Question

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.855 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80m/s2?

Answer 1

Given: Normal pull of gravity g = 9.8 m/s²; 

 g = 0.855 m/s²  (at a certain distance)

Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

Mass of the Earth Me = 5.98 x 10²⁴ Kg

Radius r = ?

g = GMe/r²

r = √GMe/g

r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)

r = 2.16 x 10⁷ m or 

r =  21,610 Km





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