# 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/R

If R = 4.19 R(Earth) we can write such expression

g'/g = (GM/(4.19R)

Thus

g' = g/4.19

g = GM/R

^{2}= 9.8 m/s^{2}If R = 4.19 R(Earth) we can write such expression

g'/g = (GM/(4.19R)

^{2})/(GM/R^{2}) = 1/4.19^{2}Thus

g' = g/4.19

^{2}= 9.8/17.5561 = 0.558 m/s^{2}Need a fast expert's response?

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