question_answer

The orbital velocity of a satellite at a height h above the surface of earth is v. The value of escape velocity from the same location is given by:

A)  \[\sqrt{2}v\]                                     

B)  v

C)  \[\frac{v}{\sqrt{2}}\]                                    

D) \[\frac{v}{2}\]

Correct Answer: A

Solution :

                The orbital velocity\[({{v}_{o}})\]of a satellite at a height h above the surface of earth is \[{{v}_{o}}=\sqrt{\frac{G{{M}_{e}}}{{{R}_{e}}+h}}\]                        ?. (i) The escape velocity\[({{v}_{e}})\]is                 \[{{v}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}+h}}\]                      ?.. (ii) From Eqs. (i) and (ii), we get                 \[{{v}_{e}}=\sqrt{2}.{{v}_{o}}\] Given,      \[{{v}_{o}}=v\] \[\therefore \]  \[{{v}_{e}}=\sqrt{2}v\]