Homeworks 2

Homework 2 due on Monday, 09/14/09

Problem 1

The star 61 Cygni has a parallax of 0.285 arcseconds. How far away is 61 Cygni in parsecs? What is its distance in light years? In centimeters?

d = 1/p" pc = 1/0.285 = 3.508 pc

3.508pc * 3.26 LY/1pc = 11.44 LY

3.508 pc * 3.08568025E18 cm/ 1 pc = 1.08E19 cm (by August Larson)

Problem 2

An electromagnetic wave traveling through space has a frequency of $5.0\times 10^{17}$ Hz. What is the wavelength of the radiation in centimeters? In °Angstroms? What is the name given to radiation with this wavelength? Where would you have to put a telescope to observe radiation with this wavelength from a star?

Because c = (frequency)(wavelength) , then the (wavelength) = c / (frequency) where c = 3*10^8 m/s so the wavelength = 6*10^-10 m or 6 * 10^-8 cm

or 6 Angstroms!

This wavelength and frequency corresponds with that of X-Rays

Earth's atmosphere absorbs most x-ray radiation so the telescopes have to be on a balloon at a very high altitude or in space (satellite) (by Paul Hellinger)

Problem 3

Procyon, the brightest star in the constellation of Canis Minor, is a double star consisting of a normal star plus a white dwarf star. The orbital period is 31.9 years and the two stars are separated by 13.3 AU. What is the sum of the masses of the two stars? The normal star has a mass of 1.70 M_⊙. What is the mass of the white dwarf?

Problem 4

The mass of the Moon is is $7.3483\times 10^{25}$ grams, which is 0.0123 times the mass of the Earth; and its radius is 1737.4 km, which is 0.272 times the radius of the Earth. If you weigh 150 pounds on the Earth, how much do you weigh on the moon.

$F_{e}=150~lbs*{\frac {4.448~N}{1~lbs}}=667.2~N$

$F_{e}={\frac {GmM_{e}}{r_{e}^{2}}}$

Solving for m,

$m={\frac {F_{e}r_{e}^{2}}{GM_{e}}}$

$={\frac {(667.2~N)(6.378136*10^{6}~m)^{2}}{(6.67*10^{-11}~N{\frac {m^{2}}{kg^{2}}})(5.9736*10^{24}~kg)}}$

$=68.121~kg$

$F_{m}={\frac {GmM_{m}}{r_{m}^{2}}}$

$={\frac {(6.67*10^{-11}~N{\frac {m^{2}}{kg^{2}}})(68.121~kg)(7.3483*10^{22}~kg)}{(1.7374*10^{6}~m)^{2}}}$

$=110.61~N$

$F_{m}=110.61~N*{\frac {1~lbs}{4.448~N}}$

$F_{m}=24.867~lbs$

where:

m is the mass of the person
F_e is the weight of the person on Earth
M_e is the mass of the Earth
r_e is the radius of the Earth
F_m is the weight of the person on the Moon
M_m is the mass of the Moon
r_m is the radius of the Moon

Problem 5

Two bodies of mass M1 and M2 orbit each other. Derive and discuss Kepler’s third law ( $P^{2}=const\cdot a^{3}$ ) in the general form by Newton. Why is the const=1 if we measure P and a in years and AU and for the solar system?

Newton's Universal Law of Gravitation : $F=G{\frac {m_{1}m_{2}}{r^{2}}}$ It's helpful to also write $F=ma'$ , where $a'={\frac {v^{2}}{r}}$ Set them equal, one r cancels, so we have $F=G{\frac {m_{1}m_{2}}{r}}=v^{2}$ Set $v={\frac {2\pi r}{P}}$ and now we can write $G{\frac {m_{1}m_{2}}{r}}={\frac {(2\pi r)^{2}}{P^{2}}}$ . Using basic algebra we can rearrange this, set $r=a$ , and get $P^{2}={\frac {4\pi ^{2}}{Gm_{1}m_{2}}}a^{3}$