Two-Body Orbits: Where's The Centre of Mass?

Second Authors: Nathan, Lauren

Introduction

Consider a problem of a planet orbiting a star.  It's easy to see from Newtonian gravitation that they exert a force on one another.  But then, how, according to Kepler's Third Law can we say the planet orbits the star alone?  The star cannot remain fixed while feeling a force.  In order for Newton's Laws to hold, we must say that they both orbit a mutual centre of mass.  Using conservation of momentum, we can determine how far each body truly is from this centre of mass.

Methods

In order to balance forces, we notice that the planet and the star must be at opposite ends of their orbits at all times as seen in the following picture (not to scale).  It follows from this that they have the same orbital period and the same angular velocity.

We know that linear momentum is equal around the centre of mass, such that:



Dividing through, we get the relationship



Rearranging and using the mean semimajor axis, a, we can see that

Conclusion
We have shown here that star does indeed orbit the centre of mass, just as the planet does.  However, looking at these equations carefully, we find that except for very massive planets, the semimajor axis of the star's orbit is roughly zero and the semimajor axis of the planet's orbit is roughly equal to the mean semimajor axis.  As a result, we find that we can in fact use the assumptions implicit in Kepler's Law.

The Death of a Star

Second Authors: Nathan, Lauren


Introduction
We know how stars are formed to a certain extent and that while they are on the main sequence, they are supported by hydrostatic equilibrium. However, when a star moves off the main sequence and can no longer support itself, what happens? We assume that at this point, the core of the sun has converted all of its mass to energy and is now undergoing gravitational collapse.

Methods
We know that the Sun generates energy throughout its lifetime at a rate of:

   

Assuming that the sun uses up the entire mass of its core (10% of a solar mass) as it undergoes fusion, and converts energy with 0.7% efficiency, we can determine the total energy it produces in its lifetime with

 

Dividing this number by the rate of energy production, we can determine the time it takes for the Sun to use all of its mass available for fusion. This time is

 

Now we know the core will collapse, but it won't collapse indefinitely.  We find that the core collapses to the point that the interparticle spacing is on the order of the De Broglie wavelength.  Since it is easy to see that electrons have greater momentum compared to protons of equal energy, electrons are the first to reach this critical density.  We can calculate this using the equations:

It is easy to tell that we have one molecule per cubic lambda.  So we have:

The actual value is 8 times this, for reasons I can't remember, but Nathan tells me Professor Johnson said the factor of 8 was okay to include in our calculations.  So multiplying this by the mass of a hydrogen atom and using T = the temperature of the sun's core, we get density

Which is more than twice the current maximum density of the sun's core.

Conclusions
We find that by converting 0.7% of the mass of the sun's core into energy, the sun's lifetime is roughly 10 billion years, which agrees with what scientists have predicted.  The density of the core after collapse will also be far greater than the current density of the sun's core, which is reasonable or it would not be able to support the sun post-collapse.

Star Formation: Timescale and Stability

Introduction
Star formation is governed by the collapse of a cloud of particles into a gravitationally bound sphere which we call a star.  The radius of the could at which this occurs is called the Jeans Length, where the gravitational force of the cloud overcomes the thermal energy causing it to expand.  Here we examine the time scale of such a collapse and also calculate the Jeans Length.

Methods
In order to determine the time it takes for this collapse to occur in terms of the mass and size of the cloud, we consider a cloud of mass M and a test particle a distance away from it.  We assume the cloud has a mass given by

where r is the length of the major axis for an elliptical orbit of eccentricity 1.  By assuming such a geometry for the free fall, we can initially approximate the orbit to a straight line with a  mass M at one end and our test particle at the other.  Since this is a free fall, we can also approximate the time t_ff to be half the orbital period we get from Kepler's 3rd law (a = 1/2 r)

 

Substituting our mass formula into this equation, we get

The implicit assumptions are that we can even call this half an orbit, as an eccentricity 1 orbit is parabolic and therefore not periodic, and that we can approximate this orbit to a straight line.  Now in order to find the Jeans Length, we equate this to the dynamical time, or the time it takes a sound wave to cross this distance.  Let's define this as

Equating the two, we get the radius at which the cloud will undergo gravitational collapse

For an isothermal gas of constant density, this length signifies the minimum radius at which it will continue to be a gas and not collapse into a much denser formation.  This is the Jeans Length to an order of magnitude.  The actual formula for the Jeans Length is

Conclusions
We have hear calculated the free fall time for star formation as well as the radius at which the gravitational force between interstellar dust particles takes over.  It is important to note that since the density is radius dependent, the Jeans Length is not constant for all star forming clouds, but varies even with the change of radius due to collapse and we have

If we consider a cloud that starts out at the Jeans Length for its particular conditions, by the time it reaches half this radius the Jeans Length has decreased by a factor of √8.  As a result, the initial Jeans Length may actually govern how far the cloud will collapse for a given mass and radius.

