Chapter 8, Problem 47Q

To determine

(a)

The angular distance between the star and its planet as seen from Earth.

Answer to Problem 47Q

The angular distance between the star and its planet as seen from Earth is $1.06\text{arcsec}$.

Explanation of Solution

Given:

The distance of the star from the Earth is $d=170\text{ly}$.

Formula Used:

The small angle formula is given by

$\alpha =\frac{D\left(206265\right)}{d}$

Here, $D$ is the angular diameter.

Calculations:

The small angle is calculated as

$\begin{array}{c}\alpha =\frac{D\left(206265\right)}{d}\\ =\frac{\left(55\text{au}\right)\left(206265\right)}{\left(170\text{ly}\times \frac{63240\text{au}}{1\text{ly}}\right)}\\ =1.06\text{arcsec}\end{array}$

Conclusion:

The angular distance between the star and its planet as seen from Earth is $1.06\text{arcsec}$.

To determine

(b)

The orbital period of the planet and whether it is possible for the astronomer to observe a complete orbit in one lifetime.

Answer to Problem 47Q

The orbital period of the planet is $2579.73\text{yr}$ and it is not possible for the astronomer to observe a complete orbit in one lifetime.

Explanation of Solution

Given:

The mass of the star is $0.025$ times that of the Sun; that is, $M=0.025M_{\text{Sun}}$.

Formula Used:

As per Kepler’s third law, the relation between the orbital period and the orbital distance is given by

$P^{\text{2}}=\left(\frac{4{\pi }^2}{GM}\right)a^3$

Here, $P$ is the period, $G$ is the gravitational constant and $a$ is the orbital distance.

Calculations:

For the Sun, the relation is given by

$P_{\text{Sun}}^2=\left(\frac{4{\pi }^2}{G{M}_{\text{Sun}}}\right)a_{\text{Sun}}^3$ …… (I)

For the star $2\text{M}1207$, the relation is given by

$P_{2\text{M}1207}^2=\left(\frac{4{\pi }^2}{G{M}_{2\text{M}1207}}\right)a_{2\text{M}1207}^3$ …… (II)

Divide equation (II) by equation (I).

$\begin{array}{c}\frac{P_{2\text{M}1207}^2}{P_{\text{Sun}}^2}=\frac{\left(\frac{4{\pi }^2}{G{M}_{2\text{M}1207}}\right)a_{2\text{M}1207}^3}{\left(\frac{4{\pi }^2}{G{M}_{\text{Sun}}}\right)a_{\text{Sun}}^3}\\ P_{2\text{M}1207}^2=\left(\frac{M_{\text{Sun}}{a}_{2\text{M}1207}^3}{M_{2\text{M}1207}{a}_{\text{Sun}}^3}\right)P_{\text{Sun}}^2\\ P_{2\text{M}1207}=\sqrt{\left(\frac{M_{\text{Sun}}{a}_{2\text{M}1207}^3}{M_{2\text{M}1207}{a}_{\text{Sun}}^3}\right)P_{\text{Sun}}^2}\\ =\sqrt{\left[\frac{M_{\text{Sun}}{\left(55\text{au}\right)}^3}{\left(0.025M_{\text{Sun}}\right){\left(1\text{au}\right)}^3}\right]{\left(1\text{yr}\right)}^2}\end{array}$

Solve further,

$\begin{array}{c}{P}_{2\text{M}1207}=\sqrt{6655000{\text{yr}}^2}\\ =2579.73\text{yr}\end{array}$

Conclusion:

The orbital period of the planet is $2579.73\text{yr}$ and it is not possible for the astronomer to observe a complete orbit in one lifetime.