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) |
| \[u=\sqrt{\frac{2gR}{1+(R/R)}}\] |
| \[=\sqrt{\frac{2\times \frac{GM}{{{R}^{2}}}\times R}{2}}=\sqrt{\frac{GM}{R}}\] |
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