A) 70 km
B) 160 km
C) 190 km
D) 220 km
Correct Answer: B
Solution :
[b]: Time period of revolution of satellite \[T=2\pi \sqrt{\frac{{{({{R}_{e}}+h)}^{3}}}{G{{M}_{e}}}}\]or\[\frac{{{T}^{2}}}{4{{\pi }^{2}}}=\frac{{{({{R}_{e}}+h)}^{3}}}{G{{M}_{e}}}\] ?(i) Centripetal acceleration, \[a=\frac{G{{M}_{e}}}{{{({{R}_{e}}+h)}^{2}}}\]or\[\frac{{{({{R}_{e}}+h)}^{2}}}{G{{M}_{e}}}=\frac{1}{a}\] ?.(ii) Divide (i) by (ii), we get \[({{R}_{e}}+h)=\frac{{{T}^{2}}}{4{{\pi }^{2}}}\times a={{\left( \frac{5.26\times {{10}^{3}}}{2\pi } \right)}^{2}}\times 9.32\] \[{{R}_{e}}+h=6.53\times {{10}^{6}}m\] \[h=6.53\times {{10}^{6}}m-6.37\times {{10}^{6}}m=0.16\times {{10}^{6}}m\] \[=160\times {{10}^{3}}m=160km\]
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