Question

A planet of small mass $m$ moves around the Sun of mass $M$ along an elliptical orbit such that its minimum and maximum distances from sun are $r$ and $R$ respectively. Its period of revolution will be:

• A. $2\pi \sqrt{\frac{(r+R{)}^3}{6GM}}$
• B. $2\pi \sqrt{\frac{(r+R{)}^3}{3GM}}$
• C. $2\pi \sqrt{\frac{(r+R{)}^3}{8GM}}$
• D. $2\pi \sqrt{\frac{(r+R{)}^3}{GM}}$

Answer 1

The correct option is C $2\pi \sqrt{\frac{(r+R{)}^3}{8GM}}$

Given,
Planet's mass$=m$
Mass of Sun$=M$


We know that if $r$ and $R$ are the shortest and longest distances respectively then, length of semi-major axis $$a=\frac{r+R}{2}$$From Kepler's law, $T=2\pi \sqrt{\frac{a^3}{GM}}$

Substituting the value of $a$ in the expression for time period.

$T=2\pi \sqrt{{\left(\frac{r+R}{2}\right)}^3\times \frac{1}{GM}}$

$\Rightarrow T=2\pi \sqrt{\frac{(r+R{)}^3}{8GM}}$

Hence, option (c) is the correct answer.

The time period of revolution of a planet in elliptical orbit is given by $$T=2\pi \sqrt{\frac{a^3}{GM}}$$Where $a$ is the length of the semi-major axis as well as the mean distance from the sun. Note: from properties of ellipse $r=a(1-e)...\left(1\right)$ and $R=a(1+e)...\left(2\right)$ From equation $\left(1\right)$ and $\left(2\right)$,$$a=\frac{r+R}{2}$$