A) 11.2
B) 5.6
C) 1.5
D) 22.4
E) 33.6
Correct Answer: D
Solution :
Escape velocity from the earth \[\left( {{V}_{e}} \right)=11.2\,\,km\text{/}s\] Let the mass, radius and density of earth be M, R and \[\rho \] respectively and for given planet mass, radius and density are M, R and \[\rho \] respectively. \[\therefore \]Escape velocity from the earth \[{{v}_{e}}=\sqrt{\frac{2G\times \left( \frac{4}{3}\pi {{R}^{3}}\rho \right)}{R}}\] \[{{v}_{e}}=\sqrt{\frac{8\,G\pi {{R}^{2}}\rho }{3}}\] ?(i) Similarly escape velocity from the given planet \[v{{}_{e}}=\sqrt{\frac{8\,G\,\pi \,R{{}^{2}}\rho }{3}}\] ?(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{v}_{e}}}{v{{}_{e}}}=\sqrt{\frac{8\,G\pi {{R}^{2}}\rho }{3}}\times \sqrt{\frac{3}{8G\pi R{{}^{2}}\rho }}\] \[=\sqrt{\frac{{{R}^{2}}}{R{{}^{2}}}}\] or \[\frac{11.2}{v{{}_{e}}}=\frac{R}{R}\] or \[\frac{11.2}{v{{}_{e}}}=\frac{R}{2R}\] \[\therefore \] \[v{{}_{e}}=22.4\,\,km\text{/}s\]
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