SCCC 115
Homework 3 - Due 22 Sept. 2005
Prof. Christina Lacey
P: 2-10, 3-5, 3-6, 3-11, 3-12, 4-4     R & D: 4-4

P 2-10:
The Sun moves in a roughly circular orbit around the center of the Milky Way Galaxy, at a distance of 26,000 ly. The orbital speed is roughly 220 km.s. Calculate the Sun's orbital period and centripetal acceleration. Calculate the Mass of the Galaxy.

Recognize that the distance from the Sun to the center of the Galaxy is the same as the radius.

$$R = 26,000 ly =26,000  ly\frac {9.46 \times 10^{12} km}{1 ly} \frac{1000 m}{1 km} =2.5\times 10^{20} m$$

Remember that velocity=distance/time. In the case of the Sun's orbit, the period is the time it takes th Sun to complete one full revolution about the center of the Galaxy. In this case the distance traveled is the circumference of the orbit and time is the period.

$$C = circumference=2\pi R$$

$$P = time=distance/velocity=\frac{d}{v}=\frac{C}{v}=\frac{2\pi R}{v}$$

$$P = \frac{2\pi ~2.5\times 10^{20} m}{220 \frac {km}{s} ~ \frac{1000~m}{1~km}}= 7.0\times 10^{15} s$$

$$P = 7.0\times 10^{15} s \frac{1 min}{60 s}\frac{1 hr}{60 min}\frac{1 day}{24 hr} \frac {1 yr}{365 days}= 2.2\times10^8 yrs$$ (1)

Now I can use either Kepler's 3rd Law or Newton's formulation to find the total mass of the Galaxy. I will use Kepler's 3rd Law here, remember to convert R to AU and use P in years. In this case $M_{Total}= M_{galaxy}+M_{Sun}\simeq M_{galaxy}$:

Newton's method:

$$v = \sqrt{GM/r} \rightarrow M=v^2r/G$$ (2)

Kepler's Method:

$$P^2 (yr) = \frac{a^3 (AU)}{M_{Total}} \Rightarrow M_{galaxy}=\frac{P^2 (yr)}{a^3 (AU)}$$

$$M_{galaxy} = \frac{(2.2\times10^8 yrs)^2}{\left (26,000 ly \frac{206,000 AU}{3.3 ly}\right )^3}$$

$$M_{galaxy} = 8.8\times10^{10}  M_{\odot} \frac{1.99 \times 10^{30}  kg}{1 M_{\odot}}=1.8\times10^{41} kg$$ (3)

Centripetal acceleration of the Sun:

$$a = \frac{v^2}{R}=\frac {(2.2\times10^5 m/s)^2}{2.5\times 10^{20} m}=2.0\times10^{-10} m/s^2=2.0\times10^{-13} km/s^2$$ (4)

P 3-5:
What is the wavelength of a wave with a frequency of 3.2 GHz?

$$c=\lambda f \rightarrow \lambda=\frac{c}{f}=\frac{3\times10^... ...mes10^9~ Hz}{1~GHz} \frac{1/s}{1Hz}}= 0.09m= 9.4 cm \nonumber$$

$\lambda=0.09$ m is a radio wave.

P 3-6:
Blackbody emission peaks at $\lambda_A=200$ nm (UV) for object A and $\lambda_B=650$ nm (red visible) for object B. Which object is hotter ad by how many times? How many more times energy/area, $F$, does the hotter body radiate per second?

Wien's Law:

$$\lambda=\frac{.0029 m}{T(K)} \rightarrow T(K)=\frac{.0029 m}{\lambda}$$ (5)

So let's look at the ratio of $T_A/T_B$:

$$\frac{T_A}{T_B}=\frac{\frac{.0029 m}{200 nm}}{\frac{.0029 m}{650 nm}} =\frac{650}{200}=3.3$$ (6)

So object A is 3.3 times hotter than object B.

Or, you could have calculated the temperaures of the objects:

$$T_A(K)=\frac{0.0029~m}{\lambda}=\frac{0.0029 m}{200\times 10^{-9}m}= 1.5\times 10^4 K$$ (7)

$$T_B(K)=\frac{0.0029~m}{\lambda}=\frac{0.0029 m}{650\times 10^{-9}m}= 4.5\times 10^3 K$$ (8)

and then determined the ratio of temperatures.

