Celestial Calculations: A Gentle Introduction to Computational Astronomy

248 Chapter 8 where j0, j1, and j2 are obtained from table 8.5 for the planet of interest. Once we have the Julian day number, it is easy to convert it to the corresponding calendar date and UT via the methods presented in section 3.6. The Julian day number obtained from equation 8.9.3 will be the date closest to the date used in equations 8.9.1 and 8.9.2, but it may well be significantly later or earlier than the date used in those 2 equations. For example, compute perihelion and aphelion for Mars closest to October 30, 1938, the date of the infamous War of the Worlds broadcast. The steps are: 1. Compute Days, the number of days into the year (see section 3.7). (Ans: Days = 303.) 2. Use equation 8.9.2 to compute Y . (Ans: Y = 1938.829569.) 3. Use equation 8.9.1 to compute K with the appropriate values for Mars taken from table 8.5. (Ans: k0 = 0.53166, k1 = 2001.78, K = −33.468226.) 4. Let Kper be the integer value closest to K, and let Kaph be the fraction ending in 0.5 that is closest to K. (Ans: Kper = −33, Kaph = −33.5.) 5. Use equation 8.9.3, the appropriate values from table 8.5, and Kper to com- pute the Julian day number corresponding to perihelion. (Ans: j0 = 2,452,195.026, j1 = 686.9957857, j2 = −1.187E-7, JDper = 2,429,524.16494.) 6. Use equation 8.9.3, the appropriate values from table 8.5, and Kaph to com- pute the Julian day number corresponding to aphelion. (Ans: JDaph = 2,429,180.66705.) 7. Convert the Julian day numbers from the previous step to calendar dates. (Ans: Dateper = 09/17.664943/1939, Dateaph = 10/9.167046/1938.) 8. Convert the fractional days from the previous step to an integer day and UT. (Ans: Dayper = 17 at 15.958623h UT, Dayaph = 9 at 4.009100h UT.) 9. Convert the UT results to HMS format. (Ans: Perihelion on 9/17/1939 at 15h58m UT, Aphelion on 10/9/1938 at 4h01m UT.) Thus, Mars passed through perihelion on September 17, 1939, at 15h58m UT and through aphelion on October 9, 1938, at 4h01m UT. Meeus indicates that his equations are not highly accurate, with errors that may range from a few hours for Mars to a month or more for Saturn and the more distant planets. Hence, computing UT in step 7 just shown implies a

Our Solar System 249 greater accuracy than is warranted; therefore, one may wish to truncate the day in step 6 to an integer and stop. Using noninteger values of K in equation 8.9.3 or that do not end in 0.5 will produce meaningless results. Furthermore, greater error is incurred for years significantly distant from 2000. In practice, one would typically consult The Astronomical Almanac or similar resource for highly accurate perihelion and aphelion times. If one knows the times at which a planet will pass through perihelion and aphelion, it is natural to ask is how far away will the planet be from Earth and from the Sun at those times. Determining a planet’s distance from Earth when the planet is at perihelion or aphelion is straightforward, although the calculations are quite lengthy: • Use the equations in this section to determine the dates and times at which a planet will pass through perihelion and aphelion. • Once the dates and times are known for perihelion and aphelion, use section 8.6 to determine the position of the planet and Earth at those times. • Finally, use section 8.8 and the positions from the prior step to determine the distance from Earth to the planet. Be careful to note that these steps determine how far away a planet is from Earth when the planet is farthest from (aphelion) or closest to (perihelion) the Sun. These steps do not determine when a planet is farthest or closest to the Earth, which is an entirely different and more difficult problem that we will not address.11 A planet’s distance from the Sun at perihelion and aphelion can be deter- mined quite easily and directly from just the characteristics of the planet’s orbit. If ep is a planet’s orbital eccentricity and ap is the length of its orbital semi-major axis, then the planet’s distance from the Sun at perihelion and aphelion are Distper = ap(1 − ep) (8.9.4) Distaph = ap(1 + ep). (8.9.5) Since ap is given in AUs, the result of both of these equations will also be in AUs. Let us determine how far away Jupiter is from the Sun at perihelion and aphelion. The required steps are: 11. It is improper to say apogee and perigee here because the planets orbit the Sun, not Earth. Hence, apogee and perigee are meaningless in this context.

