A) 20000km
B) 30000km
C) 32500km
D) 36000km
Correct Answer: D
Solution :
The period of revolution of the satellite must be exactly one day, or 86400s. The centripetal acceleration of the satellite must \[4{{\pi }^{2}}r/{{T}^{2}}\], the gravitational field must be \[g={{g}_{0}}{{\left( {{r}_{0}}/r \right)}^{2}}\]In free fall, \[a=g,\] so \[\begin{align} & r=\frac{(9.8m/{{s}^{2}})(6.4\times {{10}^{-6}}m){{(86400s)}^{2}}}{4{{\pi }^{2}}} \\ & =4.23\times {{10}^{7}}m \\ \end{align}\] To get the altitude, subtract the radius of the earth. The satellite must be at an altitude of 36000 km.
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