Chapter 22, Problem 41Q

(a)

To determine

The semi major axes of the orbits of stars $\text{S0-2}\text{}\text{and S0-19}$ having orbital periods 14.5 and 37.3 years around Sagittarius ${\text{A}}^{\text{*}}$, respectively. Given that the supermassive black hole in Sagittarius ${\text{A}}^{\text{*}}$ has a mass of $\text{4}{\text{.1×10}}^{\text{6}}{M}_{\odot }$.

Answer to Problem 41Q

Solution:

The S0-2 star has a semi-major axis of $950.20\text{au}$ and S0-19 has a semi-major axis of $1814.17\text{au}$.

Explanation of Solution

Given data:

The black hole in Sagittarius ${\text{A}}^{\text{*}}$ has a mass of $\text{4}{\text{.1×10}}^{\text{6}}{M}_{\odot }$. The stars $\text{S0-2}\text{}\text{and S0-19}$ have orbital periods of 14.5 and 37.3 years, respectively.

Formula used:

Write the expression for Newton’s form of Kepler’s third law.

$p^2=\frac{4{\pi }^2{a}^3}{GM}$

Here, $P$ is the orbital period of a revolving star, $a$ is the semi major axis of the orbit of the star, $M$ is the mass of the central object around which the star is revolving (supermassive black hole inside Sagittarius ${\text{A}}^{\text{*}}$ in this case) and $G$ is the constant of gravitation, $6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}$

Conversion from $\text{au}$ to meters is done as:

$\text{1}\text{au =1}{\text{.496×10}}^{11}\text{m}$

Conversion from seconds to years is done as:

$1\text{year =}\text{3}\text{.15}\times {\text{10}}^7\text{s}$

Conversion from mass (in kg.) to solar mass is done as:

$\text{1}{M}_{\odot }\text{=}\text{1}{\text{.99×10}}^{\text{30}}\text{kg}$

Explanation:

Recall the expression for Newton’s form of Kepler’s third law.

$p^2=\frac{4{\pi }^2{a}^3}{GM}$

Rearrange the expression in terms of semi major axis.

$\begin{array}{l}{a}^3=\frac{MG{p}^2}{4{\pi }^2}\\ a=\sqrt[3]{\frac{MG{p}^2}{4{\pi }^2}}\end{array}$

For $\text{S0-2}\text{}$ star, the orbital period is 14.5 years. Substitute $\text{14}\text{.5}\text{years}$ for $p$, $\text{4}{\text{.1×10}}^{\text{6}}{M}_{\odot }$ for $M$ and $6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}$ for $G$ in the above expression.

$\begin{array}{c}a=\sqrt[3]{\frac{\left(4.1\times {10}^6{M}_{\odot }\right)\times \left(6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(14.5\text{years}\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{\left(4.1\times {10}^6{M}_{\odot }\left(\frac{1.99\times {10}^{30}\text{kg}}{1M_{\odot }}\right)\right)\times \left(6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(14.5\text{years}\left(\frac{3.15\times {10}^7\text{s}}{1\text{year}}\right)\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{\left(8.159\times {10}^{36}\text{kg}\right)\times \left(6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(4.56\times {10}^8\text{s}\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{1.134\times {10}^{44}}{4{\pi }^2}}\end{array}$

Further calculate as,

$\begin{array}{c}a=\sqrt[3]{\frac{1.134\times {10}^{44}}{4{\pi }^2}}\\ =1.42\times {10}^{14}\text{m}\times \left(\frac{1\text{au}}{1.496\times {10}^{11}\text{m}}\right)\\ =950.20\text{au}\end{array}$

Now, for $\text{S0-19}$ star the orbital period is 37.3 years,

Substitute $\text{37}\text{.3}\text{years}$ for $p$, $\text{4}{\text{.1×10}}^{\text{6}}{M}_{\odot }$ for $M$ and $6.673·{10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}$ for $G$ in the above expression.

