JEE Mains · Physics · STD 11 - 7. gravitation
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 \(= 6.4\times10^6\, m\))
- A \(6.4\times10^6\, m\)
- B \(9.0\times10^6\, m\)
- C \(2.6\times10^6\, m\)
- D \(1.6\times10^6\, m\)
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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