250 Chapter 8 1. Use equation 8.9.4 and the appropriate ep and ap values (from table 8.4 for this example) to compute the distance at perihelion. (Ans: ep = 0.0483927, ap = 5.202887 AUs, Distper = 4.951105 AUs.) 2. Use equation 8.9.5 to compute the distance at aphelion. (Ans: Distaph = 5.45669 AUs.) 3. Convert the distances to km or miles if desired. (Ans: Distper = 7.46067E08 km, or 4.6023E08 miles, Distaph = 8.1601E08 km, or 5.0704E08 miles.) Although we will not consider the problem of determining when a planet is closest to or farthest from Earth, equations 8.9.4 and 8.9.5 can be applied to determine how far away the Moon is from Earth at perigee and apogee. In fact, those equations can be used to determine the apside distances for any object that orbits another. To determine lunar perigee and apogee distances, the required steps are: 1. Obtain ep and ap from table 7.1. (Ans: ep = 0.0549, ap = 384,400 km.) 2. Use equation 8.9.4 to compute the Moon’s distance at perigee. Since ap is given in km, the answer will be in km. (Ans: Distper = 363,296 km, or 225,742 miles.) 3. Use equation 8.9.5 to compute the Moon’s distance at apogee. The answer is in km. (Ans: Distaph = 405,504 km, or 251,968 miles.) 8.10 Planet Phases Planets go through phases just as the Moon does. Instead of computing phases and age as we did for the Moon, in this section we will deal only with percent illumination because phase, age, and percent illumination are really the same thing. The percent illumination for a planet is K% = 100 (Rp + Dist)2 − Re2 , (8.10.1) 4Rp Dist where Rp and Re are the radius vector lengths computed in section 8.6, and Dist is the distance from Earth to the planet as computed in section 8.8. To illustrate, consider Saturn on January 3, 2016. The steps required to compute how much of Saturn is illuminated as seen from Earth are:

Our Solar System 251 1. Compute the radius vector length for the planet and for Earth for the given date at 0.0h UT (see section 8.6). (Ans: Rp = 10.00361 AUs, Re = 0.983294 AUs.) 2. Compute the distance from Earth to the planet (section 8.8). (Ans: Dist = 10.833730 AUs.) 3. Apply equation 8.10.1 to calculate percent illumination. (Ans: K% = 99.9%.) 8.11 Planetary Magnitude How bright an object appears to be is obviously related to how far away it is from the viewer. Thus, as the Earth and planets move in their respective orbits, how bright a planet appears to be depends upon how far away it is from Earth. We saw how to calculate how far away a planet is from Earth in section 8.8. We can use that information along with the following equations to approximate a planet’s visual magnitude: P A = 1 + cos(λp − Lp) (8.11.1) 2 (8.11.2) mV = Vp + 5 log10 R√p Dist , PA where λp is the planet’s geocentric ecliptic longitude, Lp is the planet’s helio- centric ecliptic longitude, Rp is the planet’s radius vector length, Dist is the distance from Earth to the planet, and Vp is the visual magnitude of the planet when it is 1 AU from the Earth (see tables 8.3 and 8.4). Note that PA is the planet’s phase angle while λp − Lp is its elongation. Both concepts were introduced in section 7.6. As an example, what was Saturn’s visual magnitude on January 3, 2016? 1. Compute the planet’s position for the given date at 0.0h UT to obtain Lp, λp, and Rp. (Ans: Lp = 248.411905◦, λp = 251.311100◦, Rp = 10.000361 AUs.) 2. Compute the distance from Earth to the planet. (Ans: Dist = 10.833730 AUs.) 3. Apply equation 8.11.1 to calculate the planet’s phase angle. (Ans: PA = 0.999360◦.) 4. Obtain Vp from the appropriate table (table 8.4 for this example) and apply equation 8.11.2 to calculate the planet’s visual magnitude. (Ans: Vp = −8.88, mV = 1.29.)

252 Chapter 8 It is easy to compare the relative brightness of 2 objects from their respec- tive visual magnitudes. If m1 and m2 are the magnitudes of 2 objects, then the brightness of the first object (m1) with respect to the second (m2) is b = 100.4(m2−m1). (8.11.3) For example, how much brighter does the Sun appear to be than a Full Moon? The visual magnitude of the Sun is −26.7 while the Moon’s visual magnitude is −12.7. Using these values for m1 and m2, we have b = 100.4(m2−m1) = 100.4(−12.7−(−26.7)) = 100.4(14) = 105.6 ≈ 398,107. Thus, the Sun appears to be nearly 400,000 times brighter than a Full Moon. On January 3, 2016, how much brighter was Saturn than the Andromeda Galaxy whose visual magnitude is about 3.4? In this case, we have b = 100.4(m2−m1) = 100.4(3.4−1.29) = 100.4(2.11) = 100.844 ≈ 6.98, so Saturn was about 7 times brighter in the sky at that time than the Andromeda Galaxy. Be careful to note that an object’s apparent magnitude (or visual magni- tude) is not the same as its absolute magnitude. The apparent magnitude of an object is simply a measure of how bright it appears in the sky. An object’s absolute magnitude M is how bright the object would appear to be if it were at a standard distance of 10 parsecs (approximately 32.64 light years) away from Earth. Astronomers use absolute magnitude because it measures the intrinsic brightness of an object and therefore does not depend on the object’s distance from Earth, as apparent magnitude does. In other words, apparent magnitude indicates only how bright an object appears to be while absolute magnitude indicates how bright the object really is. For example, the apparent magnitude of the Sun is −26.7, while its absolute magnitude is +4.7. This means that if the Sun were 10 parsecs away from Earth, its visual magnitude would appear to be +4.7. The Pole Star (Polaris) has an apparent magnitude of +1.97 but an absolute magnitude of −3.64. So, although Polaris appears much dimmer in the nighttime sky than the Sun (mV of −26.7 compared to +1.97), Polaris is actually a much brighter object than the Sun (M of +4.7 compared to −3.64). Polaris only appears dimmer because it is so much farther away. Apply- ing equation 8.11.3, in terms of apparent magnitude the Sun appears to be

