Question

The presence of an unseen planet orbiting a distant star can sometimes be inferred from the motion of the star as we see it. As the star and planet orbit, the center of mass of the star-planet system, the star moves toward and away from us with what is called the line of sight velocity, a motion that can be detected. Figure 13-49 shows a graph of the line of sight velocity versus time for the star $\mathbf{14}\mathbf{}\mathbf{\text{Herculis}}$. The star’s mass is believed to be $\mathbf{\text{0.90}}$ of the mass of our Sun. Assume that only one planet orbits the star and that our view is along the plane of the orbit. Then approximate (a) the planet’s mass in terms of Jupiter’s mass ${\mathbf{\text{m}}}_{\mathbf{J}}$and

(b) the planet’s orbital radius in terms of Earth’s orbital radius${\mathbf{r}}_{\mathbf{E}}$ .

Short answer

1. The planet’s mass is${\text{3.7M}}_{\text{J}}$.
2. The radius of the planet is ${\text{r}}_{\text{2}}={\text{2.5R}}_{\text{E}}$.

Step-by-step solution

Step 1: Given

The graph of the line of sight velocity of star vs time for the star.

Determining the concept

Equating the gravitational force betweentheplanet andthestar and the centripetal force and putting the distance of the star (r₁) from C.O.M and the period of the star, find the mass of the planet. Comparing it with the mass of Jupiter, write it in terms of M_J.From r₁, find r₂using the center of mass formula.

Formulae are as follows:

${\mathrm{F}}_{\mathrm{c}}=\frac{{\mathrm{Mv}}^2}{\mathrm{r}}$

${\mathrm{F}}_{\mathrm{g}}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$

where F is force, G is gravitational constant, M and m are masses, v is velocity and r is the radius.

(a) Determining the planet’s mass in terms of Jupiter’s mass  Mj

The planet and the star are orbiting about their center of mass. The gravitational force of attraction between them provides the centripetal force to keep their motion in the circular orbit. It gives,

$\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}=\frac{{\mathrm{m}}_1{\mathrm{v}}^2}{{\mathrm{r}}_1}$

As both planet and star revolve about the center of mass,

$\begin{array}{l}{\mathrm{r}}_1=\frac{({\mathrm{m}}_1\left(0\right)+{\mathrm{m}}_2\mathrm{r})}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ {\mathrm{r}}_1=\frac{{\mathrm{m}}_2\mathrm{r}}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ \mathrm{r}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}{\mathrm{r}}_1\end{array}$

The orbital speed of m₁ in terms of T is written as,

$\mathrm{v}=\frac{2{\mathrm{\pi r}}_1}{\mathrm{T}}$

So,

${\mathrm{r}}_1=\frac{\mathrm{vT}}{2\mathrm{\pi }}$

Substituting r₁in r,

$\mathrm{r}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}\left(\frac{\mathrm{vT}}{2\mathrm{\pi }}\right)$

Putting r and r₁ in F₁,

$\begin{array}{c}\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\left[\frac{({\mathrm{m}}_1+{\mathrm{m}}_2)}{{\mathrm{m}}_2}\left(\frac{\mathrm{vT}}{2\mathrm{\pi }}\right)\right]}^2}=\frac{{\mathrm{m}}_1{\mathrm{v}}^2}{\frac{\mathrm{vT}}{2\mathrm{\pi }}}\\ \mathrm{F}=\frac{4{\mathrm{\pi }}^2{\mathrm{Gm}}_1{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2{\mathrm{v}}^2{\mathrm{T}}^2}=\frac{2{\mathrm{\pi m}}_1\mathrm{v}}{\mathrm{T}}\\ \frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{{\mathrm{v}}^3\mathrm{T}}{2\mathrm{\pi G}}\\ \frac{{\mathrm{m}}_2^3}{{(0.9{\mathrm{M}}_{\mathrm{sun}}+{\mathrm{m}}_2)}^2}=\frac{{(70\mathrm{m}/\mathrm{s})}^3\left(1500\left(86400\mathrm{s}\right)\right)}{2(3.142)(6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2)}\end{array}$

$\begin{array}{c}\mathrm{F}=\frac{{\mathrm{m}}_2^3}{{(0.9(2\times {10}^{30}\mathrm{kg})+{\mathrm{m}}_2)}^2}=1.06\times {10}^{23}\text{\hspace{0.17em}N}\\ {\text{m}}_{\text{2}}=7\times {10}^{27}\mathrm{kg}\end{array}$

This gives,

${\text{m}}_{\text{2}}=7\times {10}^{27}\text{\hspace{0.17em}kg}$

But, the mass of Jupiter is

$\begin{array}{c}{\mathrm{M}}_{\mathrm{J}}=318{\mathrm{M}}_{\mathrm{E}}\\ =318(5.98\times {10}^{24}\mathrm{kg})\\ =1.9\times {10}^{27}\mathrm{kg}\end{array}$

$\begin{array}{c}{\mathrm{m}}_2=\frac{7\times {10}^{27}\mathrm{kg}}{1.9\times {10}^{27}\mathrm{kg}}{\mathrm{M}}_{\mathrm{J}}\\ =3.7{\mathrm{M}}_{\mathrm{J}}\end{array}$

Therefore, the planet’s mass is $3.7{\mathrm{M}}_{\mathrm{J}}$.

(b) Determining the planet’s orbital radius in terms of the earth’s orbital radius  rE

From part a, the radius of the star is,

${\mathrm{r}}_1=\frac{\mathrm{vT}}{2\mathrm{\pi }}$

$\begin{array}{c}{\mathrm{r}}_1=\frac{70\mathrm{m}/\mathrm{s}(1.3\times {10}^8\mathrm{s})}{2\left(3.142\right)}\\ =1.4\times {10}^9\mathrm{m}\end{array}$

Now,

$\begin{array}{c}\mathrm{r}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}{\mathrm{r}}_1\\ {\mathrm{r}}_1+{\mathrm{r}}_2=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}{\mathrm{r}}_1\\ {\mathrm{r}}_2=\left(\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}-1\right){\mathrm{r}}_1\\ {\mathrm{r}}_2={\mathrm{r}}_1\left(\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}\right)\\ {\mathrm{r}}_2=1.4\times {10}^9\mathrm{m}\frac{0.9(2\times {10}^{30}\mathrm{kg})}{7\times {10}^{27}\mathrm{kg}}\\ {\mathrm{r}}_2=3.8\times {10}^{\text{11}}\text{\hspace{0.17em}m}\end{array}$

But, Earth’s orbital radius is${\text{R}}_{\text{E}}=1.5\times {10}^{11}\text{\hspace{0.17em}m}$

So,

$\begin{array}{c}{\mathrm{r}}_2=\frac{3.8\times {10}^{11}\mathrm{m}}{1.5\times {10}^{11}\mathrm{m}}{\mathrm{R}}_{\mathrm{E}}\\ =2.5{\mathrm{R}}_{\mathrm{E}}\end{array}$

Hence, the radius of the planet is${\mathrm{r}}_2=2.5{\mathrm{R}}_{\mathrm{E}}$.

Therefore, using the gravitational force of attraction between two objects and the centripetal force, the mass and orbital radius of one of the objects can be found.