Chapter 21, Problem 38Q

(a)

To determine

The radius of a x-ray source orbit, in kilometer, by assuming the radius of its orbit as a circle. Given that the orbital period of a binary system containing A0620-00 is 0.32 days and the radial velocity of x-ray source peaks at 457km/s as reveled by Doppler shift measurement.

Answer to Problem 38Q

Solution:

$\text{1974691}\text{.89 km}$

Explanation of Solution

Given data:

An orbital period of a binary system containing A0620-00 is 0.32 days and the radial velocity of x-ray source peaks at 457km/s as reveled by Doppler shift measurement.

Formula used:

Write the expression for the relationship between speed time and distance:

$v=\frac{D}{T}$

Here, $v$ is the velocity, $D$ is the distance and $T$ is time in seconds.

Explanation:

As the orbital period $\left(T\right)$ of a binary system containing A0620-00 is 0.32 days and its velocity $\left(v\right)$ is $457\text{ km/s}$.

Convert orbital period of 0.32 days in seconds.

$\begin{array}{c}T=0.32\text{ days}\\ \text{=0}\text{.32 days }\left(\frac{24\text{ hours}}{1\text{ day}}\right)\left(\frac{60\text{ minutes}}{1\text{ hour}}\right)\left(\frac{60\text{ second}}{1\text{ minute}}\right)\\ =2.76\times {10}^{{}^4}\text{ s}\end{array}$

Calculate radius of x-ray source orbit.

Assuming that the orbit is circular in radius $\left(r\right)$, then the time period $\left(T\right)$ and time’s velocity $\left(v\right)$ gets equal to the circumference.

Refer to the expression for the relationship between speed time and distance:

$v=\frac{D}{T}$

Substitute $2\pi r$ for $D$.

$2\pi r=Tv$

$r=\frac{Tv}{2\pi }$

Substitute $2.76\times {10}^{{}^4}\text{ s}$ for $T$, $457\text{ km/s}$ for $v$.

$\begin{array}{c}r=\frac{\left(2.76\times {10}^{{}^4}\text{ s}\right)\left(457\text{ km/s}\left(\frac{1000\text{ m/s}}{1\text{ km/s}}\right)\right)}{2\pi }\\ =1.32\times {10}^{-2}\text{ au}\left(\frac{1.496\times {10}^8\text{ km}}{1\text{ au}}\right)\\ =\text{1974691}\text{.89 km}\end{array}$

Conclusion:

Hence, the radius of x-ray source orbit is $\text{1974691}\text{.89 km}$.

(b)

To determine

The mass of x ray source, which is at least 3.1 times the mass of the Sun, by Newton’s form of Kepler’s third law. Given that the orbital period of a binary system containing A0620-00 is 0.32 days and the radial velocity of x-ray source peaks at 457km/s as reveled by Doppler shift measurement. Also, assume that the mass of K5V, a visible star, is negligible as compared to that of visible star.

Answer to Problem 38Q

Solution:

$\text{3}{\text{.1 M}}_{\odot }$

Explanation of Solution

Given data:

An orbital period of a binary system containing A0620-00 is 0.32 days and the radial velocity of x-ray source peaks at 457km/s as reveled by Doppler shift measurement.

Formula used:

Write the expression for Kepler’s third law:

$M_1+M_2=\frac{a^3}{P^2}$

Here, $M_1$ and $M_2$ are the masses of two neutron stars, $a$ is the semi major axis and $P$ is the time period.

Explanation:

As the orbital period of binary system containing A0620-00 is 0.32 days.

Convert orbital period of 0.32 days in years.

$\begin{array}{c}P=0.32\text{ days}\left(\frac{1\text{ year}}{365\text{ days}}\right)\\ =8.767\times {10}^{-4}\text{ years}\end{array}$

From part (a) of the question, orbital radius is $\text{1974691}\text{.89 km}$.

Convert orbital radius $\text{1974691}\text{.89 km}$ in au.

$\begin{array}{c}a=\text{1974691}\text{.89 km}\left(\frac{6.684\times {10}^{-9}\text{au}}{1\text{ km}}\right)\\ =1.34\times {10}^{-2}\text{ au}\end{array}$

Refer to the expression for Kepler’s third law.

$M_1+M_2=\frac{a^3}{P^2}$

Substitute $8.76\times {10}^{-4}\text{ years}$ for $P$ and $1.34\times {10}^{-2}\text{ au}$ for $a$.

$\begin{array}{c}{M}_1+M_2=\frac{{\left(1.34\times {10}^{-2}\text{ au}\right)}^3}{{\left(8.76\times {10}^{-4}\text{ years}\right)}^2}\\ =3.1{\text{ M}}_{\odot }\end{array}$

So, the total mass of x ray source binary system is $\text{3}{\text{.1 M}}_{\odot }$.

Conclusion:

Hence, the total mass of x ray source is $\text{3}{\text{.1 M}}_{\odot }$.