A) 220 km
B) 160 km
C) 70km
D) 120 km
Correct Answer: B
Solution :
Given, \[T=5.26\times {{10}^{3}}\,s,\,a=9.32\,m/{{s}^{2}}\] Centripetal acceleration \[a=\frac{{{v}^{2}}}{r}=9.32\] or \[{{v}^{2}}=9.32\,r\] or \[v=\sqrt{9.32\,({{R}_{e}}+h)}\] \[T=\frac{2\pi \,({{R}_{e}}+h)}{v}=\frac{2\pi \,({{R}_{e}}+h)}{\sqrt{9..32\,({{R}_{e}}+h)}}\] or \[T=\frac{2\pi \,\,\sqrt{{{R}_{e}}+h}}{\sqrt{9.32}}\] \[\therefore \] \[4.26\times {{10}^{3}}=\frac{2\times 3.14\,\,\sqrt{{{R}_{e}}+h}}{3.05}\] \[\sqrt{{{R}_{e}}+h}=2.55\times {{10}^{3}}\] \[{{R}_{e}}+h=6.53\times {{10}^{6}}\] \[h=6.53\times {{10}^{6}}-6.37\times {{10}^{6}}\] \[=0.16\times {{10}^{6}}m\] \[=160\times {{10}^{3}}m=160\,km\]
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