A) 9 km
B) 9m
C) 9 cm
D) 9 mm
Correct Answer: D
Solution :
Let escape velocity be \[{{v}_{e}},\]then kinetic energy is \[=\frac{1}{2}mv_{e}^{2}\] ?(i) and escape energy \[=+\,\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ?(ii) Equating Eqs. (i) and (ii),we get \[\frac{1}{2}mv_{e}^{2}\,=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] \[\Rightarrow \] \[{{v}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] \[\Rightarrow \] \[R=\frac{2G{{M}_{e}}}{v_{e}^{2}}\] Given, \[G=6.67\times {{10}^{-11}}N-{{m}^{2}}/kg,\] \[{{M}_{e}}=6\times {{10}^{24}}kg,{{v}_{e}}=3\times {{10}^{8}}m/{{s}^{2}}\] \[R=\frac{2\times 6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{(3\times {{10}^{8}})}^{2}}}\] \[R=8.89\times {{10}^{-3}}\] \[R\approx 9\times {{10}^{-3}}m=9\,mm\]You need to login to perform this action.
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