question_answer

A satellite is launched in the equatorial plane in such a way that it can transmit signals up to \[60{}^\circ \] latitude on the earth. The angular velocity of the satellite is

A) \[\sqrt{\frac{GM}{8{{R}^{3}}}}\]

B) \[\sqrt{\frac{GM}{2{{R}^{3}}}}\]

C) \[\sqrt{\frac{GM}{4{{R}^{3}}}}\]

D) \[\sqrt{\frac{3\sqrt{3}GM}{8{{R}^{3}}}}\]

Correct Answer: A

Solution :

[a] In \[\Delta AOB: cos 60{}^\circ  = R\Rightarrow  OB = 2R\] Here gravitational force will provide the required centripetal force.     Hence.  \[\frac{GMm}{{{\left( OB \right)}^{2}}}=m\left( OB \right){{\omega }^{2}}\] \[\Rightarrow \omega =\sqrt{\frac{GM}{{{(OB)}^{3}}}}=\sqrt{\frac{GM}{{{(2R)}^{3}}}}\Rightarrow \omega =\sqrt{\frac{GM}{8{{R}^{3}}}}\]