- The problem states that the diameter of the Earth becomes two times its present value, which means the radius R also becomes two times its present value (R' = 2R). - The mass of the Earth (M) remains unchanged. - The new acceleration due to gravity g' will be GM/(R')² = GM/(2R)² = GM/(4R²) = (1/4) (GM/R²) = g/4. - The new weight W' = mg' = m(g/4) = (1/4)mg = W/4. - Therefore, the weight of the object becomes one-fourth of its present value.
If the diameter of the earth becomes two times its present value and its mass remains unchanged then how would the weight of an object on the surface of the earth be affected
Asked in TNPSC CS Prelims 2019 · Topic: Mechanics (Gravitation and Central Force Motion (Kepler's Laws))
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