91Ó°ÊÓ

Since the spheres are touching each other, the distance between the centers of any two touching spheres is equal to the sum of their radii. In this case, all the spheres have radius R, so the distance between their centers is 2R.

We will use Newton's law of gravitation to find the force between two spheres of mass M separated by a distance 2R. The formula for the gravitational force between two masses is: \[F = \frac{GMm}{r^2}\] Here, G is the gravitational constant, M and m are the masses of the spheres, and r is the distance between their centers. In this case, both masses are M and the distance is 2R, so the formula becomes: \[F = \frac{GM^2}{(2R)^2}\]

Now we need to find the force on one sphere due to the other two spheres. We have an equilateral triangle shape with each sphere at a vertex. Since the forces between each pair of sphere are equal and act along the line connecting their centers (refer to step 2), we can use trigonometry to find the net force: Due to symmetry, the force vectors will make an angle of 120 degrees with each other. Now we can find the net force by adding the two force vectors. Let F1 be the force on the sphere due to one of the other spheres, and F2 be the force due to the second sphere. Since both F1 and F2 have the same magnitude from step 2, we can denote them both as F and proceed with vector addition: \[F_{net} = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos120^\circ}\] Substitute the values found in step 2: \[F_{net} = \sqrt{\left(\frac{GM^2}{(2R)^2}\right)^2 + \left(\frac{GM^2}{(2R)^2}\right)^2 + 2\left(\frac{GM^2}{(2R)^2}\right)\left(\frac{GM^2}{(2R)^2}\right)\cos120^\circ}\] Simplify the equation: \[F_{net} = \frac{GM^2}{(2R)^2}\sqrt{1 + 1 + 2(\cos120^\circ)}\] Since \(\cos120^\circ = -\frac{1}{2}\), the equation becomes: \[F_{net} = \frac{GM^2}{(2R)^2}\sqrt{1 + 1 -1}\] \[F_{net} = \frac{GM^2}{(2R)^2}\] This corresponds to option (A) \(\frac{GM^2}{4R^2}\).