An object is released from a height h₁ = 3R_E above the surface of the earth, where R_E is the radius of the earth. As it falls, it reaches a height h₂ above the earth's surface where its weight becomes 4 times its weight at the release point. The value of h₂ is:

Step-by-step solution

The acceleration due to gravity at a height h from the surface of the earth is given by g_h = g ( R_E R_E + h )^2 . At the initial height h₁ = 3R_E , the acceleration due to gravity is: g₁ = g ( R_E R_E + 3R_E )^2 = g ( 1 4 )^2 = g 16 The weight of the object at this height is W₁ = m g₁ = mg 16 . At height h₂ , the weight becomes 4 times the initial weight: W₂ = 4 W₁ = 4 mg 16 = mg 4 This means the acceleration due to gravity at h₂ is g₂ = g 4 . Using the formula for g at height h₂ : g 4 = g ( R_E R_E + h₂ )^2 Taki