Chapter 8, Problem 45Q

(a)

To determine

The Sun’s orbital speed in meters per second considers that it moves in a small orbit of radius $742,000\text{ km}$ with a period of 11.86 years because of the presence of Jupiter.

Answer to Problem 45Q

Solution:

12.5 m/s

Explanation of Solution

Given data:

The Sun moves in a small orbit of radius $742,000\text{ km}$ with a period of 11.86 years.

Formula used:

Write the expression for the circumference $C$ of a circle of radius $r$.

$C=2\pi r$

Write the expression for the speed $\nu$.

$\nu =\frac{C}{t}$

Here, $C$ is the circular distance and t is the time taken.

Write the formula for the conversion of the unit of time from years to seconds.

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

Explanation:

Recall the expression for the circumference $C$ of a circle of radius $r$.

$C=2\pi r$

The expression for the Sun’s orbital speed $\nu$ is written as,

$\begin{array}{c}\nu =\frac{C}{t}\\ =\frac{2\pi r}{t}\end{array}$

Substitute $742,000\text{ km}$ for $r$ and 11.86 years for t.

$\begin{array}{c}\nu =\frac{2\pi \left(742000\text{ km}\right)\left(\frac{1000\text{ m}}{1\text{ km}}\right)}{\left(11.86\text{ years}\right)\left(\frac{1\text{ s}}{3.15\times {10}^7\text{ years}}\right)}\\ =12.5\text{ m/s}\end{array}$

Conclusion:

Hence, the Sun’s orbital speed in meters per second is 12.5 m/s.

(b)

To determine

The angular diameter of the Sun’s orbit as seen by an alien astronomer from a hypothetical planet orbiting the star Vega, which is 25 ly from the Sun. Also explain whether, or not, the Sun’s motion is discernible if the alien astronomer could measure positions to an accuracy of 0.001 arcsec.

Answer to Problem 45Q

Solution:

$1.48\times {10}^9\text{ m}$. The motion would just barely be discernible.

Explanation of Solution

Given data:

The hypothetical planet is orbiting the star Vega, which is 25 ly from the Sun.

Formula used:

Write the small angle formula.

$\alpha =\frac{D\left(206265\right)}{d}$

Here, $\alpha$ is the small angle, $D$ is the angular diameter and $d$ is the distance from the observer.

Write the expression for the relation between the diameter D and the radius r.

$D=2r$

Write the expression for converting the unit of distance from light years to meters.

$1\text{ ly}=2.36\times {10}^{17}\text{ m}$

Explanation:

Recall the expression for the relation between diameter D and radius r.

$D=2r$

Substitute $742,000\text{ km}$ for r.

$\begin{array}{c}D=2\left(742000\text{ km}\right)\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =1.48\times {10}^9\text{ m}\end{array}$

Recall the expression for the small angle formula.

$\alpha =\frac{D\left(206265\text{ arcsec}\right)}{d}$

Substitute $1.48\times {10}^9\text{ m}$ for $D$ and 25 ly for $d$.

$\begin{array}{c}\alpha =\frac{\left(1.48\times {10}^9\text{ m}\right)\left(206265\text{ arcsec}\right)}{\left(\text{25 ly}\right)\left(\frac{2.36\times {10}^{17}\text{ m}}{1\text{ ly}}\right)}\\ =0.0013\text{ arcsec}\end{array}$

The above value tells that the motion would just barely be discernible if the alien astronomer could measure positions to an accuracy of 0.001 arcsec.

Conclusion:

Hence, the angular diameter of the Sun’s orbit is $1.48\times {10}^9\text{ m}$. The motion would just barely be discernible.

(c)

To determine

The answer, same as explained in part (b), but by considering that the astronomer is located on a hypothetical planet in the Pleiades star cluster, 360 ly from the Sun. Then provide an explanation whether the Sun’s motion is discernible to this astronomer.

Answer to Problem 45Q

Solution:

No, because the value of $\alpha$ is very small.

Explanation of Solution

Given data:

The astronomer is located on a hypothetical planet in the Pleiades star cluster, 360 ly from the Sun.

Formula used:

Write the small angle formula.

$\alpha =\frac{D\left(206265\right)}{d}$

Here, $\alpha$ is the small angle, $D$ is the angular diameter and $d$ is the distance from the observer.

Write the expression for converting the unit of distance from light years to meters.

$1\text{ ly}=2.36\times {10}^{17}\text{ m}$

Explanation:

Recall the small angle formula.

$\alpha =\frac{D\left(206265\text{ arcsec}\right)}{d}$

Substitute $1.48\times {10}^9\text{ m}$ for $D$ and 360 ly for $d$.

$\begin{array}{c}\alpha =\frac{\left(1.48\times {10}^9\text{ m}\right)\left(206265\text{ arcsec}\right)}{\left(\text{360 ly}\right)\left(\frac{2.36\times {10}^{17}\text{ m}}{1\text{ ly}}\right)}\\ =9\times {10}^5\text{ arcsec}\end{array}$

The above value of $\alpha$ is very small as compared to the value 0.001 arcsec. Hence, the Sun’s motion would not be discernible to this astronomer.

Conclusion:

Hence, the Sun’s motion would not be discernible to this astronomer.