Our Solar System 253 290 billion times brighter than Polaris, while in terms of absolute magnitude, Polaris is in reality about 2,000 times brighter than the Sun. Apparent and absolute magnitude are related by the equation M = mV + 5 − 5 log10(d), (8.11.4) where d is the distance in parsecs away from Earth. We will not consider absolute magnitude any further. 8.12 Miscellaneous Calculations It is perhaps fair to say that the Space Age began with the launch of Sputnik 1 in 1957. The progress made since then to explore the cosmos has been truly amazing. Although manned flights into space are still relatively infrequent, launching manned flights and space probes are so routine that a rocket launch rarely makes the evening news. With so many countries actively engaged in space exploration, and with serious discussions being held about manned flights to the Moon and Mars, space exploration may be even more dramatic in the very near future. Unfortunately, most of us will not have the oppor- tunity to participate in these exciting adventures. Even so, as a poor man’s substitute we can at least look at some of the interesting factors that have to be considered by those who are fortunate enough to be engaged in a space program. We will begin this section by determining what an astronaut or spacecraft would weigh on some distant celestial object and then calculate how long it takes light to reach us from that object to understand the delays inherent in radio transmissions. Just for fun, we will also calculate the length of a planet’s year and how fast a planet is moving. Finally, one of the most important calcu- lations we will make is to determine how fast a spacecraft must travel to escape a planet’s gravitational pull. 8.12.1 Weight on a Celestial Object Newton’s Law of Universal Gravitation provides a way to determine what an object will weigh on a celestial body. Newton’s law states that objects are attracted to each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, this is expressed as F = G m1m2 , (8.12.1) R2

254 Chapter 8 where G is the gravitational12 constant, m1 and m2 are the masses of the 2 objects, R is the distance between them, and F is the resulting force of attraction. To illustrate this important equation, let m be an astronaut’s mass, let me be Earth’s mass, and let mp be the mass of a celestial object such as a planet. Let re and rp be the radius of Earth and the celestial object, respectively. Also, let We and Wp be the weight of the astronaut on the Earth and celestial object, respectively. Substituting these values into equation 8.12.1, we have We = G mme (8.12.2) re2 and Wp = G mmp . (8.12.3) rp2 Now if we divide Wp by We, that will produce the ratio of the astronaut’s weight on the celestial object to the astronaut’s weight on Earth. That is, Wp = G mmp ÷ G mme = Gmmp re2 = mp re2 . (8.12.4) We rp2 re2 rp2 Gmme rp2 me Notice that this ratio depends only upon the size and mass of Earth and the celestial object. The gravitational constant and astronaut’s mass do not appear at all in the final equation. This should not be surprising because it says that the percentage by which 1 astronaut’s weight is affected by being on some celestial object is the same as the percentage by which another astronaut’s weight is affected, regardless of how much each individual astronaut weighs. Once the weight-factor ratio is known for a celestial object, one only has to multiply that ratio by an object’s weight on Earth to obtain how much the object weighs on that celestial object. For example, how much did Neil Armstrong’s spacesuit weigh on the Moon? (The weight of the multilayer space suit he wore was a hefty 180 pounds on Earth!) To answer this question, we apply equation 8.12.4 to find the ratio of weight on the Moon to weight on Earth. From table 8.3, we find that me = 1.0, mp = 0.0123, re = 6378.14 km, and rp = 1738.1 km. (Actually, table 8.3 gives masses relative to Earth, but this makes no difference in solving the problem because we are only interested in ratios.) Applying equation 8.12.4 to these 12. Although scientists have accurately determined the value of the gravitational constant G, we do not need to concern ourselves with its value.