The period of revolution of planet A around t | vedclass

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(C) According to Kepler's Third Law of Planetary Motion, the square of the period of revolution $T$ is proportional to the cube of the semi-major axis

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) According to Kepler's Third Law of Planetary Motion, the square of the period of revolution $T$ is proportional to the cube of the semi-major axis $R$ of the orbit: $T^2 \propto R^3$.Given that the period of planet $A$ $(T_A)$ is $8$ times the period of planet $B$ $(T_B)$, we have $T_A = 8 T_B$.Using the relation $\frac{T_A}{T_B} = \left(\frac{R_A}{R_B}\right)^{3/2}$, we can write:$\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3$Substituting the given values:$(8)^2 = \left(\frac{R_A}{R_B}\right)^3$$64 = \left(\frac{R_A}{R_B}\right)^3$Taking the cube root on both sides:$\frac{R_A}{R_B} = (64)^{1/3} = 4$Thus, the distance of planet $A$ from the sun is $4$ times the distance of planet $B$ from the sun.

The period of revolution of planet $A$ around the sun is $8$ times that of planet $B$. How many times is the distance of planet $A$ from the sun greater than that of planet $B$ from the sun (in $\times$)?

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