Solution:
For a transmission coverage up to latitude \(\theta = 60^\circ\), the orbital radius \(r\) satisfies \(\cos\theta = \frac{R}{r}\). This gives \(r = 2R\). The angular velocity is \(\omega = \sqrt{\frac{GM}{r^3}} = \sqrt{\frac{GM}{8R^3}}\).
For a transmission coverage up to latitude \(\theta = 60^\circ\), the orbital radius \(r\) satisfies \(\cos\theta = \frac{R}{r}\). This gives \(r = 2R\). The angular velocity is \(\omega = \sqrt{\frac{GM}{r^3}} = \sqrt{\frac{GM}{8R^3}}\).
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