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# A planet in an orbit sweeps out an angle of ${160^\circ }$ from March- May. When it is at an average distance of $140$ million km from the sun. If planet sweeps out an angle of ${10^\circ }$ from October- December , then the average distance from sun is A) $56 \times {10^5}\;{\text{km}}$B) $56 \times {10^6}\;{\text{km}}$C) $56 \times {10^7}\;{\text{km}}$D) $56 \times {10^8}\;{\text{km}}$

Last updated date: 07th Aug 2024
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Hint:- From Kepler’s law of planetary motion, the relation between the average distance of the planet in an angle at different positions can be found. It will sweep equal areas in an equal interval of time.

Given the from March to May the planet in the orbit have a average distance, ${r_1} = 140 \times {10^6}\;{\text{km}}$and the angle it sweeps is ${\phi _1} = {160^\circ }$ .
And from October to December, it sweeps an angle of ${\phi _2} = {10^\circ }$
The line segment joining the planet and the sum will sweep out equal areas in equal intervals of time. The speed at which the planet is constantly changes. Thus the planet moves faster when close to the sun and slower when far from the sun.
The circumference of the orbit is always constant. It doesn’t change according to the seasons. Therefore the product of average distance and the angle swept from March to May will be the same as that of October to December.
Therefore, the relation will be like,
$r_1^2{\phi _1} = r_2^2{\phi _2}$
Where, ${r_2}$ is the average distance swept from October to December.
Substituting the values in the relation,
${\left( {140 \times {{10}^6}\;{\text{km}}} \right)^2} \times {160^\circ } = r_2^2 \times {10^\circ } \\ r_2^2 = \dfrac{{{{\left( {140 \times {{10}^6}\;{\text{km}}} \right)}^2} \times {{160}^\circ }}}{{{{10}^\circ }}} \\ = 3.136 \times {10^{17}} \\ {r_2} = \sqrt {3.136 \times {{10}^{17}}} \\ {r_2} = 56 \times {10^7}\;{\text{km}} \\$
Thus the average distance of the planet out an angle of ${10^\circ }$ from October- December is $56 \times {10^7}\;{\text{km}}$.