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(D) Given: $T_A = 8 T_B$.According to Kepler's third law,$T^2 \propto R^3$,where $T$ is the time period and $R$ is the orbital radius.Thus,$\frac{T_A^

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$1: 2$

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(D) Given: $T_A = 8 T_B$.According to Kepler's third law,$T^2 \propto R^3$,where $T$ is the time period and $R$ is the orbital radius.Thus,$\frac{T_A^2}{T_B^2} = \frac{R_A^3}{R_B^3} \Rightarrow (8)^2 = \left(\frac{R_A}{R_B}ight)^3 \Rightarrow 64 = \left(\frac{R_A}{R_B}ight)^3$.Taking the cube root,$\frac{R_A}{R_B} = (64)^{1/3} = 4$.Orbital velocity $V$ is given by $V = \frac{2 \pi R}{T}$.Therefore,the ratio of orbital velocities is $\frac{V_A}{V_B} = \frac{R_A}{R_B} 	imes \frac{T_B}{T_A} = 4 	imes \frac{1}{8} = \frac{1}{2}$.Hence,the ratio is $1: 2$.

Two planets,$A$ and $B$,orbit around a star such that the time period of $A$ is $8$ times the time period of $B$. The ratio of the orbital velocities of planets $A$ and $B$ is:

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