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# A planet of mass $m$ moves around the sun of mass $M$ in an elliptical orbit. The maximum and minimum distance of the planet from the sun are ${r_1}$ and ${r_2}$ respectively. The time period of the planet is proportional to:(A) ${r_1}^{\dfrac{2}{5}}$(B) ${\left( {{r_1} + {r_2}} \right)^{\dfrac{3}{2}}}$(C) ${\left( {{r_1} - {r_2}} \right)^{\dfrac{3}{2}}}$(D) ${r^{\dfrac{3}{2}}}$

Last updated date: 20th Jun 2024
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Hint The time period of the planet can be determined by the Kepler’s law. This law shows the relation between the time period and the semi major axis and the semi major axis shows the maximum and minimum distance of the planet from the sun, then the solution can be determined.

Useful formula
The time period of the planet is given by the Kepler’s law,
${T^2} \propto {a^3}$
Where, $T$ is the time period of the planet to move around the sun and $a$ is the semi major axis of the planet.

Complete step by step solution
Given that,
The mass of the planet is given as, $m$,
The mass of the sun is given as, $M$,
The maximum distance of the planet from the sun is given as, ${r_1}$,
The minimum distance of the planet from the sun is given as, ${r_2}$.
Now,
The time period of the planet is given by the Kepler’s law,
${T^2} \propto {a^3}\,................\left( 1 \right)$
The semi major axis of the planet is given by,
$a = \dfrac{{{r_1} + {r_2}}}{2}$,
By substituting the values of the semi major axis in the above equation (1), then the above equation (1) is written as,
${T^2} \propto {\left( {\dfrac{{{r_1} + {r_2}}}{2}} \right)^3}$
By dividing the power of the both side by $2$, then the above equation is written as,
${T^{\dfrac{2}{2}}} \propto {\left( {\dfrac{{{r_1} + {r_2}}}{2}} \right)^{\dfrac{3}{2}}}$
Then the above equation is also written as,
$T \propto {\left( {\dfrac{{{r_1} + {r_2}}}{2}} \right)^{\dfrac{3}{2}}}$