The gravitational force between two metal spheres with the same masses are in outer space is 60 N. How large would this force be if the masses of each sphere were cut to a fourth of its original value?
F_{attractive} =_____________ N
(no rounding up)
16 times 60
Hey, this is getting old
To find the new gravitational force between the two metal spheres, we can use the formula for the gravitational force:
F = (G * m1 * m2) / r^2
where:
- F is the gravitational force
- G is the gravitational constant (approximately 6.67430 × 10^-11 Nm^2/kg^2)
- m1 and m2 are the masses of the two objects
- r is the distance between the centers of the two objects
In this case, the initial gravitational force (F_initial) between the two metal spheres is given as 60 N.
If the masses of each sphere are cut to a fourth of their original value, we can denote the new masses as m1_new = (m1_initial) / 4 and m2_new = (m2_initial) / 4.
Let's substitute these values into the formula:
F = (G * m1_new * m2_new) / r^2
F = (G * (m1_initial / 4) * (m2_initial / 4)) / r^2
Since the problem does not provide the values of the masses or the distance between the spheres, we cannot calculate the exact numerical value of the new gravitational force. However, we can express it in terms of the initial force.
F_new = (1/16) * F_initial
Therefore, the new gravitational force (F_new) between the metal spheres, when the masses of each sphere are cut to a fourth of their original value, is 1/16 times the initial force (F_initial).
F_new = (1/16) * 60 N
F_new = 3.75 N
So the new gravitational force between the metal spheres is 3.75 N.