Now use Stefan's law, which gives energy per area per second:

$$F=\sigma T^4$$ (9)

Again look at the ratio of $F_A to F_B$:

$$\frac{F_A}{F_B}=\frac{\sigma T_A^4}{\sigma T_B^4}=\frac{T_A... ...\right )^4= \left ( \frac{3.3 T_B}{T_B}\right )^4=(3.3)^4=110$$ (10)

So object A emits 110 times more energy/(m$^2$ s) than object B.

Or if you calculated F:

$$F_A=\sigma~T^4= \sigma=5.67\times 10^{-8} \frac{W}{m^2 K^4}(1.5\times 10^4 K)^4= 2.5 \times 10^9 W/m^2$$ (11)

$$F_B=\sigma~T^4= \sigma=5.67\times 10^{-8} \frac{W}{m^2 K^4}(4.5\times 10^3 K)^4 = 2.2 \times 10^7 W/m^2$$ (12)

P 3-11:
According to the Stefan-Boltzmann Law, how much energy is radiated into space per unit time by each square meter(m$^2$, of the Sun's surface? If the radius of the Sun is 696,000 km, what is the total power output of the Sun?

Stefan-Boltzmann Law:

$$F=\sigma T^4$$ (13)

where $\sigma=5.67\times 10^{-8} \frac{W}{m^2 K^4}$. Keep in mind that $1 W =1 J/s=1 kg m^2/s^3$. The temperature of the surface of the Sun is 6000 K (pg 75 in textbook).

$$F = \sigma T^4=\sigma=5.67\times 10^{-8} \frac{W}{m^2 K^4} (6000 K)^4= 7.3\times10^7 \frac{W}{m^2}$$ (14)

The luminosity (total power output is:

$$L = (Surface  area) F= 4\pi R^2 F$$

$$= 4\pi (6.96\times 10^8 m)^2 (7.3\times10^7 \frac{J}{m^2 s})=4.4 \times 10^{26} J/s=4.4 \times 10^{26} W$$

P 3-12:
Radiation from the nearby star Alpha Centauri is observed to be reduced by a factor of .999933. What is the radial velocity of Alpha Centauri relative to the Sun?

$$\lambda_{observed} = 0.999933 \lambda_{ true}$$

$$\frac{\lambda_{observed}}{\lambda_{true}} = 1+\frac{v_{recession}}{c}$$

$$v_{recession} = \left (\frac{\lambda_{observed}}{\lambda_{true}}-1 \right ) c= \left (\frac{0.999933  \lambda_{ true}}{\lambda_{true}}-1\right ) c$$

$$= \left ( 0.999933 -1\right ) 3\times 10^8  m/s= -20,100  m/s=-20 km/s$$

The negative $v_{recession}$ means that the star is moving toward us, hence the observed wavelength is reduced= blue-shifted.

P 4-4:
How many times more energy has a 1 nm gamma ray photon than a 10MHz radio photon?

$$E = hf \quad \quad h=6.63 \times10^{-34} Js$$

Let $gamma=\gamma$ then, remember that $f_\gamma=\frac{c}{\lambda_\gamma}$:

$$E_\gamma = hf_\gamma$$

$$E_{radio} = hf_{radio}$$

$$\frac{E_\gamma}{E_{radio}} = \frac{hf_\gamma}{hf_{radio}} =\frac{f_\gamma}{f_{radio}} =\frac{c}{\lambda_{\gamma} f_{radio}}$$

$$= \frac{3\times10^8m/s}{1 nm \frac{1\times10^{-9}m}{1  nm} 10 MHz \frac{1\times10^6 Hz}{1 MHz} \frac{1/s}{1 Hz}}=3\times10^{10}$$ (15)

Gamma rays have $3\times10^{10}$ times more energy than radio waves.

R& D 4-4
Why are gamma rays generally more harmful to life forms, but radio waves are generally harmless?

Gamma rays have much more energy (See problem 4-4) than radio waves. Gamma rays have enough energy to harm living tissue. In passing through cells, gamma rays can break-up important molecules, such as DNA, and also ionize atoms that causes further cellular damage.

Radio waves pass through living tissue without interacting or are reflected, which causes no tissue damage to organisms.