Question

Two planets of equal masses orbit a massive star. Planet $m_1$ moves in a circular orbit of radius $1\times {10}^8\text{km}$ with a period of $2\text{years}$ and planet $m_2$ moves in an elliptical orbit with closest distance $r_1=1\times {10}^8\text{km}$ and farthest distance $r_2=1.8\times {10}^8\text{km}$, as shown:


Using the fact that the mean radius of an elliptical orbit is the length of the semi-major axis, find the period of $m_2$'s orbit.

• A. $3.31\text{years}$
• B. $2.21\text{years}$
• C. $1.51\text{years}$
• D. $6.62\text{years}$

Answer 1

The correct option is A $3.31\text{years}$

Given,
closest distance, $r_1=1\times {10}^8\text{km}$
farthest distance, $r_2=1.8\times {10}^8\text{km}$

For the elliptical orbit $m_2$, mean radius or length of the semi-major axis,

$a_2=\frac{r_1+r_2}{2}=1.4\times {10}^8\text{km}$

From Kepler's law of periods,

$\Rightarrow {\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{a_2}{a_1}\right)}^3$

Given,
Radius of circular orbit of $m_1$,

$a_1=1\times {10}^8\text{km}$ and $T_1=2\text{years}$

$\Rightarrow T_2=T_1{\left(\frac{a_2}{a_1}\right)}^{\frac{3}{2}}$

$\Rightarrow T_2=2\times {\left(\frac{1.4\times {10}^8}{1\times {10}^8}\right)}^{\frac{3}{2}}$

$\Rightarrow T_2\approx 3.31\text{years}$.

Hence, option (a) is the correct answer.