Determine the force of attraction between two lead balls with a diameter of 1 m each, located at a distance of 1.5 m

Determine the force of attraction between two lead balls with a diameter of 1 m each, located at a distance of 1.5 m from each other. The density of lead is 11.3 g / cm ^ 3

universal gravitation:
F = G * m1 m * 2 / R2
Ftyag = Gm2R2 = Gρ2V2R2 = Gρ2 (4πR3 / 3) 2R2
F≈0.001H
find out the mass of the ball (the same)
m = p * V, the volume of the ball = 4 / 3ПR ^ 3 = 0.523598 cubic meters, then the mass
m = 11300 * 0.523598 = 5916.6 kg
F = G * ((m1 * m2) / R ^ 2), where G – gravitational constant = 6.67 * 10 ^ (- 11)
F = 6.67 * 10 ^ (- 11) * ((5916.6 * 5916.6) / 2.25)
F = 6.67 * 10 ^ (- 11) * 15558291.36
F = 6.67 * 10 ^ (- 11) * 0.1555829136 * 10 ^ 8
F = 1.037 * 10 ^ (- 3) = 1.037 miles Newtons
Answer: 1.037mN.