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Physics · GRAVITATION

हिन्दी

At what height \( h \) above the surface of the Earth does the acceleration due to gravity become \( \frac{g}{9} \), where \( g \) is the value of acceleration due to gravity on the Earth's surface and \( R_E \) is the Earth's radius?

  1. A
    \( h = 2R_E \)
  2. B
    \( h = 3R_E \)
  3. C
    \( h = 8R_E \)
  4. D
    \( h = \frac{R_E}{2} \)

Correct answer · Option A

\( h = 2R_E \)

Explanation

Using \( g(h) = \frac{g}{(1 + \frac{h}{R_E})^2} = \frac{g}{9} \), we get \( 1 + \frac{h}{R_E} = 3 \implies h = 2R_E \).
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