Question

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

Answer 1

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Answer 2

Answer:

The angular velocity of the satellite 0.0004387 m/s.

Explanation:

According to figure,

In Δ BOA,

$OA= \dfrac{R}{cos60^{\circ}}$

$OA = \dfrac{R}{\dfrac{1}{2}}$

$OA=2R$

Now, The gravitational force will provide the required centripetal force.

The gravitational force = centripetal force

$\dfrac{GmM}{(2R)^2}=m(OA)^2\omega^2$

$\omega^2=\dfrac{GM}{(2R)^3}$

$\omega=\sqrt{\dfrac{GM}{8R^3}}$

$\omega=\sqrt{\dfrac{6.67\times10^{-11}\times5.972\times10^{24}}{8(6371\times10^{3})^3}}$

$\omega=0.0004387\ m/s$

Hence, The angular velocity of the satellite 0.0004387 m/s.