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(B) According to Kepler's second law of planetary motion,the areal velocity of a planet revolving around the sun is constant.This means the planet swe

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$t_1 > t_2$

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(B) According to Kepler's second law of planetary motion,the areal velocity of a planet revolving around the sun is constant.This means the planet sweeps out equal areas in equal intervals of time.In the given ellipse,the sun $S$ is at one of the foci.The path $ACB$ covers more than half of the area of the ellipse,while the path $BDA$ covers less than half of the area of the ellipse.Since the area swept by the radius vector in path $ACB$ (Area $S-ACB$) is greater than the area swept in path $BDA$ (Area $S-BDA$),and the areal velocity is constant,it follows that the time taken $t_1$ must be greater than $t_2$.Therefore,$t_1 > t_2$.

The figure shows the orbit of a planet $P$ around the sun $S.$ $AB$ and $CD$ are the minor and major axes of the ellipse,respectively.
If $t_1$ is the time taken by the planet to travel along the path $ACB$ and $t_2$ is the time taken to travel along the path $BDA,$ then:

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The figure shows the orbit of a planet $P$ around the sun $S.$ $AB$ and $CD$ are the minor and major axes of the ellipse,respectively.If $t_1$ is the time taken by the planet to travel along the path $ACB$ and $t_2$ is the time taken to travel along the path $BDA,$ then: