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Physics · GRAVITATION
हिन्दीAt what height \( h \) above the surface of the Earth does the acceleration due to gravity become equal to the acceleration due to gravity at a depth \( d = \frac{R_E}{4} \) below the surface of the Earth? (where \( R_E \) is the radius of the Earth)
Correct answer · Option 4
\( (\frac{2}{\sqrt{3}} - 1) R_E \)
Explanation
At depth \( d = R_E/4 \), \( g_d = g(1 - d/R_E) = \frac{3}{4}g \). At height \( h \), \( g_h = \frac{g}{(1 + h/R_E)^2} \). Equating gives \( (1 + h/R_E)^2 = \frac{4}{3} \implies h = (\frac{2}{\sqrt{3}} - 1) R_E \).
