Chapter 17, Problem 80Q

To determine

Whether the orbital speed of earth or the speed of the earth about its axis or both is accountable for the study of radial velocities of the stars or not. Also, determine the value of the orbital speed and the speed of the earth about its own axis. Also, explain how astronomers compensated the effect of earth orbital speed and its speed about its axis.

Answer to Problem 80Q

Solution:

$v_{\text{orbital}}=29.8\text{ km/s}$, $v_{\text{rotational}}=463\text{ m/s}$, and astronomers feed these data into the modern telescope to compensate the effect of orbital and rotational speed of the earth.

Explanation of Solution

Given data:

The stars radial velocity is to be measured to an accuracy of $1\text{ km/s}$ or better.

Formula used:

The formula for radius of a sphere in term of diameter is:

$r=\frac{D}{2}$

Here, $D$ is the diameter.

The expression used for the circumference of a circle C of radius $r$ is,

$C=2\pi r$

The expression used for the speed $v$ of an object is,

$v=\frac{\text{distance}}{\text{time}}$

Explanation:

Refer to the appendix 1, the value of the orbital radius of the earth is,

$r=149.6\times {10}^6\text{ km}$

The expression for the circumference at the orbit of the earth is,

$C=2\pi r$

Substitute $149.6\times {10}^6\text{ km}$ for $r$.

$\begin{array}{c}C=2\pi \left(149.6\times {10}^6\text{ km}\right)\\ =9.4\times {10}^8\text{ km}\end{array}$

Use the expression of speed to write the expression of orbital speed of the earth.

$v_{\text{orbital}}=\frac{C}{t}$

Substitute $9.4\times {10}^8\text{ km}$ for $C$ and $1\text{ yr}$ for $t$.

$\begin{array}{c}v=\frac{9.4\times {10}^8\text{ km}}{1\text{ yr}}\\ =\frac{9.4\times {10}^8\text{ km}}{1\text{ yr}\left(\frac{3.154\times {10}^7\text{ s}}{1\text{ yr}}\right)}\\ =29.8\text{ km/s}\end{array}$

Refer to the appendix 2 to obtain the value of the diameter of the earth:

$D=12756\text{ km}$

The formula for radius of earth is:

$r^{\prime }=\frac{D}{2}$

Substitute $12756\text{ km}$ for $D$.

$\begin{array}{c}{r}^{\prime }=\frac{12756\text{ km}}{2}\\ =6378\text{ km}\end{array}$

Therefore, the radius of the earth is $6378\text{ km}$.

The expression for the circumference of the earth at equator is:

$C^{\prime }=2\pi r^{\prime }$

Substitute $6378\text{ km}$ for $r^{\prime }$.

$\begin{array}{c}{C}^{\prime }=2\pi \left(6378\text{ km}\right)\\ =4.0\times {10}^4\text{ km}\end{array}$

Use the expression of speed to write the expression of rotational speed of the earth.

$v_{\text{rotational}}=\frac{C^{\prime }}{t^{\prime }}$

Substitute $4.0\times {10}^4\text{ km}$ for $C^{\prime }$ and $1\text{ day}$ for $t^{\prime }$.

$\begin{array}{c}{v}_{\text{rotational}}=\frac{4.0\times {10}^4\text{ km}}{1\text{ day}}\\ =\frac{4.0\times {10}^4\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)}{1\text{ day}\left(\frac{86400\text{ s}}{1\text{ day}}\right)}\\ =463\text{ m/s}\end{array}$

The orbital speed of the earth is $29.8\text{km/s}$. So, it will impact the calculation of the radial velocity of the star if the observation is done from the surface of the earth. Also, the rotational speed of the earth is $463\text{ m/s}$, which is also large enough to impact the observation for the calculation of radial velocity of the earth. So, to obtain the value of the radial velocity with accuracy upto $1\text{ km}$ or better, the details of the orbital and rotational speed of the earth is feed to the telescope.

Conclusion:

The orbital speed of the earth is $29.8\text{km/s}$ and its rotational speed is $463\text{ m/s}$, which is sufficient to impact the calculation for the radials velocity of the stars. So, these are taken into consideration for the calculation purpose.