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  • The minimum time period of revolution of an artifical satellite is equal to <img alt="\mathrm{\left(M=\right. \text{mass of earth}, R= \text{radius of earth}, g=9.8 \mathrm{~m} / \mathrm{s}^2}" src

The minimum time period of revolution of an artifical satellite is equal to \mathrm{\left(M=\right. \text{mass of earth}, R= \text{radius of earth}, g=9.8 \mathrm{~m} / \mathrm{s}^2}
 

Option: 1

\mathrm{2 \pi \sqrt{R^2 / G M}}
 


Option: 2

\mathrm{82 \mathrm{~min}}
 


Option: 3

\mathrm{2 \pi \sqrt{R / g}}
 


Option: 4

All of the above


Answers (1)

best_answer

The time period of revolution \mathrm{T}=2 \pi / \omega where \omega can be given as

\mathrm{\mathrm{mr} \omega^2 =\frac{\mathrm{GMm}}{r^2} }

\mathrm{\Rightarrow \omega =\sqrt{\frac{\mathrm{GM}}{\mathrm{r}^3}} }

\mathrm{\Rightarrow \mathrm{T} =\frac{2 \pi}{\omega} \omega=2 \pi \sqrt{\frac{r^3}{\mathrm{GM}}} }

\mathrm{T} is minimum when \mathrm{r} is minimum. \mathrm{\gamma_{\mathrm{min}}=\mathrm{R}}

\mathrm{ \Rightarrow T_{\min }=2 \pi \sqrt{\frac{r^3}{G M}}=2 \pi \sqrt{\frac{R}{g}} }

\mathrm{=2 \pi \sqrt{\frac{6400 \times 10^3}{9.8}} \text { sec. }=82 \mathrm{~min} .}

Hence option 4 is correct.



 

Posted by

sudhir kumar

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