the original distance, is {{{F2 = (m1*m2)/((2/3)d)^2 = (m1*m2)/((4/9)d^2) = (9/4)*((m1*m2)/d^2) = (9/4)*F1}}}<br>
So the force F2 is F1, increased by a factor of 9/4.