The figure shows an elliptical orbit of a pla | vedclass

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(B) According to Kepler's Second Law of Planetary Motion,a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of

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$t_1 = 2t_2$

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(B) According to Kepler's Second Law of Planetary Motion,a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.This means the areal velocity $\frac{dA}{dt}$ is constant.Therefore,the time taken to sweep an area is directly proportional to the area swept.Given that the area $SCD = 2 	imes 	ext{Area } SAB$.Since $t_1$ is the time taken to sweep area $SCD$ and $t_2$ is the time taken to sweep area $SAB$,we have:$\frac{t_1}{t_2} = \frac{	ext{Area } SCD}{	ext{Area } SAB} = \frac{2 	imes 	ext{Area } SAB}{	ext{Area } SAB} = 2$.Thus,$t_1 = 2t_2$.

The figure shows an elliptical orbit of a planet of mass $m$ about the sun $S.$ The shaded area $SCD$ is twice the shaded area $SAB.$ If $t_1$ is the time taken for the planet to move from $C$ to $D$ and $t_2$ is the time taken to move from $A$ to $B,$ then:

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