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: Total energy of satellite is equal to sum of kinetic energy and potential energy in circular orbit around the 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, \[{{m}_{e}}=\]mass of earth, \[m=\]mass of satellite \[{{R}_{e}}=\]radius of earth and \[G=\]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, as energy is conserved, is \[E=U+K=\frac{G{{M}_{e}}m}{{{R}_{e}}}+\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ?.. (iii) But we know that necessary centripetal force to the satellite is provided by the gravitational force, i.e., \[\frac{m{{v}^{2}}}{{{R}_{e}}}=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] Or \[m{{v}^{2}}=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ??. (iv) Hence,, from Eqs. (iii) and (iv), we get \[E=-\frac{1}{2}m{{v}^{2}}\] Note: If positive value of this energy is given to the satellite, then it will escape the earth's orbit and never returns.
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