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(A) According to Kepler's third law of planetary motion,the square of the orbital period $T$ is proportional to the cube of the semi-major axis $R$ of

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$10\sqrt{10}$

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(A) According to Kepler's third law of planetary motion,the square of the orbital period $T$ is proportional to the cube of the semi-major axis $R$ of the orbit,i.e.,$T^2 \propto R^3$.Therefore,the ratio of the orbital periods is given by $\frac{T_1}{T_2} = \left(\frac{R_1}{R_2}ight)^{3/2}$.Given $R_1 = 10^{11} \ m$ and $R_2 = 10^{10} \ m$.Substituting the values: $\frac{T_1}{T_2} = \left(\frac{10^{11}}{10^{10}}ight)^{3/2} = (10)^{3/2}$.This simplifies to $10^1 	imes 10^{1/2} = 10\sqrt{10}$.

The distances of two planets from the Sun are $10^{11} \ m$ and $10^{10} \ m$ respectively. Find the ratio of their orbital periods.

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