Question 8.7: Find the distance between earth and Mars at 12 h UT on Augus...

Step 1:
According to Eq. (5.48), the Julian day number J_0 for midnight (0h UT) of this date is

J_0=367 y-\operatorname{INT}\left\{\frac{7\left[y+\operatorname{INT}\left(\frac{m+9}{12}\right)\right]}{4}\right\}+\operatorname{INT}\left(\frac{275 m}{9}\right)+d+1,721,013.5                           (5.48)

J_0=367 \cdot 2003-\operatorname{INT}\left\{\frac{7\left[2003+\operatorname{INT}\left(\frac{8+9}{12}\right)\right]}{4}\right\}+\operatorname{INT}\left(\frac{275 \cdot 8}{9}\right)+27+1,721,013.5

= 735,101 – 3507 + 244 + 27 + 1,721,013.5

= 2,452,878.5

At UT = 12, the Julian day number is

J D=2,452,878.5+\frac{12}{24}=2,452,879.0

Step 2:
The number of Julian centuries between J2000 and this date is

T_0=\frac{J D-2,451,545}{36,525}=\frac{2,452,879-2,451,545}{36,525}=0.036523  Cy

Step 3:
Table 8.1 and Eq. (8.93b) yield the orbital elements of earth and Mars at 12 h UT on August 27, 2003:

Q=Q_0+\dot{Q} T_0                        (8.93b)

Table 8.1 Planetary orbital elements and their centennial rates

Step 4:

h_{\text {earth }}=4.4451\left(10^9\right)  km ^2 / s

h_{\text {Mars }}=5.4760\left(10^9\right)  km ^2 / s

Step 5:

\omega_{\text {earth }}=(\varpi-\Omega)_{\text {earth }}=102.95-0=102.95^{\circ}

\omega_{\text {Mars }}=(\varpi-\Omega)_{\text {Mars }}=336.07-49.549=286.52^{\circ}

M_{\text {earth }}=(L-\varpi)_{\text {earth }}=335.27-102.95=232.32^{\circ}

M_{\text {Mars }}=(L-\varpi)_{\text {Mars }}=334.51-336.07=-1.56^{\circ}\left(358.43^{\circ}\right)

Step 6:

E_{\text {earth }}-0.016710 \sin E_{\text {earth }}=232.32^{\circ}(\pi / 180) \Rightarrow E_{\text {earth }}=231.57^{\circ}

E_{\text {Mars }}-0.093397 \sin E_{\text {Mars }}=358.43^{\circ}(\pi / 180) \Rightarrow E_{\text {Mars }}=358.27^{\circ}

Step 7:

\theta_{\text {earth }}=2 \tan ^{-1}\left(\sqrt{\frac{1+0.016710}{1-0.016710}} \tan \frac{231.57^{\circ}}{2}\right)=-129.18 \Rightarrow \theta_{\text {earth }}=230.8^{\circ}

\theta_{\text {Mars }}=2 \tan ^{-1}\left(\sqrt{\frac{1+0.093397}{1-0.093397}} \tan \frac{358.27^{\circ}}{2}\right)=-1.8998^{\circ} \Rightarrow \theta_{\text {Mars }}=358.10^{\circ}

Step 8:
From Algorithm 4.5,

R _{\text {earth }}=(135.59 \hat{ I }-66.803 \hat{ J }-0.00056916 \hat{ K })\left(10^6\right)  ( km )

V _{\text {earth }}=(12.680 \hat{ I }-26.610 \hat{ J }-0.00022672 \hat{ K })  ( km / s )

R _{\text {Mars }}=(185.95 \hat{ I }-89.959 \hat{ J }-6.4534 \hat{ K })\left(10^6\right)  ( km )

V _{\text {Mars }}=11.478 \hat{ I }-23.881 \hat{ J }-0.21828 \hat{ K }  ( km )

The distance d between the two planets is therefore

or

d=55.80\left(10^6\right)  km

The positions of earth and Mars are illustrated in Fig. 8.26. It is a rare event for Mars to be in opposition (lined up with earth on the same side of the sun) when Mars is at or near perihelion. The two planets had not been this close in recorded history.