Question

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

• A. $\sqrt{\frac{GM}{8R^3}}$
• B. $\sqrt{\frac{GM}{2R^3}}$
• C. $\sqrt{\frac{GM}{4R^3}}$
• D. $\sqrt{\frac{3GM}{8R^3}}$

Answer 1

The correct option is A $\sqrt{\frac{GM}{8R^3}}$


Here, $\text{A}$ is the farthest point on earth where the signal can reach from satellite and $\text{AB}$ will be a tangent at point $\text{A}$.

So, distance of satellite from the centre of the earth will be

$r=OB=\frac{OA}{\cos {60}^{\circ }}=\frac{R}{1/2}=2R$

Since gravitational force provides the centripetal force,

$\therefore \frac{GMm}{r^2}=m{\omega }^{2}r$

where, $\omega$ is the angular velocity of the satellite.

$\Rightarrow \omega =\sqrt{\frac{GM}{r^3}}=\sqrt{\frac{GM}{8R^3}}(\because r=2R)$

Hence, option $\left(a\right)$ is correct.