Question

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?

Answer 1

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