An earth satellite S has an orbit radius whic | vedclass

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(B) 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

$8$

Vedclass · Answer

(B) 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., $T^2 \propto r^3$.Given that the orbital radius of satellite $S$ is $r_s = 4r_c$, where $r_c$ is the orbital radius of the communication satellite $C$.The time period of a communication satellite is $T_c = 1 	ext{ day}$.Using the relation $\frac{T_s}{T_c} = \left(\frac{r_s}{r_c}ight)^{3/2}$, we substitute the given values:$\frac{T_s}{1} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.Therefore, the time period of satellite $S$ is $8 	ext{ days}$.

An earth satellite $S$ has an orbit radius which is $4$ times that of a communication satellite $C$. The period of revolution of $S$ is ........ $days$.

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