At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.995 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80 m/s2?

Question

Bonilla

Physics

13 December, 04:15

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.995 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80 m/s2?

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Amiyah Petersen · Answer 1

If we assume the Earth to be of uniform density and spherical, then Newton's inverse square law of g-field can be applied.

i. e. g = GM/r², which means g is inversely proportional to r²

Or, gr² = constant.

The Earth's radius is approximately 6.4x10^6 m

So, 0.975 x (R') ² = 9.80 x (6.4x10^6) ², where R' is the distance from Earth's centre.

R' = 2.03x10^7 m

Distance above Earth's surface = 2.03x10^7 - 6.4x10^6 = 1.39x10^7