A) \[9\,\,km\]
B) \[9\,\,m\]
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,\[G=6.67\times {{10}^{-11}}N\text{-}{{m}^{2}}/kg\] \[R=\frac{2\times 6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{(3\times {{10}^{8}})}^{2}}}\] \[R=8.89\times {{10}^{-3}}\] \[R\approx 9\times {{10}^{-3}}m=9mm\]
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