A) 1 hour
B) 2 hour
C) 24 hour
D) 36 hour
Correct Answer: B
Solution :
From Keplers law\[{{T}^{2}}\propto {{r}^{3}}\] \[\therefore \,{{T}^{2}}\propto {{(36000)}^{3}}\] and \[T{{}^{2}}\propto {{(6400+h)}^{3}}\] There fore \[{{(T)}^{2}}={{T}^{2}}{{\left( \frac{6400+h}{36000} \right)}^{3}}\] \[>{{T}^{2}}{{\left( \frac{6400}{36000} \right)}^{3}}\,\,(\therefore h<<R)\] \[>{{(24)}^{2}}{{\left( \frac{8}{45} \right)}^{3}}\,\] \[\therefore \] \[T>\frac{24\times 8\times \sqrt{8}}{45\times \sqrt{45}}\] \[>1.8\,\text{hour}\] So, \[T\approx 2\,\text{hour}\]
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