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Physics · GRAVITATION
हिन्दीFor a small distance \( x \ll R_E \), let \( \Delta g_h \) be the decrease in acceleration due to gravity at a height \( h = x \) above the Earth's surface and \( \Delta g_d \) be the decrease at a depth \( d = x \) below the Earth's surface. The ratio \( \frac{\Delta g_h}{\Delta g_d} \) is:
Correct answer · Option 1
\( 2 \)
Explanation
\( \Delta g_h = g - g(1 - 2x/R_E) = \frac{2x}{R_E}g \), and \( \Delta g_d = g - g(1 - x/R_E) = \frac{x}{R_E}g \). Thus, \( \frac{\Delta g_h}{\Delta g_d} = 2 \).
