A) 10 hour
B) \[\text{5}.\text{54}\times \text{1}{{0}^{-\text{7}}}\text{rad}\] hour
C) 6 hour
D) \[\text{2}.\text{54}\times \text{1}{{0}^{-\text{4}}}\text{ rad}\] hour
Correct Answer: D
Solution :
By Keplers law \[{{T}^{2}}\propto {{R}^{3}}\] Hence, \[{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2}}={{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3}}\] \[={{\left( \frac{2.5R+R}{6R+R} \right)}^{3}}\] \[={{\left( \frac{1}{2} \right)}^{3}}\] \[{{T}_{2}}=\frac{{{T}_{1}}}{{{(2)}^{3/2}}}\] For a geostationary satellite \[{{T}_{1}}=24\,\,hour\] So, \[{{T}_{2}}=\frac{24}{2\sqrt{2}}\] \[=6\sqrt{2}\,\,hour\]
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