Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.
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Hey mate,
◆ Answer-
F = √3/4 Gm^2 / R^2
◆ Explaination-
Consider 3 spheres with same mass m and radius R connected to each other.
[Refer to the figure]
Three masses are m1 = m2 = m3 = m.
Distance between centre of any two spheres r = 2R
The 3 centres will form equilateral triangle.
Individual forces on m1-
F12 = Gm^2 / 4R^2
F13 = Gm^2 / 4R^2
Resultant force on m1-
F = √[F12^2 + F13^2 + 2×F12×F13×cos60°]
F = √[(Gm^2/4R^2)^2 + (Gm^2/4R^2)^2 + 2×(Gm^2/4R^2)(Gm^2/4R^2)×1/2]
F = √3/4 (Gm^2/R^2)
Thus, resultant force on one sphere by other two is √3/4 Gm^2 / R^2.
Hope that is useful...
◆ Answer-
F = √3/4 Gm^2 / R^2
◆ Explaination-
Consider 3 spheres with same mass m and radius R connected to each other.
[Refer to the figure]
Three masses are m1 = m2 = m3 = m.
Distance between centre of any two spheres r = 2R
The 3 centres will form equilateral triangle.
Individual forces on m1-
F12 = Gm^2 / 4R^2
F13 = Gm^2 / 4R^2
Resultant force on m1-
F = √[F12^2 + F13^2 + 2×F12×F13×cos60°]
F = √[(Gm^2/4R^2)^2 + (Gm^2/4R^2)^2 + 2×(Gm^2/4R^2)(Gm^2/4R^2)×1/2]
F = √3/4 (Gm^2/R^2)
Thus, resultant force on one sphere by other two is √3/4 Gm^2 / R^2.
Hope that is useful...
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