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

Question

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.

Answer 1

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