A) \[\sqrt{\frac{GM}{R}}\]
B) \[\sqrt{\frac{2GM}{R}}\]
C) \[\sqrt{\frac{5}{4}\,\frac{GM}{R}}\]
D) \[\sqrt{\frac{3GM}{R}}\]
Correct Answer: A
Solution :
Key Idea: According to the conservation of energy, total energy at the surface of earth must equal to the total energy at the maximum height. As from key idea, energy at surface of earth = energy at maximum height \[\therefore \frac{1}{2}m{{u}^{2}}-\frac{GMm}{R}=\frac{1}{2}m\times {{(0)}^{2}}-\frac{GMm}{R+h}\] \[or\frac{1}{2}m{{u}^{2}}=\frac{GMm}{R}-\frac{GMm}{R+R}\,\,\,\,(\because \,h=R)\] \[or{{u}^{2}}=\frac{2GM}{R}-\frac{2GM}{2R}\] \[or{{u}^{2}}=\frac{GM}{R}\] \[\therefore u=\sqrt{\frac{GM}{R}}\] Alternative: The expression for the speed with which a body should be projected so as to reach a height h is \[u=\sqrt{\frac{2gh}{1+(h/R)}}\] Here, \[h=R\] (given) \[u=\sqrt{\frac{2gR}{1+(R/R)}}\] \[=\sqrt{\frac{2\times \frac{GM}{{{R}^{2}}}\times R}{2}}=\sqrt{\frac{GM}{R}}\]You need to login to perform this action.
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