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(A) The angular velocity of a satellite is given by $\omega = \sqrt{\frac{GM_e}{r^3}}$.For the first satellite with radius $R$,$\omega_1 = \sqrt{\frac

Vedclass · Accepted Answer

$\frac{8\pi}{7}\sqrt{\frac{R^3}{GM_e}}$

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(A) The angular velocity of a satellite is given by $\omega = \sqrt{\frac{GM_e}{r^3}}$.For the first satellite with radius $R$,$\omega_1 = \sqrt{\frac{GM_e}{R^3}}$.For the second satellite with radius $4R$,$\omega_2 = \sqrt{\frac{GM_e}{(4R)^3}} = \sqrt{\frac{GM_e}{64R^3}} = \frac{1}{8}\sqrt{\frac{GM_e}{R^3}} = \frac{\omega_1}{8}$.Since they start on the same radial line,they have maximum separation when the difference in their angular positions is $\pi$ radians.Let $	heta_1 = \omega_1 t$ and $	heta_2 = \omega_2 t$. The condition for maximum separation is $	heta_1 - 	heta_2 = \pi$.Substituting the values: $(\omega_1 - \omega_2)t = \pi$.$t = \frac{\pi}{\omega_1 - \omega_2} = \frac{\pi}{\omega_1 - \frac{\omega_1}{8}} = \frac{\pi}{\frac{7\omega_1}{8}} = \frac{8\pi}{7\omega_1}$.Substituting $\omega_1 = \sqrt{\frac{GM_e}{R^3}}$,we get $t = \frac{8\pi}{7}\sqrt{\frac{R^3}{GM_e}}$.

Two satellites are launched simultaneously into orbits of radius $R$ and $4R$. At an instant,the two satellites are on the same radial line. The time after which they have maximum separation is (mass of Earth = $M_e$). Assume the same sense of rotation.

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