A) \[\frac{1}{2}h\]
B) \[1\,h\]
C) \[2\,h\]
D) \[4\,h\]
Correct Answer: C
Solution :
\[{{T}_{2}}={{T}_{1}}{{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{3/2}}\] \[=24{{\left( \frac{6400}{36000} \right)}^{3/2}}\] \[=1.7h\] For spy satellite h is slightly greater than\[{{R}_{e}}\]. So, \[{{T}_{1}}>{{T}_{2}}\] hence,\[{{T}_{s}}=2h\]You need to login to perform this action.
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