JEE Mains · Physics · STD 11 - 7. gravitation
The percentage decrease in the weight of a rocket, when taken to a height of \(32 km\) above the surface of earth will, be\(.....\%\) (Radius of earth \(=6400\,km\) )
- A \(1\)
- B \(3\)
- C \(4\)
- D \(0.5\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
Acceleration due to gravity at a height \(h << R\) is \(g ^{\prime}= g \left(1-\frac{2 h }{ R }\right)\) \(\therefore \frac{\Delta g }{ g }=\frac{2 h }{ R }\) \(\Rightarrow \frac{\Delta g }{ g } \times 100=\frac{2 h }{ R } \times 100\)…
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