Question

At what height above the surface of the earth does the acceleration due to gravity reduces by 64% of its value on the surface of the earth . Radius of the earth =6400 km.
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Answer 1

Answer:-

4266.7 km

Explanation:-

We know that:-

=> g' = g/(1+h/r)²

Now, g reduces by 64%, thus the value of g' becomes:-

=> g' = g-64/100×g

=> g' = g - 64/100 g

=> g' = 36/100 g

Thus:-

=> 36/100 g = g/(1+h/r)²

=> 36/100 = 1/(1+h/r)²

=> (10/6)² = (1+h/r)²

=> 10/6 = 1+h/r

=> 10/6-1 = h/r

=> 2/3 = h/r

=> h = 2r/3

•Value of 'r' (Radius of Earth ) = 6400km

=> h = 2×6400/3

=> h = 12800/3

=> h = 4266.7km

Thus, at a height of 4266.7km above the surface of earth, acceleration due to gravity reduces by 64% of it's value on the surface of earth.

Some Extra Information:-

•Value of g at the surface of earth is

9.8m/s² on an average.

•The value of g decreases with height.

•The value of g decreases with depth.

•The value of g is more at poles than equator.

•The value of g is zero at the centre of the earth.

Answer 2

Answer:

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