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(A) According to Kepler's Third Law,the orbital period $T$ is related to the orbital radius $R$ as $T \propto R^{3/2}$.Also,the orbital speed $v$ is g

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$R / 4$

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(A) According to Kepler's Third Law,the orbital period $T$ is related to the orbital radius $R$ as $T \propto R^{3/2}$.Also,the orbital speed $v$ is given by $v = \frac{2\pi R}{T}$.Substituting $T \propto R^{3/2}$ into the speed formula,we get $v \propto \frac{R}{R^{3/2}} = R^{-1/2}$,which means $v \propto \frac{1}{\sqrt{R}}$.Let $v_1$ and $R_1$ be the speed and radius of the Earth,and $v_2$ and $R_2$ be the speed and radius of the planet.Given $v_2 = 2v_1$,we have $\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}}$.Substituting the values: $\frac{v_1}{2v_1} = \sqrt{\frac{R_2}{R}}$.$\frac{1}{2} = \sqrt{\frac{R_2}{R}}$.Squaring both sides: $\frac{1}{4} = \frac{R_2}{R}$.Therefore,$R_2 = \frac{R}{4}$.

Suppose a planet goes around the Sun with a linear speed twice as fast as that of the Earth. What will be its orbit size compared to that of the Earth? (Radius of Earth $= R$)

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