A) \[2.4\,{{r}_{e}}\]
B) \[3.6\,{{r}_{e}}\]
C) \[4.8\,{{r}_{e}}\]
D) \[6.6\,{{r}_{e}}\]
Correct Answer: D
Solution :
Time period of satellite \[T=2\pi \sqrt{\frac{{{r}^{3}}}{G{{M}_{e}}}}\] or \[{{T}^{2}}=\frac{4{{\pi }^{2}}{{r}^{3}}}{G{{M}_{e}}}\] ... (i) Also \[g=\frac{G{{M}_{e}}}{r_{e}^{2}}\] ?. (ii) From Eqs. (i) and (ii), we get \[{{T}^{2}}=\frac{4{{\pi }^{2}}{{r}^{3}}}{gr_{e}^{2}}\] ... (iii) Substituting the given values in Eq. (iii), we get \[{{(24\times 60\times 60)}^{2}}=\frac{4\times {{(3.14)}^{2}}{{r}^{3}}}{9.8r_{e}^{2}}\] or \[r=6.6\,{{r}_{e}}\]You need to login to perform this action.
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