An uniform sphere of mass M and radius R exerts a force of F on a small mass m placed at a distance of 3R from the centre of the sphere. A spherical portion of diameter R is cut from the sphere as shown in the fig. The force of attraction between the remaining part of the disc and the mass m is

Step-by-step solution

The force exerted by the original sphere of mass M and radius R on a mass m at a distance d = 3R from its center is given by F = GMm (3R)^2 = GMm 9R^2 . The mass of the original sphere is M = 4 3 R^3 , where is the density. The spherical portion cut out has a radius r = R/2 . Its mass M' is M' = 4 3 (R/2)^3 = 4 3 R^3 8 = M 8 . The center of the cut-out sphere is at a distance x = R/2 from the center of the original sphere. The mass m is at a distance d = 3R from the center of the original sphere. The distance of m