A planet takes 200 days to complete one revol | vedclass

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Question

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

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$: $T^2 \propto r^3$.Let the initial time period be $T_1 = 200 	ext{ days}$ and the initial distance be $r_1 = r$.Let the new time period be $T_2$ and the new distance be $r_2 = \frac{r}{4}$.Using the relation $\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$,we get:$\frac{(200)^2}{r^3} = \frac{T_2^2}{(\frac{r}{4})^3}$$T_2^2 = (200)^2 	imes \frac{(\frac{r}{4})^3}{r^3}$$T_2^2 = (200)^2 	imes \frac{r^3}{64 	imes r^3}$$T_2^2 = \frac{(200)^2}{64}$Taking the square root of both sides:$T_2 = \frac{200}{\sqrt{64}}$$T_2 = \frac{200}{8}$$T_2 = 25 	ext{ days}$.

$A$ planet takes $200$ days to complete one revolution around the Sun. If the distance of the planet from the Sun is reduced to one-fourth of the original distance,how many days will it take to complete one revolution?

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