$\begin{array}{c}a=\sqrt[3]{\frac{\left(4.1\times {10}^6{M}_0\right)\times \left(6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(37.3\text{years}\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{\left(4.1\times {10}^6{M}_0\left(\frac{1.99\times {10}^{30}\text{kg}}{1M_0}\right)\right)\times \left(\text{6}{\text{.673×10}}^{\text{-11}}{\text{ m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(37.3\text{years}\left(\frac{3.15\times {10}^{7}s}{1\text{year}}\right)\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{\left(8.159\times {10}^{36}\text{kg}\right)\times \left(6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(1.174\times {10}^9\text{s}\right)}^2}{4{\pi }^2}}\\ =\sqrt[3]{\frac{7.90\times {10}^{44}}{4{\pi }^2}}\end{array}$

Further calculate as,

$\begin{array}{c}a=\sqrt[3]{\frac{7.90\times {10}^{44}}{4{\pi }^2}}\\ =2.714\times {10}^{14}\text{m}\times \left(\frac{1\text{au}}{1.496\times {10}^{11}\text{m}}\right)\\ =1814.17\text{au}\end{array}$

Conclusion:

Therefore, the semi major axes of the orbits of stars $\text{S0-2}\text{}\text{and S0-19}$ are $\text{950}\text{.20}\text{au}$ and $\text{1814}\text{.17}\text{au}$, respectively.

(b)

To determine

The angular size of the semi major axes of the orbits of stars $\text{S0-2}\text{}\text{and S0-19}$ around Sagittarius ${\text{A}}^{\text{*}}$ and the reason for the use of very high resolution infrared images for the study of the motion of these stars. Given that the orbital periods of S0-2 and S0-19 are 14.5 and 37.3 years, respectively and that the supermassive black hole in Sagittarius ${\text{A}}^{\text{*}}$ has a mass of $\text{4}{\text{.1×10}}^{\text{6}}{M}_{\odot }$.

Answer to Problem 41Q

Solution:

The angular size of the orbit of S0-2 star is $0.23\text{arcsec}$ and that of S0-19 is $0.453\text{arcsec}$. These are very small angles and hence infrared images of very high resolution are used to study the motion of these stars.

Explanation of Solution

Given data:

Sagittarius ${\text{A}}^{\text{*}}$ has a mass of $\text{4}{\text{.1×10}}^{\text{6}}{M}_{\odot }$. The stars $\text{S0-2}\text{}\text{and S0-19}$ have orbital periods of 14.5 and 37.3 years, respectively. The distance of Earth from the center of the galaxy is $8000\text{pc}$.

Formula used:

Write the small-angle formula.

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

Here, α is the angle subtended by the object (in arcseconds), d is the distance between the observer and the object and D is the linear size of the object.

Conversion formula for 1 pc to au is,

$1\text{pc}\text{=}\text{206265}\text{au}$

Explanation:

The linear size of an orbit can be taken as twice its semi major axis.

$D=2a$

The angular size of the orbit of star $\text{S0-2}$ is calculated by using the small-angle formula. Here the subscript S0-2 denotes that the corresponding values are for S0-2 star.

${\alpha }_{S0-2}=\frac{206265D_{S0-2}}{d_{S0-2}}$

Substitute $2\times 950.20\text{au}$ for $D_{S0-2}$ and $8000\text{pc}$ for $d_{S0-2}$.

$\begin{array}{c}{\alpha }_{S0-2}=\frac{206265\left(2\times 950.20\text{au}\right)}{\left(\text{8000 pc}\right)\left(\frac{\text{206265}\text{au}}{\text{1 pc}}\right)}\\ =0.23\text{arcsec}\end{array}$

Similarly, the angular size of the orbit of star $\text{S0-19}$ is calculated by using the small-angle formula. Here the subscript S0-19 denotes that the corresponding values are for S0-2 star.

${\alpha }_{S0-19}=\frac{206265D_{S0-19}}{d_{S0-19}}$

Substitute $2\times \text{1814}\text{.17}\text{au}$ for $D_{S0-19}$ and $8000\text{pc}$ for $d_{S0-19}$.

$\begin{array}{c}{\alpha }_{S0-19}=\frac{206265\times \left(2\times \text{1814}\text{.17}\text{au}\right)}{\left(\text{8000 pc}\right)\times \left(\frac{\text{206265}\text{au}}{\text{1 pc}}\right)}\\ =0.453\text{arcsec}\end{array}$

These angle are quite small. To study the motion of such stars with very small angular sizes, high resolution infrared imaging is adopted because for the observation of far off tiny objects, radiation of large wavelengths must be used.

Conclusion:

The angular size of each orbit of stars $\text{S0-2}\text{}\text{and S0-19}$ are $0.23\text{arcsec}$ and $0.453\text{arcsec}$, respectively. Due to the very small angular size of their orbits, very high resolution infrared images are used to study their motion.