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Question

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ is directly proportional to the cube of the orbital radius

Vedclass · Accepted Answer

$4R$

Vedclass · Answer

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ is directly proportional to the cube of the orbital radius $(r)$: $T^2 \propto r^3$.Given for the first satellite: $T_1 = 1 	ext{ day}$,$r_1 = R$.Given for the second satellite: $T_2 = 8 	ext{ days}$,$r_2 = ?$.Using the ratio: $\left( \frac{T_2}{T_1} ight)^2 = \left( \frac{r_2}{r_1} ight)^3$.Substituting the values: $\left( \frac{8}{1} ight)^2 = \left( \frac{r_2}{R} ight)^3$.$64 = \left( \frac{r_2}{R} ight)^3$.Taking the cube root on both sides: $\sqrt[3]{64} = \frac{r_2}{R}$.$4 = \frac{r_2}{R}$,which implies $r_2 = 4R$.

$A$ satellite moves around the Earth in a circular orbit of radius $R$ making one revolution per day. $A$ second satellite moving in a circular orbit moves around the Earth once in $8$ days. The radius of the orbit of the second satellite is

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