A satellite is launched in the equatorial plane in such a way that it can transmit signals upto $60^{0}$ latitude on the earth. Then the angular velocity of the satellite is :

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$\sqrt{\frac{3 \sqrt{3} G M}{8 R^{3}}}$

$\sqrt{\frac{G M}{8 R^{3}}}$

$\sqrt{\frac{G M}{2 R^{3}}}$

$\sqrt{\frac{G M}{4 R^{3}}}$

answer is A.

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Detailed Solution

In $Δ A O B : - \cos 60^{0} = \frac{R}{O B} \Rightarrow O B = 2 R$

Here gravitational force will provide the required centripetal force. Hence $\frac{G M m}{{(O B)}^{2}} = m (O B) ω^{2}$ $\Rightarrow ω = \sqrt{\frac{G m}{{(O B)}^{3}}} = \sqrt{\frac{G M}{{(2 R)}^{3}}}, ω = \sqrt{\frac{G M}{8 R^{3}}}$

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