A) \[\frac{1}{2}m{{v}^{2}}\]
B) \[-\frac{1}{2}m{{v}^{2}}\]
C) \[m{{v}^{2}}\]
D) \[\frac{3}{2}\,m{{v}^{2}}\]
Correct Answer: B
Solution :
Key Idea: The necessary centripetal force to the satellite is provided by gravitational force of earth. Let a satellite is revolving around earth with orbital velocity v. The gravitational potential energy of satellite is \[U=-\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ?. (i) where, \[{{M}_{e}}\]is mass of earth, m is mass of satellite \[{{R}_{e}}\] is radius of earth and G is gravitational constant. The kinetic energy of satellite is \[K=\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ... (ii) \[\therefore \] Total energy of satellite \[E=U+K=-\frac{G{{M}_{e}}m}{{{R}_{e}}}+\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] \[=-\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ... (iii) But we know that necessary centripetal force to the satellite is provided by the gravitational force, ie., \[\frac{m{{v}^{2}}}{{{R}_{e}}}=\frac{G{{M}_{e}}m}{R_{e}^{2}}\] or \[m{{v}^{2}}=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ?. (iv) Hence, from Eqs. (iii) and (iv), we get \[E=-\frac{1}{2}m{{v}^{2}}\]
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