question_answer

A remote sensing satellite of earth revolves in a circular orbit at a height of \[0.25\times {{10}^{6}}\] m above the surface of earth. If earth's radius is \[6.38\times {{10}^{6}}\] and \[g=9.8m{{s}^{-2}}\], then the orbital speed of the satellite is

A)  7.76 \[km{{s}^{-1}}\]       

B)  8.5\[km{{s}^{-1}}\]        

C)  9.13 \[km{{s}^{-1}}\]    

D)  6.67 \[km{{s}^{-1}}\]

Correct Answer: A

Solution :

Given, height of a satellite \[h=0.25\times {{10}^{6}}m\] Earth's radius, \[{{R}_{e}}=6.38\times {{10}^{6}}m\] For the satellite revolving around the earth, orbital velocity of the satellite. \[{{v}_{0}}=\sqrt{\frac{G{{M}_{e}}}{{{R}_{e}}}}=\sqrt{\frac{G{{M}_{e}}}{{{R}_{e}}\left[ 1+\frac{h}{{{R}_{e}}} \right]}}\]\[{{v}_{0}}\sqrt{\frac{g{{R}_{e}}}{1+\frac{h}{{{R}_{e}}}}}\] Substitutes the value of \[g,{{\operatorname{R}}_{e}}\] and h, we get                                 \[{{v}_{0}}=\sqrt{60\times {{10}^{6}}}m/s\]                                 \[{{v}_{0}}=7.76\times {{10}^{3}}m/s=7.76km/s\]