The period of a planet around the sun is 8 ti | vedclass

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(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

$4$

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 $(r)$ of the orbit: $T^2 \propto r^3$. This implies that $\frac{T_p^2}{T_e^2} = \frac{r_p^3}{r_e^3}$,where $T_p$ and $r_p$ are the time period and orbital radius of the planet,and $T_e$ and $r_e$ are those of the earth. Given that $T_p = 8 T_e$,we substitute this into the ratio: $\frac{(8 T_e)^2}{T_e^2} = \frac{r_p^3}{r_e^3}$ $64 = \frac{r_p^3}{r_e^3}$ Taking the cube root on both sides: $\frac{r_p}{r_e} = (64)^{1/3} = 4$. Therefore,the ratio of the radius of the planet's orbit to the radius of the earth's orbit is $4$.

The period of a planet around the sun is $8$ times that of earth. The ratio of the radius of the planet's orbit to the radius of the earth's orbit is:

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