If both masses are increased by 60% and the distance between them is halved, by what percent will the force of attraction increase?
Since the force is given by
F = (constant)*M1*M2/R^2 ,
The new force, after the changes, will be higher by a factor (1.6)^2/(0.5)^2 = 10.24
That means there is a 924% increase.
You do not have to know the actual masses and distances to make that conclusion.
M1 and M2 are the two masses, and R is the distance between them.
To find the percent increase in the force of attraction, we need to compare the initial force of attraction with the final force of attraction after the changes.
Let's assume the initial masses are m1 and m2, and the initial distance between them is d. The initial force of attraction, F1, can be calculated using the formula of gravitational force:
F1 = G * (m1 * m2) / d^2 (where G is the gravitational constant)
Now, when both masses are increased by 60%, the new masses will be (1 + 0.6) * m1 = 1.6 * m1 and (1 + 0.6) * m2 = 1.6 * m2. Additionally, when the distance is halved, the new distance will be d/2.
So, the final force of attraction, F2, can be calculated using the same formula:
F2 = G * ((1.6 * m1) * (1.6 * m2)) / (d/2)^2
Now, let's calculate the percent increase in the force of attraction by comparing F1 and F2:
Percent Increase = ((F2 - F1) / F1) * 100
Substituting the values and simplifying the formula, we get:
Percent Increase = ((1.6 * 1.6) - 1) * 100
Percent Increase = (2.56 - 1) * 100
Percent Increase = 1.56 * 100 = 156%
Therefore, the force of attraction will increase by 156%.