Chapter 8, Problem 47Q

(a)

To determine

The angular distance in arcseconds between the star 2M1207 and its planet, seen from Earth, considering this star is 170 light years from Earth.

Answer to Problem 47Q

Solution:

$1.1\text{ arcsec}$.

Explanation of Solution

Given data:

The star 2M1207 is 170 light years from Earth.

Formula used:

Write the expression for the small angle formula.

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

Here, $\alpha$ is the small angle, $D$ is the orbital distance, and $d$ is the distance from the observer.

Explanation:

Recall the expression for the small angle formula.

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

Substitute 55 au for $D$, 170 ly for $d$.

$\begin{array}{c}\alpha =\frac{\left(55\text{ au}\right)\left(206265\text{ arcsec}\right)}{\left(170\text{ ly}\right)\left(\frac{63,240\text{ au}}{1\text{ ly}}\right)}\\ =1.1\text{ arcsec}\end{array}$

Conclusion:

Hence, the angular distance in arcseconds between the star 2M1207 and its planet is $1.1\text{ arcsec}$.

(b)

To determine

The orbital period of the orbiting star 2M1207, whose mass is 0.025 times that of the Sun, by considering that the distance between the star and its planet is the semi-major axis of the orbit.

Answer to Problem 47Q

Solution:

$2580\text{ yrs}$.

Explanation of Solution

Given data:

The mass of star 2M1207 is 0.025 times that of the Sun.

Formula used:

Write the formula for the relation between orbital period and orbital distance according to Kepler’s third law.

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

Here, $P$ is the period, G is the gravitational constant, M is the mass of planet or star, and $a$ is the orbital distance.

Explanation:

The formula for the relation between orbital period and orbital distance for Sun, according to Kepler’s third law is written as,

$P_{\text{Sun}}^2=\left(\frac{4{\pi }^2}{G{M}_{\text{Sun}}}\right)a_{\text{Sun}}^3$ …… (1)

Here, subscript ‘Sun’ is used for the respective quantities of the Sun.

The formula for the relation between orbital period and orbital distance for star 2M1207, according to Kepler’s third law is written as,

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

Here, subscript ‘2M1207’ is used for the respective quantities of the star 2M1207.

Divide equation (2) by equation (1).

$\begin{array}{l}\frac{P_{2\text{M}1207}^2}{P_{\text{Sun}}^2}=\frac{\left(\frac{4{\pi }^2}{G{M}_{2\text{M}1207}}\right)a_{2\text{M}1207}^3}{\left(\frac{4{\pi }^2}{G{M}_{\text{Sun}}}\right)a_{\text{Sun}}^3}\\ P_{2\text{M}1207}^2=\frac{\left(M_{\text{Sun}}\right)\left(a_{2\text{M}1207}^3\right)}{\left(M_{2\text{M}1207}\right)\left(a_{\text{Sun}}^3\right)}\left(P_{\text{Sun}}^2\right)\end{array}$

Substitute 1 yr for $P_{\text{Sun}}$, $0.025M_{\text{Sun}}$ for $M_{2\text{M1207}}$, 1 au for $a_{\text{Sun}}$ and 55 au for $a_{\text{2M1207}}$.

$\begin{array}{c}{P}_{2\text{M}1207}=\sqrt{\frac{\left(M_{\text{Sun}}\right){\left(55\text{ au}\right)}^3}{\left(0.025M_{\text{Sun}}\right){\left(1\text{ au}\right)}^3}{\left(1\text{ yr}\right)}^2}\\ =2580\text{ yrs}\end{array}$

Conclusion:

Hence, the orbital period for star 2M1207 is $2580\text{ yrs}$.