| (1) The value of gravitational acceleration\[g=10\,m\,{{s}^{-2}}\]. |
| (2) Radius of earth \[{{R}_{E}}=6400\text{ }km\]. Take \[\pi =3.14\]. |
A) \[83.73\text{ }minutes~\]
B) \[85\text{ }minutes\]
C) \[90minutes\]
D) \[156\text{ }minutes\]
Correct Answer: A
Solution :
: Period of revolution of earth satellite is \[T=2\pi \sqrt{\frac{{{({{R}_{E}}+h)}^{3}}}{gR_{E}^{2}}}\] Neglecting h, then \[T=2\pi \sqrt{\frac{{{R}_{E}}}{g}}\] Substituting the given values, we get \[T=2\times 3.14\sqrt{\frac{6400\times {{10}^{3}}}{10}}s\] \[=5024s=\frac{5024}{60}\]minutes \[=83.73\] minutes
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