A) \[2.4\times {{10}^{4}}m/s\]
B) \[1.4\times {{10}^{5}}m/s\]
C) \[3.8\times {{10}^{4}}m/s\]
D) \[2.8\times {{10}^{5}}m/s\]
Correct Answer: D
Solution :
\[\frac{1}{2}m{{v}^{2}}+\frac{2(-GMm)}{r}=0\] |
\[{{V}^{2}}=\frac{4GM}{r}\] |
\[=\frac{4\times 6.67\times {{10}^{-11}}\times 3\times {{10}^{31}}}{2\times {{10}^{11}}}\] |
\[V=20\sqrt{2}\times {{10}^{4}}m/s\] |
\[=2.828\times {{10}^{5}}m/s.\] |
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