Question

A planet of mass $m$ moves around the sun of mass $M$ in elliptical orbit. The maximum and minimum distances of the planet from the sun are $r_{1}$ and $r_{2}$ respectively.The time period of the planet is proportional to :

• A. $ r_{1}^{2/5} $
• B. $\left(r_{1}+r_{2}\right)^{3 / 2}$
• C. $\left(r_{1}-r_{2}\right)^{3 / 2}$
• D. $ r_{1}^{3/2} $

Answer 1

Answer: (B)

In astronomy the point: of closest approach is called the pericentre, ạnd that of farthest excursion is called apocentre: When $a$ is semimajor axis we have
Periapsis $=(1-e) a$.
Apoapsis $=(1+e) a-$
From condition given in question,
$r_{1} =a(1+e)$
$r_{2} =a(1-e)$
$\Rightarrow r_{1}+r_{2} =2 a $
$\Rightarrow a =\frac{r_{1}+r_{2}}{2}$
Also from Kepler's third law of planetary motion we have, the square of the period of revolution $(T)$ of any planet around the sun is directly proportional to the cube of the semimajor axis of its elliptic orbit $(a)$.
i.e., $T^{2} \propto a^{3}4$
$\Rightarrow T=a^{3 / 2} $
$\propto\left(\frac{r_{1}+r_{2}}{2}\right)^{3 / 2} $
$\Rightarrow T \propto\left(r_{1}+r_{2}\right)^{3 / 2}$