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SCCC 115
Homework 2 - Due 18 Sept. 2003
Prof. Christina Lacey

P:2-6,3-2, 3-7     R & D:3-11

P 2-6: Halley's comet has a perihelion distance of 0.6 AU and an orbital period of 76 years. What is the aphelion distance from the Sun?

$\displaystyle \mbox{perihelion}$ $\textstyle =$ $\displaystyle a(1-e)=0.6 AU$  
$\displaystyle \mbox{aphelion}$ $\textstyle =$ $\displaystyle a(1+e)=?$  

First, using the period find the the semi-major axis, $a$. Then use the perihelion equation to solve for eccentricity, $e$. Finally, solve for the aphelion. Remember that $M_{\odot}$ is the $M_{sun}$.
$\displaystyle M_{total}$ $\textstyle =$ $\displaystyle M_{\odot}+M_{Halley's Comet} \simeq 1 M_{\odot}$  
$\displaystyle P(yrs)^2$ $\textstyle =$ $\displaystyle \frac{a (AU)^3}{M_{total} (M_\odot)}$  
$\displaystyle (a (AU))^3$ $\textstyle =$ $\displaystyle (P(yrs))^2  1 M_\odot$  
$\displaystyle a (AU)$ $\textstyle =$ $\displaystyle \left ( (P(yrs))^2 \right) ^{1/3}=\left ( (76 yr)^2\right)^{1/3}
=\left ( 5776\right)^{1/3}=17.9 AU$  
$\displaystyle \mbox{perihelion}$ $\textstyle =$ $\displaystyle a(1-e)=0.6 AU$  
$\displaystyle (1-e)$ $\textstyle =$ $\displaystyle \frac{0.6 AU}{a}$  
$\displaystyle e$ $\textstyle =$ $\displaystyle 1-\frac{0.6 AU}{a}=1- \frac{0.6 AU}{17.9 AU}=.97$  
$\displaystyle \mbox{aphelion}$ $\textstyle =$ $\displaystyle a(1+e)= 0.6 AU(1+.97)=35.2 AU$  

P 3-2

What is the wavelength ($\lambda$) of a 100 MHz (``FM 100'') radio signal?

$\displaystyle f$ $\textstyle =$ $\displaystyle 100 MHz =100\times 10^6 Hz = 100 \times 10^6 /s =1\times 10^8 /s$  
$\displaystyle c$ $\textstyle =$ $\displaystyle \lambda f \Rightarrow \lambda=\frac{c}{f}=\frac{3\times 10^8
\frac{m}{s}}{ 1\times 10^8 \frac{}{s}}= 3m$ (1)

P 3-7
Normal human body temperature is 37$^\circ$ C. What is the temperature in Kelvins? What is the peak wavelength emitted by a person with this temperature? In what part of the spectrum is this wavelength?

\begin{displaymath}
T(K)= 273K + T(C)= 273K + 37= 310 K= T_{human}
\end{displaymath} (2)


$\displaystyle \lambda_{peak}$ $\textstyle =$ $\displaystyle \frac{.0029 m}{T(K)}=\frac{.0029 m}{310}=9.4\times
10^{-6} m$  
$\displaystyle \lambda_{peak}$ $\textstyle =$ $\displaystyle 9.4\times
10^{-6} m = 9.4\times10^{-4} cm$  
  $\textstyle =$ $\displaystyle 9400 nm \quad (1  nm = 1 \times 10^{-9}m)$ (3)

The wavelength peaks in the infrared part of the spectrum.

R & D 3-11
What do radio waves, X-rays, and gamma rays have in common? How do they differ?

All the above are photons, part of the electromagnetic (EM) spectrum. Photons are massless particles that all travel at the speed of light in vacuum.

Each photon in different parts of the electromagnetic spectrum has a different frequency and corresponding wavelength ($\lambda f
=c$). Each photon has a specific energy: $E=h f=hc/\lambda$. Radio waves have the longest wavelengths, lowest frequencies, and smallest energies; gamma ray waves have the shortest wavelengths, highest frequencies, and highest energies.




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Christina Lacey 2005-09-21