Chapter 22, Problem 42Q

(a)

To determine

The orbital period of a star, revolving around Sagittarius ${\text{A}}^{\text{*}}$ in years. It is given that the star has a circular orbit of radius $530\text{au}$ and an orbital speed of $2500\text{km/s}$.

Answer to Problem 42Q

Solution:

$6.3\text{years}$

Explanation of Solution

Given data:

Orbital velocity of the star is $2500\text{km/s}$ and the radius of its circular orbit is $530\text{au}$.

Formula used:

The expression for the orbital period of a star:

$P=\frac{2\pi r}{v}$

Here, $P$ is the orbital period of a star, $r$ is the radius of its orbit and $v$ is its orbital speed.

Conversion from $\text{au}$ to km,

$1\text{au }=1.496\times {10}^8\text{km}$

Conversion from seconds to year,

$1\text{year }=3.15\times {10}^7\text{s}$

Explanation:

Recall the expression for the period of an orbiting object.

$P=\frac{2\pi r}{v}$

Substitute $2500\text{km/s}$ for $v$ and $530\text{au}$ for $r$.

$\begin{array}{c}P=\frac{2\pi \left(530\text{au}\right)}{2500\text{km/s}}\\ =\frac{2\pi \left(530\text{au}\left(\frac{1.496\times {10}^8\text{km}}{1\text{au}}\right)\right)}{2500\text{km/s}}\\ =1.99\times {10}^8\text{s}\\ \text{=}6.3\text{ years}\end{array}$

Conclusion:

Therefore, the orbital period of the star is $\text{6}\text{.3 years}$.

(b)

To determine

The sum of the masses of Sagittarius A* and the star in terms of solar mass. It is given that the star orbits around Sagittarius ${\text{A}}^{\text{*}}$ in a circular orbit of radius $530\text{au}$ with an orbital speed of $2500\text{km/s}$.

Answer to Problem 42Q

Solution:

$\text{3}{\text{.76×10}}^{\text{6}}{M}_{\odot }$

Explanation of Solution

Given data:

Orbital velocity of the star is $2500\text{km/s}$ and the radius of its circular orbit is $530\text{au}$.

Formula used:

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

$P^2=\frac{4{\pi }^2{a}^3}{G\left(M+m\right)}$

Here, $P$ is the orbital period of a star, $r$ is the distance of the star from Sagittarius ${\text{A}}^{\text{*}}$(radius of its orbit), $v$ is the orbital speed of the star, $\text{M}$ is the mass of Sagittarius ${\text{A}}^{\text{*}}$, $\text{m}$ is the mass of the star and $G$ is the constant of gravitation, $6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}$.

From the previous subpart, the orbital period of the star is $\text{1}{\text{.99×10}}^{\text{8}}\text{s}$.

Conversion from $\text{au}$ to m,

$1\text{au}=1.496\times {10}^{11}\text{m}$

Conversion from mass (in kg.) to solar mass,

$1M_{\odot }=1.99\times {10}^{30}\text{kg}$

Explanation:

In the previous subpart (b), the orbital period of the star was calculated as $\text{1}{\text{.99×10}}^{\text{8}}\text{s}$.

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

$P^2=\frac{4{\pi }^2{a}^3}{G\left(M+m\right)}$

Rearrange the expression in terms of the masses of the star and Sagittarius A*,

$\left(M+m\right)=\frac{4{\pi }^2{a}^3}{G{P}^2}$

Substitute $\text{1}{\text{.99×10}}^{\text{8}}\text{s}$ for $p$, $530\text{au}$ for $a$ and $6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}$ for $G$,

$\begin{array}{c}M+m=\frac{4{\pi }^2\times {\left(530\text{au}\right)}^3}{\left(6.673\times {10}^{-11}{\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(1.99\times {10}^8\text{s}\right)}^2}\\ =\frac{4{\pi }^2\times {\left(530\text{au}\left(\frac{1.496\times {10}^{11}\text{m}}{1\text{au}}\right)\right)}^3}{\left(6.673\times {10}^{-11} {\text{m}}^{\text{3}}{\text{/kg-s}}^{\text{2}}\right)\times {\left(1.99\times {10}^8\text{s}\right)}^2}\\ =\frac{1.967\times {10}^{43}{\text{m}}^{\text{3}}}{2.642\times {10}^6{\text{m}}^{\text{3}}\text{/kg}}\\ =7.44\times {10}^{36}\text{kg}\end{array}$

The sum of masses in terms of solar mass of the star and Sagittarius A*.

$\begin{array}{c}M+m=7.44\times {10}^{36}\text{kg}\left(\frac{1M_{\odot }}{1.99\times {10}^{30}\text{kg}}\right)\\ =3.76\times {10}^6{M}_{\odot }\end{array}$

Conclusion:

Hence, the sum of masses in terms of solar mass of Sagittarius A* and the star is $3.76\times {10}^6{M}_{\odot }$.