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

Vedclass · Answer

(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: $T^2 \propto r^3$ or $\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$.Given:$r_1 = 1.5 	imes 10^{11} \ m$ (Earth's orbital radius)$T_1 = 1 	ext{ year}$$r_2 = 6 	imes 10^{10} \ m$ (Mercury's orbital radius)Substituting the values:$\frac{1^2}{(1.5 	imes 10^{11})^3} = \frac{T_2^2}{(6 	imes 10^{10})^3}$$T_2^2 = \left( \frac{6 	imes 10^{10}}{1.5 	imes 10^{11}} ight)^3$$T_2^2 = \left( \frac{6}{15} ight)^3 = \left( 0.4 ight)^3 = 0.064$$T_2 = \sqrt{0.064} \approx 0.25 	ext{ years}$.Thus,Mercury revolves around the sun in nearly $\frac{1}{4}$ of a 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 $6 \times 10^{10} \ m$. Mercury will revolve around the sun in:

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