A) 44.8 km/s
B) 22.4 km/s
C) 11.2 km/s (remain unchanged)
D) 5.6 km/s
Correct Answer: B
Solution :
Escape velocity on earths surface is given by \[{{v}_{es}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] where, G = gravitational constant, \[{{M}_{e}}\] and \[{{R}_{e}}\] are the mass and radius of earth respectively. \[\therefore \] \[\frac{v_{es}^{}}{{{v}_{es}}}\sqrt{\frac{M_{e}^{}}{{{M}_{e}}}\times \frac{{{R}_{e}}}{R_{e}^{}}}\] but \[M_{e}^{}=2{{M}_{e}}\], and \[R_{e}^{}=\frac{{{R}_{e}}}{2},\]\[{{v}_{es}}=11.2km/s.\] \[\therefore \] \[\frac{v_{es}^{}}{{{v}_{es}}}=\sqrt{\frac{2{{M}_{e}}}{{{M}_{e}}}\times \frac{{{R}_{e}}}{{{R}_{e}}/2}}=\sqrt{4}=2\] \[\therefore \] \[v_{es}^{}=2{{v}_{es}}=2\times 11.2\] \[=22.4km/s\] NOTE: The escape velocity on moons surface is only 2.5 km/s. This is the basic fundamental on which, absence of atmosphere on moon can be explained.You need to login to perform this action.
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