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(B) According to Kepler's third law,$T^2 \propto R^3$. Given $T_1 = 2 	ext{ days}$,$R_1 = R$,$v_{01} = v_0$,and $T_2 = 16 	ext{ days}$. Using the ra

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its radius of orbit is $4 R$ and orbital velocity is $\frac{v_0}{2}$

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(B) According to Kepler's third law,$T^2 \propto R^3$. Given $T_1 = 2 	ext{ days}$,$R_1 = R$,$v_{01} = v_0$,and $T_2 = 16 	ext{ days}$. Using the ratio: $\frac{T_2}{T_1} = \left(\frac{R_2}{R_1}ight)^{3/2}$. $\frac{16}{2} = \left(\frac{R_2}{R}ight)^{3/2} \implies 8 = \left(\frac{R_2}{R}ight)^{3/2}$. Taking the power of $2/3$ on both sides: $8^{2/3} = \frac{R_2}{R} \implies 4 = \frac{R_2}{R} \implies R_2 = 4R$. Orbital velocity is given by $v_0 = \sqrt{\frac{GM}{R}}$. Therefore,$\frac{v_{02}}{v_{01}} = \sqrt{\frac{R_1}{R_2}} = \sqrt{\frac{R}{4R}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$. Thus,$v_{02} = \frac{v_0}{2}$.

The time period of an artificial satellite in a circular orbit of radius $R$ is $2$ days and its orbital velocity is $v_0$. If the time period of another satellite in a circular orbit is $16$ days,then :-

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