Question

Two planets of equal masses, orbit a much more 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:


What is the mass of the star? (Take $G=\frac{20}{3}\times {10}^{-11}{\text{N m}}^2{\text{kg}}^{-2}\text{and}{\pi }^2\approx 10$)

• A. $5.29\times {10}^{20}\text{kg}$
• B. $1.49\times {10}^{25}\text{kg}$
• C. $1.49\times {10}^{29}\text{kg}$
• D. $1.49\times {10}^{31}\text{kg}$

Answer 1

The correct option is C $1.49\times {10}^{29}\text{kg}$

From Kepler's law
$$T^2=\frac{4{\pi }^2{r}^3}{GM}$$$\Rightarrow M=\frac{4{\pi }^2\times r^3}{G{T}^2}....\left(1\right)$

$M\to$ mass of the star,
$r\to$ mean radius of elliptical path,

Given, for planet $m_1$,

Time period, $T=2\text{years}=2\times 365\times 24\times 60\times 60\text{seconds}$,
$r={10}^8\text{km}={10}^{11}\text{m}$.
$G=\frac{20}{3}\times {10}^{-11}{\text{N m}}^2{\text{kg}}^{-2}$
${\pi }^2\approx 10$

Substituting the values, in equation $\left(1\right)$,

$\Rightarrow M=\frac{4\times 10\times ({10}^{11}{)}^3}{\frac{20}{3}\times {10}^{-11}\times (2\times 365\times 24\times 3600{)}^2}$

$\Rightarrow M\approx 1.49\times {10}^{29}\text{kg}$

Hence, option (c) is the correct answer.