Chapter 8, Problem 45Q

To determine

(a)

The Sun’s orbital speed.

Answer to Problem 45Q

The Sun’s orbital speed is $12.48\text{m}/\text{s}$.

Explanation of Solution

Given:

The radius of the orbit of the Sun is $r=742000\text{km}$.

The orbital period of the Sun is $t=11.86\text{years}$.

Formula Used:

The length of the orbital path of the Sun is given by

$C=2\pi r$

The Sun’s orbital speed is given by

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

Calculations:

The Sun’s orbital speed is calculated as

$\begin{array}{c}v=\frac{2\pi r}{t}\\ =\frac{2\pi \left(742000\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)}{\left(11.86\text{yrs}\times \frac{3.15\times {10}^7\text{s}}{1\text{yr}}\right)}\\ =12.48\text{m}/\text{s}\end{array}$

Conclusion:

The Sun’s orbital speed is $12.48\text{m}/\text{s}$.

To determine

(b)

The angular diameter of the Sun’s orbit as seen by the astronomer and to check whether the Sun’s motion would be discernible if the alien astronomer is able to measure positions to an accuracy of $0.001\text{arcsec}$.

Answer to Problem 45Q

The angular diameter of the Sun’s orbit as seen by the astronomer is $1.484\times {10}^9\text{m}$ and the small angle is calculated to be $0.0013\text{arcsec}$, which indicates that the Sun’s motion would barely be discernible if the alien astronomer is able to measure positions to an accuracy of $0.001\text{arcsec}$.

Explanation of Solution

Given:

The hypothetical planet is orbiting around Vega.

The distance from the Sun is $d=25\text{ly}$.

Formula Used:

The small angle formula is given by

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

Here, $D$ is the angular diameter.

Calculations:

The angular diameter is calculated as

$\begin{array}{c}D=2r\\ =2\left(742000\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)\\ =1.484\times {10}^9\text{m}\end{array}$

The small angle is calculated as

$\begin{array}{c}\alpha =\frac{D\left(206265\right)}{d}\\ =\frac{\left(1.484\times {10}^9\text{m}\right)\left(206265\right)}{\left(25\text{ly}\times \frac{9.46\times {10}^{15}\text{m}}{1\text{ly}}\right)}\\ =0.0013\text{arcsec}\end{array}$

Conclusion:

The angular diameter of the Sun’s orbit as seen by the astronomer is $1.484\times {10}^9\text{m}$ and the small angle is calculated to be $0.0013\text{arcsec}$, which indicates that the Sun’s motion would barely be discernible if the alien astronomer is able to measure positions to an accuracy of $0.001\text{arcsec}$.

To determine

(c)

The angular diameter of the Sun’s orbit as seen by the astronomer and to check whether the Sun’s motion would be discernible if the alien astronomer is able to measure positions to an accuracy of $0.001\text{arcsec}$.

Answer to Problem 45Q

The angular diameter of the Sun’s orbit as seen by the astronomer is $1.484\times {10}^9\text{m}$ and the small angle is calculated to be $9\times {10}^{-5}\text{arcsec}$, which is very small as compared to $0.001\text{arcsec}$. This indicates that the Sun’s motion would not be discernible if the alien astronomer is able to measure positions to an accuracy of $0.001\text{arcsec}$.

Explanation of Solution

Given:

The hypothetical planet is in the Pleiades star cluster.

The distance from the Sun is $d=360\text{ly}$.

Formula Used:

The small angle formula is given by

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

Here, $D$ is the angular diameter.

Calculations:

The angular diameter is calculated as

$\begin{array}{c}D=2r\\ =2\left(742000\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)\\ =1.484\times {10}^9\text{m}\end{array}$

The small angle is calculated as

$\begin{array}{c}\alpha =\frac{D\left(206265\right)}{d}\\ =\frac{\left(1.484\times {10}^9\text{m}\right)\left(206265\right)}{\left(360\text{ly}\times \frac{9.46\times {10}^{15}\text{m}}{1\text{ly}}\right)}\\ =9\times {10}^{-5}\text{arcsec}\end{array}$

Conclusion:

The angular diameter of the Sun’s orbit as seen by the astronomer is $1.484\times {10}^9\text{m}$ and the small angle is calculated to be $9\times {10}^{-5}\text{arcsec}$, which is very small as compared to $0.001\text{arcsec}$. This indicates that the Sun’s motion would not be discernible if the alien astronomer is able to measure positions to an accuracy of $0.001\text{arcsec}$.