Chapter 18, Problem 28Q

To determine

(a)

The radius of the circumsteller accretion disc using ruler.

Answer to Problem 28Q

The radius of circumstellar accretion disc is $166.6\text{au}$ and The radius of circumstellar accretion disc in km is $248.2\times {10}^8\text{km}$.

Explanation of Solution

Calculation:

The radius of circumstellar accretion disc is one-sixth of the measuring ruler which has a total length of $1000\text{au}$.

The radius of circumstellar accretion disc is calculated as,

$\begin{array}{c}R=\frac{1}{6}\times 1000\text{au}\\ =166.6\text{au}\end{array}$

The radius of circumstellar accretion disc in km is calculated as,

$\begin{array}{c}R=166.6\text{au}\\ =\left(166.6\text{au}\times \frac{1.49\times {10}^8\text{km}}{1\text{au}}\right)\\ =248.2\times {10}^8\text{km}\end{array}$

Conclusion:

The radius of circumstellar accretion disc is $166.6\text{au}$ and The radius of circumstellar accretion disc in km is $248.2\times {10}^8\text{km}$.

To determine

(b)

The orbital period of the particle at outer edge of the disc.

Answer to Problem 28Q

The orbital period of particle is $7.67\times {10}^6\text{years}$.

Explanation of Solution

Given:

Mass of the young star is $M=1{\text{M}}_{\odot }$.

Formula used:

The expression of orbital period is given by,

$T=\sqrt{\frac{4{\pi }^2{R}^3}{G\cdot M}}$

Calculation:

The orbital period of the particle at the edge of disc is calculated as,

$\begin{array}{c}T=\sqrt{\frac{4{\pi }^2{R}^3}{G\cdot M}}\\ =\sqrt{\frac{4{\pi }^2{\left(248.23\times {10}^8\text{km}\times \frac{\text{1}\text{m}}{{\text{10}}^{-3}\text{km}}\right)}^3}{\left(6.673\times {10}^{-11} {\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\cdot \left(\text{1}{\text{M}}_{\odot }\times \frac{2\times {10}^{30}\text{kg}}{\text{1}{\text{M}}_{\odot }}\right)}}\\ =\left(6723.03\times {10}^7\text{hours}\times \frac{\text{1}\text{year}}{\text{8760}\text{hours}}\right)\\ =7.67\times {10}^6\text{years}\end{array}$

Conclusion:

The orbital period of particle is $7.67\times {10}^6\text{years}$.

To determine

(c)

The length of the jet that extends to the right of the circumstellar accretion disc and time taken by the star to traverse the entire visible range of jet.

Answer to Problem 28Q

The distance of the jet extends to the right of the disc is $496.617\times {10}^8\text{km}$ and time taken by the star to traverse the entire visible range of jet is $6.88\times {10}^4\text{hours}$.

Explanation of Solution

Given:

Speed of the gas is $v=200\text{km}/\text{s}$

Formula used:

The expression of time taken is given by,

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

Calculation:

The length of the jet which extends at the right side of the circumstellar accretion disc is one third of the total length of about $1000\text{au}$.

The length of jet is calculated as,

$\begin{array}{c}L=\frac{1}{3}\times 1000\text{au}\\ =\left(3\text{33}\text{.3}\text{au}\times \frac{1.49\times {10}^8\text{km}}{1\text{au}}\right)\\ =496.617\times {10}^8\text{km}\end{array}$

The time taken by the gas is calculated as,

$\begin{array}{c}T=\frac{D}{v}\\ =\frac{\left(496.617\times {10}^8\text{km}\right)}{\left(200\text{km}/\text{s}\right)}\\ =\left(2.48\times {10}^8\text{s}\times \frac{\text{1}\text{hour}}{\text{3600}\text{s}}\right)\\ =6.88\times {10}^4\text{hours}\end{array}$

Conclusion:

The distance of the jet extends to the right of the disc is $496.617\times {10}^8\text{km}$ and time taken by the star to traverse the entire visible range of jet is $6.88\times {10}^4\text{hours}$.