Chapter 17, Problem 68Q

To determine

(a)

The distance to 70 Ophiuchi in parsecs.

Answer to Problem 68Q

$5\text{ Parsecs}$

Explanation of Solution

Given:

$\text{Parallax}=0.2\text{arcsec}$

Formula used:

$P=\frac{1}{d}$

$$

Calculation:

$\begin{array}{l}d=\frac{\text{1}}{\text{0}\text{.2 arcsec}}\\ d=5\text{ pc}\end{array}$

Conclusion:

The distance to 70 Ophiuchi in parsecs = $5\text{ pc}$

To determine

(b)

The actual length of the semimajor axis in AU.

Answer to Problem 68Q

Actual length of the semimajor axis in astronomical units = $22.5\text{ au}$

Explanation of Solution

Given:

$$

The distance to 70 Ophiuchi in parsecs = $5\text{ Parsecs}$ (calculated in Part a)

$\text{Apparent length of semimajor axis}\text{=}4.5\text{arcsec}$

Formula Used:

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

$1\text{ pc = 206265 au}$

Calculation:

$\begin{array}{l}\alpha =\frac{D}{\text{d}}\times 206265\\ D=\frac{\alpha d}{206265}\\ D=\frac{4.5\times 5}{206265}\text{ pc}\\ D=\frac{4.5\times 5}{206265}\times 206265\text{ au}\\ D=22.5\text{ au}\end{array}$

Conclusion:

Thus, semi major axis is $22.5\text{ au}$

To determine

(c)

The sum of the mass of the two stars.

Answer to Problem 68Q

The sum of masses of the 2 stars in 70 Ophiuchi = $1.495{\text{ M}}_{\odot }$

Explanation of Solution

Given:

$\text{Period}=87.7\text{years}$

Formula Used:

$\begin{array}{l}{T}^2=\frac{4{\pi }^2}{G\left(M_1+M_2\right)}{a}^3\\ M_1+M_2=\frac{4{\pi }^2}{G{T}^2}{a}^3\end{array}$

Where, M represents mass, G is the gravitational constant, T is the period and a is the semi-major axis.

$1{\text{M}}_{\odot }=1.989\times {10}^{30}\text{ kg}$

$\begin{array}{l}\text{1 au = }1.496\times {10}^8\text{ km}\\ \text{1 year = 365}\times \text{24}\times \text{60}\times \text{60 s}\end{array}$

Calculation:

$\begin{array}{l}{M}_1+M_2=\frac{4{\pi }^2}{G{T}^2}{a}^3\\ M_1+M_2=\frac{4{\pi }^2}{\text{6}\text{.67259 }\times {\text{10}}^{\text{-}}{}^{\text{11}}\text{ N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{\text{-2}}{\left(\text{87}\text{.7}\times 3\text{65}\times 24\times 60\times 60\right)}^{\text{2}}{\text{ s}}^2}{(}^{22.5}{\text{ m}}^3\\ M_1+M_2=2.9735\times {10}^{30}\text{kg}\\ M_1+M_2=2.9735\times {10}^{30}\text{kg}\\ M_1+M_2=\frac{2.9735\times {10}^{30}}{1.989\times {10}^{30}}{\text{M}}_{\odot }\\ M_1+M_2=1.495{\text{ M}}_{\odot }\end{array}$

Conclusion:

The sum of masses of the 2 stars in 70 Ophiuchi = $1.495{\text{ M}}_{\odot }$