ANSWERS TO EVEN-NUMBERED PROBLEMS | (^633)
Chapter 1:
2. 3475 km; 4. 1.1 × 10^8 km; 6. about 1.2 seconds; 8. 75,000 years;
10. about 27
Chapter 2:
2. 2; 4. 631; 6. A is brighter than B by a factor of 170; 8. 66.5°; 113.5°
Chapter 3:
2. (a) full; (b) fi rst quarter; (c) waxing gibbous; (d) waxing crescent;
4. 29.5 days later on about March 30th; 27.3 days later on about March
24th; 6. 6850 arc seconds or about 1.9°; 8. (a) Th e moon won’t be full
until Oct. 17; (b) Th e moon will no longer be near the node of its orbit;
10. August 12, 2026 [July 10, 1972 + 3 × (6585^1 ⁄ 3 days)]. In order to get
August 12 instead of August 11, you must take into account the number
of leap days in the interval.
Chapter 4:
2. Retrograde motion: Jupiter, Saturn, Uranus, Neptune; Never seen as
crescents: Jupiter, Saturn, Uranus, Neptune; 4.^ √
27 = 5.2 years; 6. No. Th e
ratio in the diagram is about 1.5.
Chapter 5:
2. Th e force of gravity on the moon is about^1 ⁄ 6 the force of gravity on
Earth; 4. 7350 m/s; 6. 5070 s (1 hr and 25 min); 8. Th e cannonball would
move in an elliptical orbit with Earth’s center at one focus of the ellipse;
10. 6320 s (1 hr and 45 min)
Chapter 6:
2. 3m; 4. Either Keck telescope has a light-gathering power that is
1.56 million times greater than the human eye; 6. No, his resolving
power should have been about 5.8 seconds of arc at best; 8. 0.013 m
(1.3 cm or about 0.5 in.); 10. about 50 cm (From 400 km above, a
human is about 0.25 second of arc from shoulder to shoulder.)
Chapter 7:
2. 150 nm; 4. by a factor of 16; 6. 250 nm; 8. (a) B; (b) F; (c) M;
(d) K; 10. about 0.58 nm
Chapter 8:
2. 730 km; 4. 9 × 10^16 J; 6. 0.222 kg; 8. 400,000 years; 10. about
3.6 times
Chapter 19:
2. It will look 206,265^2 = about 4.3 × 10^10 times fainter, which is 26.6
magnitudes fainter; +22.6 mag; 4. about 2.3 half-lives, or 3.0 billion
years; 6. large amounts of methane and water ices; 8. about 1300
impacts per hour
Chapter 20:
2. about 17 percent; 4. about 8.2 × 10^7 yr (82 million yr); they have been
subducted; 6. 0.22 percent
Chapter 21:
2. Th e rate at which an object radiates energy and cools depends on its
surface area, proportional to its radius squared (r^2 ). Th e energy an object
contains as heat depends on its mass and therefore its volume, and that is
proportional to its radius cubed (r^3 ). So, the cooling time depends on the
amount of stored energy divided by the rate of cooling, which is
proportional to the radius cubed divided by the radius squared (r^3 /r^2 ),
that equals the radius (r). Th is means that the bigger an object is, the
longer it takes to cool; 4. No. Th eir angular width would be only about
0.5 arc seconds. Th ey would be easily visible, but are not seen, in photos
taken from orbit around the moon; 6. about 0.5 arc seconds (assuming
an astronaut seen from above is about 0.5 meter in diameter); no,
because that is smaller than the angular resolution of the human eye;
8. 10.0016 cm (at western elongation the planet will be moving away
from Earth, so the signal will be redshifted); 10. Mercury, Ve = 4250 m/s;
moon, Ve = 2370 m/s; Earth, Ve = 11,200 m/s
Chapter 22:
2. 33,400 km (39,500 km from the center of Venus); 4. 61 arc seconds;
6. 380 km; 8. 120 arc seconds (corresponding to 12 km)
Chapter 23:
2. 3.0 × 10^5 sec (about 35 Earth days); 4. 1.57 × 10^4 arc seconds (about
4.3°); 6. about 0.056 nm; 8. 5.2 m/s
Chapter 24:
2. 42 arc seconds; 4. 256 m/s; 6. 8.8 s; yes; 8. about 410 m/s (0.41 km/s);
10. 1.03 × 10^26 kg (17.2 Earth masses)
Chapter 25:
2. 1 × 10^9 (one billion); 4. about 330 m/s; no; 6. About 4.7 yr;
8. 2.5 × 10^5 s (about 2.9 days); 10. 300 m/s at r = 10,000 AU, about
100 m/s at r = 100,000 AU
Chapter 26:
2. 8.8 cm; 0.88 mm; 4. approximately 1.5 solar masses, spectral type F2;
6. 380 km; 8. No correct answer; perhaps somewhere between pessimistic
and optimistic estimates in Table 26-1, 2 × 10−5 and 10^7 , respectively
Answers to Even-Numbered Problems