Chapter 22, Problem 41Q

To determine

(a)

The semi-major axis of the star Sagittarius A* with orbital period $\text{S0-2}$ and $\text{S0-19}$ years.

Answer to Problem 41Q

The length of the semi major axis of $\text{S0-19}$ is $1814.17\text{au}$ and $\text{S0-2}$ is $950.20\text{au}$.

Explanation of Solution

Given:

The orbital period of $\text{S0-2}$ is, $p=14.5$.

The orbital period of $\text{S0-19}$ is, $p=37.3$.

The mass of Sagittarius $A$ is, $M=4.1\times {10}^6{\text{M}}_{\odot }$.

Formula Used:

The expression for the length of the semi major axis is given by,

$a=\sqrt[3]{\frac{MF{p}^2}{4{\pi }^2}}$

Calculation:

The value of one solar mass is $1{\text{M}}_{\odot }=1.99\times {10}^{30}\text{kg}$.

The length of the semi major axis of $\text{S0-2}$ is calculated as,

$\begin{array}{c}a=\sqrt[3]{\frac{MG{p}^2}{4{\pi }^2}}1{\text{M}}_{\odot }\\ =\sqrt[3]{\frac{\left(4.1\times {10}^6{M}_{\odot }\right)\left(6.673\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^2\right){\left(14.5\text{years}\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{\begin{array}{l}\left(4.1\times {10}^6\left(1.99\times {10}^{30}\text{kg}\right)\right)\left(6.673\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^2\right)\\ {\left(14.5\text{years}\left(\frac{3.15\times {10}^7\text{s}}{1\text{year}}\right)\right)}^2\end{array}}{4{\pi }^2}}\\ =\sqrt[3]{\frac{1.134\times {10}^{44}{\text{m}}^3}{4{\pi }^2}}\end{array}$

Solve further as,

$\begin{array}{c}a=1.42\times {10}^{14}\text{m}\left(\frac{1\text{au}}{1.496\times {10}^{11}\text{m}}\right)\\ =950.20\text{au}\end{array}$

The length of the semi major axis of $\text{S0-19}$ is calculated as,

$\begin{array}{c}a=\sqrt[3]{\frac{MG{p}^2}{4{\pi }^2}}1{\text{M}}_{\odot }\\ =\sqrt[3]{\frac{\left(4.1\times {10}^6{M}_{\odot }\right)\left(6.673\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^2\right){\left(37.3\text{years}\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{\begin{array}{l}\left(4.1\times {10}^6\left(1.99\times {10}^{30}\text{kg}\right)\right)\left(6.673\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^2\right)\\ {\left(37.3\text{years}\left(\frac{3.15\times {10}^7\text{s}}{1\text{year}}\right)\right)}^2\end{array}}{4{\pi }^2}}\\ =\sqrt[3]{\frac{7.90\times {10}^{44}{\text{m}}^3}{4{\pi }^2}}\end{array}$

Solve further as,

$\begin{array}{c}a=2.714\times {10}^{14}\text{m}\left(\frac{1\text{au}}{1.496\times {10}^{11}\text{m}}\right)\\ =1814.17\text{au}\end{array}$

Conclusion:

Therefore, the length of the semi major axis of $\text{S0-19}$ is $1814.17\text{au}$ and $\text{S0-2}$ is $950.20\text{au}$.

To determine

(b)

The angular size of each orbits semi major axis as seen from Earth to the center of the galaxy. Also the reason for extremely high-resolution infrared images is required to observe the motions of stars.

Answer to Problem 41Q

The angular size of orbit of star $\text{S0-2}$ is $0.23\text{arcsec}$ and $\text{S0-19}$ is $0.453\text{arcsec}$. These small angles are very small and to observe the far and tiny object high resolution infrared imaging is required.

Explanation of Solution

Given:

The distance from the Earth to the center of the galaxy is $206265\text{au}$.

Formula Used:

The expression for the small angle formula is given by,

$\alpha =\frac{206265D}{d}$

Here, $\alpha$ is the angle subtended by the object, $d$ is the distance between the observer and $D$ is the linear size of the object.

The formula to calculate the linear size of the orbit is given by,

$D=2a$

Calculation:

The linear size of the orbit for $\text{S0-2}$ is calculated as,

$\begin{array}{c}D=2a\\ =2\left(950.20\text{au}\right)\\ =1900.4\end{array}$

The linear size of the orbit for $\text{S0-19}$ is calculated as,

$\begin{array}{c}D=2a\\ =2\left(1814.17\text{au}\right)\\ =3628.34\text{au}\end{array}$

The angular size of the orbit of $\text{S0-2}$ is calculated as,

$\begin{array}{c}\alpha =\frac{206265D}{d}\\ =\frac{206265\left(1900.4\right)}{800\text{pc}}\\ =\frac{206265\left(1900.4\right)}{8000\text{pc}\left(\frac{206265}{1\text{pc}}\right)}\\ =0.23\text{arcsec}\end{array}$

The angular size of the orbit of $\text{S0-19}$ is calculated as,

$\begin{array}{c}\alpha =\frac{206265D}{d}\\ =\frac{206265\left(3628.34\text{au}\right)}{800\text{pc}}\\ =\frac{206265\left(3628.34\text{au}\right)}{8000\text{pc}\left(\frac{206265}{1\text{pc}}\right)}\\ =0.453\text{arcsec}\end{array}$

The above given angles are very small and make it difficult to study the motion of the stars with very small angular sizes they require high resolution infrared imaging so that the far and tiny objects large wavelengths of radiation are used.

Conclusion:

The angular size of orbit of star $\text{S0-2}$ is $0.23\text{arcsec}$ and $\text{S0-19}$ is $0.453\text{arcsec}$. These small angles are very small and to observe the far and tiny object high resolution infrared imaging is required.