Gravitational field intensity of a carved out spherical body

From a uniform sphere of radius 2R2R, a spherical cavity of radius RR is cut in such a way that the sphere and the spherical cavity share a common tangent, as shown in the diagram. The mass of the new body is MM. Find the gravitational field intensity at point AA which is at a distance of 6R6R from the center of the sphere.

If this value can be expressed as abGMR2\dfrac{a}{b} \cdot \dfrac{GM}{R^2}, where aa and bb are coprime positive integers, then evaluate a+ba+b.

Details and Assumptions:

  • Point AA is at a distance of 6R6R from the geometrical center of the original, larger sphere, not from the center of mass of the newly formed body.
  • Point AA lies such that it is collinear with the centers of the sphere and the spherical cavity and nearer to the common tangent shared by them.
  • GG denotes the universal gravitational constant: G=6.674×1011 m3 kg1 s2.G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}.
  • Neglect Earth's gravitational field and deformities within the sphere.
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