Chapter 3, Problem 3.1P

(a)

To determine

The distance between earth and mars in meter and astronomical units.

Answer to Problem 3.1P

The distance between earth and mars in meter is $7.831\times {10}^6\text{m}$ and astronomical units is $0.526\text{AU}$.

Explanation of Solution

Write the expression for distance between earth and mars.

    $d=\frac{R_e}{\tan p}$        (1)

Here, $R_e$ is radius of earth and angle subtended by the earth is $p$.

Conclusion:

Substitute $6.378\times {10}^6\text{m}$ for $R_e$ and $\frac{33.6^{″}}{2}$ for $p$ in equation (1).

    $\begin{array}{c}d=\frac{6.378\times {10}^6\text{m}}{\tan \left(\frac{33.6^{″}}{2}\right)\left(\frac{1\text{rad}}{206264.806^{″}}\right)}\\ =\frac{6.378\times {10}^6\text{m}}{\tan \left(8.15\times {10}^{-5}\text{rad}\right)}\\ =7.831\times {10}^6\text{m}\end{array}$

Convert $7.831\times {10}^6\text{m}$ into astronomical unit.

    $\begin{array}{c}d=7.831\times {10}^6\text{m}\\ =\left(7.831\times {10}^6\text{m}\right)\left(\frac{1\text{AU}}{1.496\times {10}^{11}\text{m}}\right)\\ =0.526\text{AU}\end{array}$

Thus, the distance between earth and mars in meter is $7.831\times {10}^6\text{m}$.

The distance between earth and mars in astronomical units is $0.526\text{AU}$.

(b)

To determine

The clocks used by the two observers must be synchronized.

Answer to Problem 3.1P

The clocks used by the two observers must be synchronized at $225\text{}\text{μs}$.

Explanation of Solution

Write the expression for actual distance between the earth and the mars.

    $d_{act}=\frac{R_e+0.5v_{rel}\Delta t}{\tan p}$        (2)

Here, $v_{rel}$ is relative velocity of earth and $\Delta t$ is time difference.

Write the expression for relative velocity.

    $v_{rel}=v_e-v_m$        (3)

Here, $v_e$ is velocity of earth and $v_m$ is velocity of mars.

Write the expression for difference between actual distance and calculated distances.

    $\begin{array}{c}\frac{d_{act}-d}{d}=0.1\\ \frac{R_e+0.5v_{rel}\Delta t}{\tan p}-\frac{R_e}{\tan p}=0.1\\ \frac{0.5v_{rel}\Delta t}{\tan p}=0.1\end{array}$        (4)

Conclusion:

Substitute $29.79\text{}\text{km}/\text{sec}$ for $v_e$ and $24.13\text{}\text{km}/\text{sec}$ for $v_m$ in equation (3).

    $\begin{array}{c}{v}_{rel}=29.79\text{}\text{km}/\text{sec}-24.13\text{}\text{km}/\text{sec}\\ =5.66\text{}\text{km}/\text{sec}\end{array}$

Substitute $5.66\text{}\text{km}/\text{sec}$ for $v_{rel}$, $6.378\times {10}^6\text{m}$ for $R_e$ and $\frac{33.6^{″}}{2}$ for $p$ in equation (4).

    $\begin{array}{c}\frac{0.5\left(5.66\text{}\text{km}/\text{sec}\right)\Delta t}{\tan \left(\frac{33.6^{″}}{2}\right)}=0.1\\ \frac{0.5\left(5.66\text{}\text{km}/\text{sec}\right)\Delta t}{\tan \left(\frac{33.6^{″}}{2}\right)\left(\frac{1\text{rad}}{206264.806^{″}}\right)}=0.1\\ \Delta t=\left(225\times {10}^{-6}\text{s}\right)\left(\frac{1\text{μs}}{{10}^{-6}\text{s}}\right)\\ \Delta t=225\text{}\text{μs}\end{array}$

Thus, the clocks used by the two observers must be synchronized at $225\text{}\text{μs}$.