• question_answer When a satellite is in the synchronous orbit above the equator, it stays in one place with reference to the earth by making each revolution in just the same time as it takes the earth to rotate once. What is the altitude of the synchronous orbit? A)  20000km                     B)  30000kmC)  32500km          D)         36000km

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.