Problems

#02

(a) $1\times 10{^3}$; $1\times 10{^-6}$; $1.001\times 10{^3}$; $1\times 10{^15}$; $1.23\times 10{^5}$; $4.56\times 10{^-4}$
(b) 3,160,000; 299,800; 0.0000000000667; 2
(c) $2.00001\times 10{^3}$; $3.33\times 10{^5}$; $9.47\times 10{^12}$

#07

Speed is distance over time. If the Moon orbits on a roughly circular path of radius 384,000 km, then the distance traveled in one full orbit is $d=2\pi r = 2.41\times 10^6$ km. The orbital period is 27.3 days (i.e., its sidereal period to move through 360 degrees). We want km/s, so 1 day has 24 hours, and each hour has 60 minutes, and each minute has 60 seconds, hence the orbital period is $P = 27.3\times 24 \times 60 \times 60 = 2.36\times 10^6$ seconds. The speed is then $v=d/P = 1.02$ km/s. (How accurate is this? Well, each day is not exactly 24 hours, but more like 23.9 hours, and the orbit of the Moon is not an exact circle, but is slightly elliptical. These are about 1% effects, so a typical speed of 1.0 km/s is probably reasonably accurate.)

#09

Use the formula $s = d \times \theta$, where $s$ is the baseline, $d$ is the distance, and $\theta$ is the parallax angle. Note that $\theta$ must be in radians, and that there are about 57.3 degrees in 1 radian.
(a) $1^\circ = 1/57.3 = 0.0175$ rad, so $d=s/\theta= 57.3 \times 1000 =5.73\times 10^4$ km.
(b) The angle is 60 times smaller, so the distance for a fixed baseline is 60 times farther, or $3.44\times 10^7$ km.
(c) Now the angle is 60 times smaller than in (b), or 3600 times smaller than in (a), and thus the distance to the object is correspondingly larger at $2.06\times 10^8$ km.

#10

Again use the geometrical formula $s = d \times \theta$, only now $s$ is the diameter of Venus and $\theta$ is its angular size. There are 206,265" in 1 radian, so $\theta = 55/206265 = 2.7\times 10^{-4}$ rad. The diameter must then be $(4.5\times 10^7 ) (2.7\times 10^{-4}) = 12,000$ km (which is slightly smaller than the Earth).

#13

This is a proportions problem. We are given the distances of the Sun ($d_1$) and the Moon ($d_2$). We seek the ratio of their sizes, $s_1$ and $s_2$. We know that the apparent angular size of the Sun and Moon are the same. This can only be true if

$$\theta_1 = \theta_2$$

$$\frac{s_1}{d_1} = \frac{s_2}{d_2}$$

which upon re-arrangement (i.e., cross multiply) gives

$$\frac{s_1}{s_2} = \frac{d_1}{d_2} = 390$$

(Why is the answer 390 instead of 391?)