ay16-CDeMeo

  Planets do not orbit stars. Rather, the star and planet both orbit the center of mass of the system. The stars therefore "wobble" due to the tug of gravity from their planet. This wobble allows astronomers to detect planets orbiting stars other than the sun using the doppler technique. The light coming from the planet and star shifts when the star's line of sight velocity changes. 1. (a) The center of mass of a planet star system in general is given by $x_{com }= \sum_{i} m_i x_i/  \sum_{i} m_i$ Set up the problem by drawing an x-axis with the star at $x = -́a_*$ with mass $M_*$, and the planet at $x = a_p$ and mass $m_p$. Also, set $x_{com}= 0$. How do $a_p$ and $a_*$ depend on the masses of the star and planet? $x_{com }= \sum_{i} m_i x_i/  \sum_{i} m_i$ $m_pv_p=M_*v_*$ $\frac{m_pv_p^2}{2}= \frac{m}{2}- \frac{M_*^2v_*^2}{m_p^2}=\frac{M_*^2v_*^2}{2m_p}$ $x_{com }= \sum_{i} m_i x_i/  \sum_{i} m_i$ $0=\frac{m_pa_p-M_*a_*}{m_p+M_*}$ $M_*a_*=m_pa_p$ $...

Introduction Accurate descriptions of low mass stars are difficult to obtain and the most accurate data comes from eclipsing binary stars. Measuring these systems can allow us to calculate the radii, masses, and separation of the stars in the system. This lab focuses on double-lined spectroscopic binary NSVS01031772.  The blue star is brighter than the yellow star in the binary system represented in Figure 1.   Figure 1b is the secondary eclipse and Figure 1c is the primary eclipse. The primary eclipse causes a greater decrease in brightness, represented by the larger peak in the light wave curve and the secondary eclipse represents the smaller peak in the light wave curve ( Figure 1d) . Observing these stars through the Clay telescope and using data from the paper  rvcurve.pdf (harvard.edu)  will allow us to determine the mass, separation, radius, and period of these stars.                      ...

The sun is a little less than halfway through its lifetime and in about 5 million years it will become a white dwarf. Currently it is on the main sequence when stars undergo nuclear fusion and convert Hydrogen into Helium. When there is no more hydrogen in the sun's core to be converted, the force of gravity will cause it to collapse until his is balanced out by some outward force due to pressure. One possible outward force is the force due to degeneracy pressure. White dwarfs are very dense so the fermions in the sun's core do not have space between them. They cannot occupy the same quantum state as their neighbors so they must constantly be in motion to avoid other fermions. The Broglie wavelength $\lambda $ is when the space between the particles is on the order of the particles themselves. $\lambda $ is related to the momentum p of the particle by the formula: $\lambda= \frac{h}{p}$ where h is the Planck constant, $h = 6.6 \times 10^{27} erg s$ (a) For a stellar core of a ...

The virial theorem, maybe better described as bound body energetics, requires that an object has an inward central force $r^{-a}$ such as a gravitational force or an electromagnetic force. The theorem also requires a constant moment of inertia on average so that the object maintains its rough shape over whatever timescale is relevant. Most structures can be virialized such as Earth, the sun, and the galaxy. For virialized structures we can say that Kinetic energy (K)$= \frac{-1}{2}$ Potential Energy (U) Worksheet 8.1 #1 For a planet of mass m orbiting a star of mass M, at a distance a, start with the Virial Theorem and derive Kepler’s Third Law of motion. Assume that m is much less than M  $U_g = \frac{-GmM}{a}$ $K=\frac {U_g}{2}=\frac{-GmM}{2a}$=$\frac {1}{2} mv_p^2$ $v_p= \frac{2 \pi a}{T}= (\frac{GM}{a})^{\frac{1}{2}}$ $\frac{4 \pi ^2 a^2}{T^2}= \frac{GM}{a}$ $T^2=\frac{4 \pi ^2a^3}{GM}$ #2 Consider a spherical distribution of particles, each with a mass $m_i$ and a total (coll...