Two weights with masses of 50 kg, being at a distance of 4 m from each other, are attracted to each other with a force F. With what force will they be attracted to each other if the distance between them decreases 2 times?
m1 = 50 kg.
m2 = 50 kg.
r1 = 4 m.
F1 = F.
G = 6.67 * 10 ^ -11H * m ^ 2 / kg ^ 2.
r2 = r1 / 2.
According to the law of universal gravitation, between bodies of masses m1 and m2, which are at a distance r from each other, there is an attractive force F, the value of which is determined by the formula: F = G * m1 * m2 / r ^ 2.
G is the gravitational constant.
F1 = G * m1 * m2 / r1 ^ 2.
F1 = 6.67 * 10 ^ -11 N * m ^ 2 / kg ^ 2 * 50 kg * 50 kg / (4 m) ^ 2 = 1042.2 * 10 ^ -11 N.
F2 = G * m1 * m2 / r2 ^ 2 = 4 * G * m1 * m2 / r1 ^ 2 = 4 * F1.
F2 = 4 * 1042.2 * 10 ^ -11 H = 4186.75 * 10 ^ -11 N.
Answer: the force of mutual attraction between the weights will increase 4 times and become F2 = 4186.75 * 10 ^ -11 N.
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