Question

Figure shows elliptical path $abcd$ of a planet around the sun $S$ such that the area of triangle csa is $\frac{1}{4}$ the area of the ellipse. (See figure) With $db$ as the major axis, and $ca$ as the minor axis. If $t_1$ is the time taken for planet to go over path $abc$ and $t_2$ for path taken over $cda$ then

• A. $t_1=4t_2$
• B. $t_1=2t_2$
• C. $t_1=t_2$
• D. $t_1=3t_2$

Answer 1

The correct option is D $t_1=3t_2$

Since area of triangle $csa$ is $\frac{1}{4}$ of total area of ellipse, therefore:

Area of $cdas$ = $\frac{1}{3}$Area of $abcs$

Now that from Kepler's second law areal velocities of the planets are constant which essentially means planets cover equal area in equal time interval.

Hence,

Time taken in covering path $abc$ and path $cda$ will be in proportion to their respective enclosed areas.

$⟹$ $t_1=3t_2$