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?

The distance above the surface of the earth is .

Explanation:

Given that,

Acceleration due to the gravity g'= 0.995 m/s² (at height )

Acceleration due to the gravity g= 9.80 m/s² (at surface )

We know that,

The formula of gravity at height

Where, h = height

r= radius of the earth

We substitute the value into the formula

Hence, The distance above the surface of the earth is .

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.