JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2022
- Medium
- Numerical
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 \(=6400\,km\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
Acceleration due to gravity at a height \(h << R\) is
\(g ^{\prime}= g \left(1-\frac{2 h }{ R }\right)\)
\(\therefore \frac{\Delta g }{ g }=\frac{2 h }{ R }\)
\(\Rightarrow \frac{\Delta g }{ g } \times 100=\frac{2 h }{ R } \times 100\)
\(=2 \times \frac{32}{6400} \times 100=1 \%\)
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