Question

At what height above the surface of earth will the value of 'g' be reduced to 81% of
its value at the surface ? (Take the radius of the earth as 6300 km).
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Answer 1

Answer:

711.04km

Explanation:

Let g' be height to which acceleration due to gravity decreases to 81% of its value at the surface

g' = (81/100)g

According to formula g' = g ( R/(R + h))^2

(81/100) g = g×(R/(R + h))^2

Taking square root of both sides

9/10 = R/(R+h)

0.9 = R/(R + h)

0.9 × (R + h) = R

0.9 R + 0.9 h =R

0.9 h = 0.1 R

h = (1/9) R

h = 0.1111 R

h = 0.1111 × 6400 km

h = 711.04 km from surface of earth

Hope this may help