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  • A particle is launched from the surface of earth with speed For the particle to move as a satellite,

A particle is launched from the surface of earth with speed \mathrm{v} For the particle to move as a satellite, which statement is correct?
 

Option: 1

\frac{\mathrm{v}_{\mathrm{e}}}{2}<\mathrm{v}<\mathrm{ve}
 


Option: 2

\mathrm{\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}<\mathrm{v}<\mathrm{ve}}

 


Option: 3

\quad \mathrm{Ve}_{\mathrm{}}<\mathrm{v}<\sqrt{2 \mathrm{v}_{\mathrm{e}}}


Option: 4

\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}<\mathrm{v}<\frac{\mathrm{v}_{\mathrm{e}}}{2}


Answers (1)

best_answer

From SOLUTION to PROBLEM 8, we have

\mathrm{ \mathrm{v}_0<\mathrm{v}<\mathrm{v}_{\mathrm{e}} }

\mathrm{\Rightarrow \frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}<\mathrm{v}<\mathrm{v}_{\mathrm{e}} }

\mathrm{\left\{\because \mathrm{v}_0=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}} \text { andv }_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{r}}}\right\} }

Hence option 2 is correct.



 

Posted by

vinayak

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