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  • Two planets of equal mass are having their period of revolutions <img alt="\mathrm{T_{A} \: and\: T

Two planets \mathrm{A \: and\: B} of equal mass are having their period of revolutions \mathrm{T_{A} \: and\: T_{B} \: such \: that \: T_{A}=2 T_{B}}. These planets are revolving in the circular orbits of radii \mathrm{r_{A} \: and \: r_{B}} respectively. Which out of the following would be the correct relationship of their orbits?
 

Option: 1

2 \mathrm{r}_{\mathrm{A}}^{2}=\mathrm{r}_{\mathrm{B}}^{3}


Option: 2

\mathrm{r}_{\mathrm{A}}^{3}=2 \mathrm{r}_{\mathrm{B}}^{3}


Option: 3

\mathrm{r}_{\mathrm{A}}^{3}=4 \mathrm{r}_{\mathrm{B}}^{3}


Option: 4

\mathrm{T}_{\mathrm{A}}^{2}-\mathrm{T}_{\mathrm{B}}^{2}=\frac{\pi^{2}}{\mathrm{GM}}\left(\mathrm{r}_{\mathrm{B}}^{3}-4 \mathrm{r}_{\mathrm{A}}^{3}\right)


Answers (1)

best_answer

According to Kepler's law

\mathrm{T^{2} \propto r^{3}}

\mathrm{T \rightarrow\text{time period} }

\mathrm{r \rightarrow\text{radius of orbit }}

\mathrm{\frac{T_{A}^{2}}{T_{B}^{2}}=\frac{r_{A}^{3}}{r_{B}^{3}}, }                      \mathrm{\frac{T_{A}}{T_{B}}=2} \\

\mathrm{2^{2}=\left(\frac{r_{A}}{r_{B}}\right)^{3}} \\

\mathrm{4=\frac{r_{A}^{3}}{r_{B}^{3}}} \\

\mathrm{4 r_{B}^{3}=r_{A}^{3} }

Hence correct option is 3

 

Posted by

vinayak

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