The mean radius of the earth's orbit around t | vedclass

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(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$ o

Vedclass · Accepted Answer

Nearly $\frac{1}{4}$ year

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(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.Therefore,the ratio of the time periods is given by: $\frac{T_{	ext{mercury}}}{T_{	ext{earth}}} = \left( \frac{r_{	ext{mercury}}}{r_{	ext{earth}}} ight)^{3/2}$.Given: $r_{	ext{earth}} = 1.5 	imes 10^{11} \ m$ and $r_{	ext{mercury}} = 6 	imes 10^{10} \ m$.Substituting the values: $\frac{T_{	ext{mercury}}}{T_{	ext{earth}}} = \left( \frac{6 	imes 10^{10}}{1.5 	imes 10^{11}} ight)^{3/2} = \left( \frac{0.6 	imes 10^{11}}{1.5 	imes 10^{11}} ight)^{3/2} = \left( \frac{0.6}{1.5} ight)^{3/2} = \left( \frac{2}{5} ight)^{3/2} = \left( 0.4 ight)^{1.5} \approx 0.252$.Since $T_{	ext{earth}} = 1$ year,$T_{	ext{mercury}} \approx 0.25$ years or $\frac{1}{4}$ year.

The mean radius of the earth's orbit around the sun is $1.5 \times 10^{11} \ m$. The mean radius of the orbit of mercury around the sun is $6 \times 10^{10} \ m$. The mercury will rotate around the sun in

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