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:

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

Where:
- \( g_h \) is the acceleration due to gravity at height \( h \).
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