Two masses are attracted by a gravitational force of 0.48 N. What will the force of attraction be if the distance between the two masses is halved?
F_{attraction} =
4 times
1/(.5r)^2 = 1/.25 r^2 = 4/r^2
To calculate the force of attraction between two masses when the distance between them is halved, we can use Newton's law of universal gravitation. The formula is:
F_{attraction} = (G * m1 * m2) / r^2
Where:
F_{attraction} is the force of attraction between the two masses,
G is the gravitational constant (approximately 6.674 x 10^-11 N(m/kg)^2),
m1 and m2 are the masses of the two objects, and
r is the distance between the centers of the masses.
In this case, the original force of attraction is given as 0.48 N. Let's assume that this force is the result of two masses, m1 and m2, and a distance r.
Now, if the distance between the two masses is halved, the new distance will be r/2. Let's calculate the new force of attraction:
F_{attraction} = (G * m1 * m2) / (r/2)^2
= (G * m1 * m2) / (r^2 / 4)
= (4 * G * m1 * m2) / r^2
Since G and the masses, m1 and m2, remain constant, the force of attraction when the distance is halved will simply be four times the original force:
F_{attraction} = 4 * 0.48 N
= 1.92 N
Therefore, the force of attraction when the distance between the two masses is halved will be 1.92 N.