Three galaxies,each of mass M = 4.0 *10⁴¹ kg,lie in a plane at the corners of an equilateral triangle with sides of 5.0 * 10²² m length.The magnitude of the force the other two galaxies exert on each galaxy is A)4.3 * 10²⁷ N. B)6.4 * 10²⁷ N. C)7.4*10²⁷ N. D)8.6 * 10²⁷ N. E)4.3 * 10²⁸ N.

Question

Quizplus · Accepted Answer

We can use Newton's Law of Gravitation to calculate the force between two masses:

F = G*m1*m2/r^2

Where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

In this case, each galaxy is acting on the other two galaxies. Since the triangle is equilateral, each side has a length of 5.0 x 10^10 m, and the distance between each galaxy is the length of a side divided by the square root of 3 (which is the ratio of the height of an equilateral triangle to its side length):

r = (5.0 x 10^10 m)/sqrt(3)

Plugging in the numbers, we get:

F = G*m1*m2/r^2 = (6.67 x 10^-11 N m^2/kg^2)*(4.0 x 10^10 kg)^2/[(5.0 x 10^10 m)/sqrt(3)]^2

Simplifying this expression, we get:

F = 7.4 x 10^27 N

This is closest to choice C, 7.4 x 10^27 N.

Three Galaxies,each of Mass M = 4