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(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$

Vedclass · Accepted Answer

$1032$

Vedclass · Answer

(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.Let $T_E$ and $R_E$ be the time period and orbital radius of the Earth,and $T_P$ and $R_P$ be those of the new planet.Given $R_P = 2R_E$ and $T_E = 365$ days.Using the ratio formula:$\left( \frac{T_P}{T_E} ight)^2 = \left( \frac{R_P}{R_E} ight)^3 = \left( \frac{2R_E}{R_E} ight)^3 = 2^3 = 8$.Taking the square root on both sides:$\frac{T_P}{T_E} = \sqrt{8} = 2\sqrt{2}$.Substituting the value of $T_E$:$T_P = 2\sqrt{2} 	imes 365 \approx 2.8284 	imes 365 \approx 1032.37$ days.Rounding to the nearest whole number,the time period is $1032$ days.

If a new planet is discovered rotating around the Sun with an orbital radius double that of Earth,what will be its time period (in Earth's days)?

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