From what height above the surface of the earth should an object be dropped to initially experience an acceleration of 0.54g? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg,R_earth = 6.38 × 10⁶ m) A)2300 km B)1700 km C)5400 km D)2900 km

Question

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The acceleration due to gravity at a height h above the surface of the Earth is given by:$$g' = \frac{GM_e}{(R_e + h)^2}$$where G is the gravitational constant, M_e is the mass of the Earth, and R_e is the radius of the Earth.We want to find the height h at which g' = 0.54g. Substituting the given values into the equation, we get:$$0.54g = \frac{GM_e}{(R_e + h)^2}$$$$0.54(9.81 m/s^2) = \frac{(6.67 × 10^{-11} N ∙ m^2/kg^2)(5.97 × 10^{24} kg)}{(6.38 × 10^6 m + h)^2}$$Solving for h, we get:$$h = 2.3 × 10^6 m = 2300 km$$Therefore, the correct answer is A) 2300 km.

From What Height Above the Surface of the Earth Should