Answer to Question #159936 in Classical Mechanics for Yolande

Question #159936

When a falling meteoroid is at a distance above the Earth's surface of 3.00 times the Earth's radius, what is its accleration due to the Earth's gravitation?

Expert's answer

"a=\\frac{GM}{(3R)^2}=\\frac{GM}{9R^2}=\\frac{g}{9}\\\\a=\\frac{9.8}{9}=1.1\\frac{m}{s^2}"

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