Which of the following is the variation of acceleration due to gravity at a depth “d” below the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth. (Assume the density of the earth to be constant)(a) g = (G*M1)/(R – d)(b) g = [(G*M1) x density]/d(c) g = (G x M1/R^3) / (R – d)(d) g = (G x M1/R^3) x (R – d)The question was posed to me in an interview.Query is from Acceleration due to Gravity below and above the Surface of Earth topic in section Gravitation of Physics – Class 11

Which of the following is the variation of acceleration due to gravity

1.	Which of the following is the variation of acceleration due to gravity at a depth “d” below the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth. (Assume the density of the earth to be constant)(a) g = (GM1)/(R – d)(b) g = [(GM1) x density]/d(c) g = (G x M1/R^3) / (R – d)(d) g = (G x M1/R^3) x (R – d)The question was posed to me in an interview.Query is from Acceleration due to Gravity below and above the Surface of Earth topic in section Gravitation of Physics – Class 11
Answer» The CORRECT option is (d) g = (G x M1/R^3) x (R – d) EASY explanation: g = (G*M)/r^2; M = MASS contained in an enclosed volume r = distance from centre of the earth to the DEPTH “d” Therefore; r = R – d; where, R = Radius of the earth M = (density) x (volume) = (density) x [(4/3) x (pi) x (r^3)] = (density x 4 x pi x r^3) / 3 = (M1 x r^3) / R^3 Therefore; g = (G x density x 4 x pi x r^3) / (3 x r^2) = (G x density x 4 x pi x r^1) / 3 Therefore; g = [(G x density x 4 x pi) / 3] x (R – d) = (G x M1/R^3) x (R – d).

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