Angular Velocity of a Communication Satellite – Rankers Physics

Planet and Satellite: Practice Problem & Solution

A satellite is launched in the equatorial plane in such a way that it can transmit signals upto \(60^\circ\) latitude on the earth. Then the angular velocity of the satellite is :
\(\sqrt{\frac{GM}{8R^3}}\)
\(\sqrt{\frac{GM}{2R^3}}\)
\(\sqrt{\frac{GM}{4R^3}}\)
\(\sqrt{\frac{3\sqrt{3}GM}{8R^3}}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Planet and Satellite. Understanding the underlying formula is key to arriving at the correct answer below:

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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