A) 0.0084 N
B) 0.084 N
C) 0.84 N
D) 8.40 N
Correct Answer: B
Solution :
Let angular speed of spaceship and astronaut is co. |
Forces on astronaut provides necessary centripetal pull. |
\[\Rightarrow \]\[\frac{G{{M}_{e}}m}{{{(r+L)}^{2}}}+T=m(r+L){{\omega }^{2}}\] |
\[\Rightarrow \]\[mg{{\left( 1-\frac{L}{r} \right)}^{2}}+T=\frac{mg}{r}(r+L)\] |
Taking \[r=R\](radius of earth) |
\[\Rightarrow \]\[mg-\frac{2mgL}{R}+T=mg+\frac{mgL}{R}\] |
\[\Rightarrow \]\[T=\frac{3mgL}{R}=\frac{3\times 200\times 10\times 90}{6400\times {{10}^{3}}}=0.084N\] |
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