At what altitude, the acceleration due to gravity reduces to one fourth of its value as that on the surface of the earth ? Take radius of earth as `6.4 xx 10^(6) m`, g on the surface of the earth as `9.8 ms^(-2)`.

Explanation

To find the altitude at which the acceleration due to gravity reduces to one-fourth of its value on the surface of the Earth, we can use the formula for gravitational acceleration at a height \( h \) above the Earth's surface:

\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]

Where:
- \( g' \) is the acceleration due to gravity at height \( h \)
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