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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2019
  • Medium
  • MCQ

The value of acceleration due to gravity at Earth’s surface is \(9.8\, m\,s^{-2}\). The altitude above its surface at which the acceleration due to gravity decreases to \(4.9\, m\,s^{-2}\), is close to: (Radius of earth \(= 6.4\times10^6\, m\))

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Answer & Solution

Correct Answer

(C) \(2.6\times10^6\, m\)

Step-by-step Solution

Detailed explanation

\(\frac{{GM}}{{{{\left( {R + h} \right)}^2}}} = \frac{{GM}}{{2{R^2}}}\)

\(R + h = \sqrt 2 R\)

\(h = \left( {\sqrt 2  - 1} \right)R\)

\( \simeq 2.6 \times {10^6}m\)

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