A) \[{{v}_{e}}\]
B) \[2{{v}_{e}}\]
C) \[1.5{{v}_{e}}\]
D) \[\sqrt{1.5}{{v}_{e}}\]
Correct Answer: A
Solution :
[a] : Taking the gravitational potential at a large distance from the earth as zero, the gravitational potential at the surface of the planet \[=-\frac{GM}{R}.\] From law of conservation of energy, if v is the velocity of particle while reaching the surface of the planet and m is its mass, then \[\frac{1}{2}m{{v}^{2}}+\left( -\frac{GMm}{R} \right)=0\]or\[{{v}^{2}}=\frac{2GM}{R}=v_{e}^{2}\] \[\left( \because {{v}_{e}}=\sqrt{\frac{2GM}{R}} \right)\]or\[v={{v}_{e}}.\]
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