91Ó°ÊÓ

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

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.

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,

$\mathrm{v}=\frac{2{\mathrm{Ï€°ù}}_1}{\mathrm{T}}$

So,

${\mathrm{r}}_1=\frac{\mathrm{vT}}{2\mathrm{Ï€}}$

Substituting r₁in r,

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

Putting r and r₁ in F₁,

$\begin{array}{c}\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\left[\frac{({\mathrm{m}}_1+{\mathrm{m}}_2)}{{\mathrm{m}}_2}\left(\frac{\mathrm{vT}}{2\mathrm{Ï€}}\right)\right]}^2}=\frac{{\mathrm{m}}_1{\mathrm{v}}^2}{\frac{\mathrm{vT}}{2\mathrm{Ï€}}}\\ \mathrm{F}=\frac{4{\mathrm{Ï€}}^2{\mathrm{Gm}}_1{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2{\mathrm{v}}^2{\mathrm{T}}^2}=\frac{2{\mathrm{Ï€³¾}}_1\mathrm{v}}{\mathrm{T}}\\ \frac{{\mathrm{m}}_2^3}{{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2}=\frac{{\mathrm{v}}^3\mathrm{T}}{2\mathrm{Ï€³Ò}}\\ \frac{{\mathrm{m}}_2^3}{{(0.9{\mathrm{M}}_{\mathrm{sun}}+{\mathrm{m}}_2)}^2}=\frac{{(70\text{ }\mathrm{m}/\mathrm{s})}^3\left(1500\left(86400\text{ }\mathrm{s}\right)\right)}{2(3.142)(6.67×{10}^{−11}\text{ }\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2)}\end{array}$

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

This gives,

${\text{m}}_{\text{2}}=7×{10}^{27}\text{ k²µ}$

But, the mass of Jupiter is

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

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

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

From part a, the radius of the star is,

${\mathrm{r}}_1=\frac{\mathrm{vT}}{2\mathrm{Ï€}}$

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

Now,

$\begin{array}{c}\mathrm{r}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}{\mathrm{r}}_1\\ {\mathrm{r}}_1+{\mathrm{r}}_2=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}{\mathrm{r}}_1\\ {\mathrm{r}}_2=\left(\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}−1\right){\mathrm{r}}_1\\ {\mathrm{r}}_2={\mathrm{r}}_1\left(\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}\right)\\ {\mathrm{r}}_2=1.4×{10}^9\text{ }\mathrm{m}\frac{0.9(2×{10}^{30}\text{ }\mathrm{kg})}{7×{10}^{27}\text{ }\mathrm{kg}}\\ {\mathrm{r}}_2=3.8×{10}^{\text{11}}\text{″¾}\end{array}$

But, Earth’s orbital radius is${\text{R}}_{\text{E}}=1.5×{10}^{11}\text{″¾}$

So,

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

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.