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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

its radius of orbit is $4 \, R$ and orbital velocity is $\frac{v_0}{2}$

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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.,$T^2 \propto R^3$.Given $T_1 = 2 \, days$,$R_1 = R$,$T_2 = 16 \, days$,and $R_2 = ?$.Using the ratio: $\frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \Rightarrow \frac{16^2}{2^2} = \left(\frac{R_2}{R}\right)^3$.$\frac{256}{4} = 64 = \left(\frac{R_2}{R}\right)^3$.Taking the cube root on both sides: $\frac{R_2}{R} = 4$,so $R_2 = 4R$.For orbital velocity $v = \sqrt{\frac{GM}{R}}$,we have $v \propto \frac{1}{\sqrt{R}}$.Therefore,$\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{\frac{R}{4R}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,$v_2 = \frac{v_1}{2} = \frac{v_0}{2}$.

The time period of an artificial satellite in a circular orbit of radius $R$ is $2 \, days$ and its orbital velocity is $v_0$. If the time period of another satellite in a circular orbit is $16 \, days$,then:

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