Chapter 17, Problem 7Q

(a)

To determine

The tangential velocity of the star if the Kapteyn star has a parallax angle of $0.255\text{arcsec}$ and the proper motion of the star is $8.67\text{arcsec per year}$ with a radial velociy of $+246\text{ km/s}$.

Answer to Problem 7Q

Solution:

$161\text{km/s}$.

Explanation of Solution

Given data:

Parallax angle of the star is $0.255\text{arcsec}$. The proper motion of the star is $8.67\text{arcsec per year}$. The radial velocity of the star is $+246\text{ km/s}$.

Formula used:

The expression for the distance of a star in the form of parallax angle is written as,

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

Here, $d$ is the distance to the star in parsec and $p$ is the parallax angle in arcsec.

The expression for the tangential velocity of a star is written as,

$v_t=4.74\mu d$

Here, $v_t$ is the tangential velocity of the star and $\mu$ is the proper motion in arcsec per year.

Explanation:

Recall the expression for the distance of a star in the form of parallax angle.

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

Substitute $0.255\text{arcsec}$ for $p$,

$\begin{array}{c}d=\frac{1}{0.255\text{arcsec}}\\ =3.92\text{parsec}\end{array}$

Recall the expression for the tangential velocity of a star.

$v_t=4.74\mu d$

Upon substituting $8.67\text{arcsec per year}$ for $\mu$ and $3.92\text{pc}$ for $d$,

$\begin{array}{c}{v}_t=4.74\left(8.67\text{arcsec per year}\right)\left(3.92\text{pc}\right)\\ =161\text{km/s}\end{array}$

Conclusion:

Thus, the tangential velocity of the star is $161\text{km/s}$.

(b)

To determine

The actual speed of the star relative to the Sun if the Kapteyn star has a parallax angle of $0.255\text{arcsec}$ and the proper motion of the star is $8.67\text{arcsec per year}$ with a radial velociy of $+246\text{ km/s}$.

Answer to Problem 7Q

Solution:

$294\text{km/s}$.

Explanation of Solution

Given data:

Parallax angle of the star is $0.255\text{arcsec}$. The proper motion of the star is $8.67\text{arcsec per year}$. The radial velocity of the star is $+246\text{ km/s}$.

Formula used:

The expression for the calculation of the tangential velocity of a star is written as,

$v^2=v_t{}^2+v_r{}^2$

Here, $v_t$ is the tangential velocity, $v_r$ is the radial velocity and $v$ is the velocity of the star relative to the Sun.

Explanation:

Recall the expression for the velocity of a star relative to the Sun.

$v^2=v_t{}^2+v_r{}^2$

Upon substituting $246\text{km/s}$ for $v_r$ and $161\text{km/s}$ for $v_t$,

$\begin{array}{c}{v}^2={\left(161\text{km/s}\right)}^2+{\left(246\text{km/s}\right)}^2\\ =25921{\text{km}}^2/{\text{s}}^2+60516{\text{km}}^2/{\text{s}}^2\\ =86437{\text{km}}^2/{\text{s}}^2\end{array}$

Further solve,

$\begin{array}{c}{v}^2=86437{\text{km}}^2/{\text{s}}^2\\ v=\sqrt{86437{\text{km}}^2/{\text{s}}^2}\\ =294\text{km/s}\end{array}$

Conclusion:

Thus, the star’s speed, relative to the sun, is $294\text{km/s}$.

(c)

To determine

Whether the Kapteyn star is moving towards the sun or away from the sun.

Answer to Problem 7Q

Solution:

The star is moving away from the sun.

Explanation of Solution

Introduction:

The component of the velocity of a star in our line of sight, away from us or towards us, is called radial velocity.

Explanation:

Radial velocity can be determined by measuring the Doppler shift in the spectral lines of the star’s spectrum. If the star is moving towards us, the wavelength of the spectral lines will decrease (blueshifted) and if the star going away from us, the wavelength of the spectral line will increase (redshifted).

As the value of radial velocity of the Sun is positive, which corresponds to redshift in the wavelength, the star is moving away from the sun.

Conclusion:

So, by measuring the shift in wavelength, the motion of a star, whether it is towards or away from us, can be determined.