Problems

#01

The question is, what length subtends an angle of 1' for the different distances. Make use of $s = d \times \theta$, for $\theta$ the fixed angle, $d$ the different distances, and solve for $s$.
(a) $d=384,000$ km, and $\theta = (1/57.3) \times (1/60) = 1/3440$ rad, so $s=112$ km.
(b) $d=1.50\times 10^8$ km, so $s=43,600$ km.
(c) Saturn has an orbit of 9.54 AU. At closest approach it is on the same side of the Earth, about 9.54 AU from the Sun. However the Earth is 1 AU from the Sun in the direction of Saturn, so the separation is just 8.54 AU. The answer to (b) was for a distance of 1AU. The length $s$ is linear in the distance $d$, so if $d$ is increased by 8.54, so is $s$, giving $s=372,000$ km.

#10

The gravity of the Milky Way keeps the Sun in orbit. The expression for centripetal acceleration is

$$a_c = \frac{v^2}{r}$$

where for this problem $v$ is the Sun's speed, and $r$ is the Sun's orbital distance. We first must convert $r$ to m and $v$ to m/s, giving $r=26000 \times 3.15\times 10^7\times 2.998 \times 10^8 = 2.5\times 10^{20}$ m and $v=2.2\times 10^5$ m/s. The resultant acceleration is thus $a_c = 2.0\times 10^{-10}$ m/s$^2$.

That is pretty small, so we might expect the orbital period to be pretty long. We can use Kepler's third law to estimate the mass of the Galaxy interior to the Sun's orbit, with

$$\frac{r^3}{P^2} = \frac{GM}{4\pi^2}$$

First we need the orbital period, which for a circular path is just $P = 2\pi r/v = 7.1\times 10^{15}$ s. (In years, this is about 230 million years.)

Cross multiplying the above expression gives $M=(4\pi^2 r^3)/(GP^2)$. Plugging in all the numbers gives $M = 1.8 \times 10^{41}$ kg, or about 100 billion times the mass of the Sun.

#12

Recall that the force of gravity is

$$g = \frac{GM}{r^2}$$

where for this problem $M$ is the Earth's mass (at $5.67 \times 10^{24}$ kg) and $r$ is the distance of an object from the center of the Earth. The altitudes given are distances above the ground, and the Earth's surface already lies a distance of 6400 km or $6.4\times 10^6$ m from the center of the Earth. One has only to add the radius of the Earth to the altitude to get the correct value of $r$. The answer is expected to be in m/s$^2$ (or MKS units), so keep distances in meters, and use $G=6.67\times 10^{-11}$ m$^3$/s$^2$/kg.
(a) $r=6400 + 100 = 6500$ km giving $g=8.95$ m/s$^2$.
(b) $r=6400 + 1000 = 7400$ km giving $g=6.91$ m/s$^2$.
(c) $r=6400 + 10000 = 16400$ km giving $g=1.41$ m/s$^2$.

#14

My weight is about 160 lbs. At the Earth's surface, 2.2 lbs is about 1 kg, so my mass is 73 kg. (This conversion factor works only at the Earth's surface!) The force of gravity is $F=Gm_1m_2/r^2$, where in this case $m_1$ will be the Earth mass, $m_2$ my mass, and $r$ the radius of the Earth. Using numbers from the previous problem, we could crank all of this out. However, we also know that $F=mg$, and $g=9.80$ m/s$^2$. Taking $m$ to be my mass, I get a force of gravitational attraction between the Earth and myself (or vice versa) of 720 N (this is a MKS unit). At 4.45 N per pound, the force can be expressed as $F=720/4.45 = 160$ lbs. That is my weight! Thus, this force is normally called ``weight''.