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 $semimajor$ axis, and $ca$ as the $semiminor$ 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. $t_1 = t_2$
• B. $t_1 = 2 t_2$
• C. $t_1 = 3 t_2$
• D. $t_1 = 4 t_2$

Answer 1

The Correct Option is C

To solve this problem, we must make use of Kepler's Second Law of Planetary Motion, which states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. Given this, the ratio of the times $t_1$ and $t_2$ must be equal to the ratio of the areas swept out by the line segment during those times.

1. The problem states that the area of triangle \(csa\) is \(\frac{1}{4}\) of the total area of the ellipse. Since the elliptical path is symmetric, the area of sector \(cda\) is \(\frac{1}{4}\) of the ellipse.
2. Thus, the area swept by the line segment \(S\) during \(t_2\) (over the path \(cda\)) is \(\frac{1}{4}\) of the total ellipsoidal area.
3. Therefore, the remaining area (the area swept out during \(t_1\) over path \(abc\)) must be the complementary \(\frac{3}{4}\) of the total area of the ellipse. This is calculated by subtracting the \(\frac{1}{4}\) area swept during \(t_2\) from the entire area of the ellipse.
4. Now, according to Kepler's Second Law, the time \(t_1\) taken to sweep an area of \(\frac{3}{4}\) must be three times the time \(t_2\) takes to sweep an area of \(\frac{1}{4}\).
5. Thus, we conclude that \(t_1 = 3 t_2\).

Hence, the correct answer is \(t_1 = 3 t_2\).