A) \[3\times {{10}^{-2}}\]
B) \[{{10}^{-2}}\]
C) \[2\times {{10}^{-2}}\]
D) \[6\times {{10}^{-2}}\]
Correct Answer: C
Solution :
\[T=2\pi \sqrt{\frac{{{r}^{3}}}{G{{M}_{e}}}}\] \[{{T}^{2}}=\frac{4{{(\pi )}^{2}}{{r}^{3}}}{GM}\] \[\ln ({{T}^{2}})=ln(\frac{4{{(\pi )}^{2}}{{r}^{3}}}{GM})\] \[\ln ({{T}^{2}})=\ln (\frac{4{{(\pi )}^{2}}{{r}^{3}}}{G})+\ln (M)\] \[2\ln (T)=\ln (K)+\ln (M)\] \[2\frac{\Delta \Tau }{T}=\frac{\Delta M}{M}\] \[\frac{\Delta \Mu }{M}=2\times {{10}^{-2}}\]
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