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Answer to Question #1331 in Mechanics | Relativity for Jon Allen

Question #1331
When a falling meteor is at a distance 4.19 times the radius of the Earth above the Earth’s surface, what is its free fall acceleration? The acceleration of gravity is 9.8 m/s2, the universal gravitational constant is 6.67259×10−11 N · m2/kg2, and the Earth’s radius is 6.37 × 106 m. Answer in units of m/s2.
Expert's answer
The gravitational acceleration on the Earth is
g = GM/R2 = 9.8 m/s2
If R = 4.19 R(Earth) we can write such expression
g'/g = (GM/(4.19R)2)/(GM/R2) = 1/4.192
Thus
g' = g/4.192 = 9.8/17.5561 = 0.558 m/s2

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