Hints for Problem Set 1

Last revised 1999/08/25

HRW pp. 8-10 #1, 16, 39
Sheet # I and II

1. Don't spend a lot of time with this one.

16. In figure 1-8, a perpendicular to the side of the triangle which is 1 AU in length is very, very nearly 1pc in length. The 1-AU side of the triangle is also essentially the arc of a circle 1 pc in radius. So you can either use trigonometry or the fact that, if angles are measured in radians, the length of an arc of a circle of radius r is given by d = r theta. Alternatively, you might want to use the approximations sin (theta) = theta or tan (theta) = theta for small angles in radians. 2 pi radians = 360 degrees.

39. If you need an equation involving density, construct it using units. The constant that cannot be found by units will be 1.

When you pack spheres on top of each other to make a mass (or a solid), there are of course gaps between the spheres. The total volume of the gaps is small compared to the total volume of the spheres, however, if the spheres are properly packed. For the purpose of getting estimates of the size of an atom, you can neglect the volume of the gaps and assume that spheres pack as tightly as cubes would.

I. The parts lead you through the problem.

II. Warning: the answer to this problem is not unique. There is more than one expression for R that satisfies the condition that the units must work out correctly. Look for as general an expression as you can find.

Answers to Selected Practice Problems
Problem # Answer
6 (a) Circumference 4 x 104 km
(b) Surface area 5.10 x 108 km2
(c) Volume 1.08 x 1012 km3
8 (a) 1 yd2 = 9 ft 2
(b) 1 in2 = 6.45 cm2
(c) 1 mi2 = 2.59 km2
(d) 1 m3 = 106 cm3
19 (a) c = 0.984 ft/nsec
(b) c = 0.30 mm/psec
20 RHW answer: 52.6 minutes, 5.2% away from Fermi's estimate.
21 1 year = 3.16 x 107 sec which is very close to pi x 107 sec
24 RHW answers:
(a) Yes
(b) 8.6 sec
33 (a) A water molecule has a mass of 2.99 x 10 -26 kg
(b) There are about 4.68 x 1046 molecules of water in the earth's oceans