Hydrostatic Equilibrium and the Sun

Abstract
We would like to know how the sun is being "supported".  We assume that this mechanism is hydrostatic equilibrium, but to be sure we work through the derivation.

Introduction
We know that the sun is somehow being prevented from gravitational contraction.  Our theory is that it is supported by hydrostatic equilibrium, which means that the internal pressure provides an opposing support force.  We calculate the gravitational force on a mass shell, the pressure required to balance it, and then derive the force equation for hydrostatic equilibrium.

Methods and Results
We first assume the Sun to be a spherical gas cloud with density ρ(r).  We consider a differential mass shell of this sphere with radius r.  We recall that the volume of a sphere is ⁴/₃πr³ and that the differential volume is its derivative. Then we get a differential mass dM:

We know the equation for universal gravitation:

Here we let M be the total mass enclosed by the mass shell and m be the differential mass element.  As a result, we get the differential gravitational force to be:

We know that pressure is equal to force divided by area.  So we can say:

Now dividing by dr on both sides of the equation we arrive at the equation of hydrostatic equilibrium:

Conclusions

We have derived from simple physical laws that the equation for hydrostatic equilibrium is a plausible explanation for the way the sun is supported.  A quick search shows that we are indeed correct.  Hooray!

Stellar Properties From Afar (Problem 1)

Abstract
Considering the angular diameter of the sun and the astronomical unit, we can estimate the radius of the sun, the AU in solar diameters, and the mass of the sun using Kepler's 3rd law.

Methods
Applying basic trigonometric identities and taking the astronomical unit a to be the distances from us to the closest point of the sun to us (i.e., the centre of the circle we see from earth), we can see that:

Multiplying through by a we get a value for the radius of the sun.  It is clear from here that if we divide a by twice the solar radius we can easily determine the answer to the second part of our question.  Finally, we have Kepler's 3rd law:

Where P is the period of the earth and G = 6.7 x 10^-8 dyne-cm²/g².  From here we can solve for the mass of the sun.

Results
Solving the first equation using a = 1.5 x 10¹³ cm we get the radius of the sun equal to 6.545 x 10¹⁰ cm which is very close to the actual value of 6.955 x 10¹⁰7 s.  Dividing, we get 1 AU = 114.6 solar diameters.  Then, solving for the mass of the sun in Kepler's 3rd law with P = 3.154 x 10⁷ s, we have the mass of the sun equal to 2.007 x 10³³ g which is a surprisingly accurate number.

Surface Temperature of Planets

by Eric S. Mukherjee, Nathan Baskin, and I forget who else (sorry).


Abstract
In this problem we consider the how the temperature of the sun affects the temperature of the earth.  This is possible to estimate by assuming both the sun and the earth to behave like perfect blackbodies.

Introduction
Assuming the Earth has constant surface temperature and that it behaves like a blackbody, we can estimate the surface temperature using the energy emitted by the sun.  We also assume the sun to be a perfect blackbody.  Under these assumptions we can find the surface temperature of the Earth by knowing the temperature of the sun, the radius of the sun, the mass of the sun, the mass of the earth^*, and the radius of the earth.

Methods
We start with the equation for flux at the surface of a blackbody (σ is the Stefan-Boltzmann constant):

From this we derive the luminosity of the sun by multiplying through by the surface area:

Then the flux of the sun at the surface of the earth is (where a is the astronomical unit):

If we consider the area of the earth through which the flux passes, it is the circle of area π R_⊕².  Multiplying through by this quantity we get the power input to the earth from the sun.  We then realise that this is necessarily equal to the power output of the earth due to energy conservation which, at the surface of the earth, is equal to σT⁴π R_⊕².   Thus we have an equation of the form:

Using this equation with R_☉= 695,500 km and T_☉= 5778 K, we get T_⊕ = 279 K.

Conclusions
This temperature that we calculate is around 5.5°C which sounds reasonable for an earth without accounting for atmospheric greenhouse effects and allowing for the temperature at the poles.  The true average temperature of the earth is around 16°C but that is measured with the warming effect of the atmosphere.  The sun is not a perfect blackbody which also contributes to the difference between our calculation and the true value.

Acknowledgements
I'd like to thank the entire Ay 20 class and teachers for collective brainpower due to the fact that I can't remember who exactly worked on this problem and I'm sure we drew from the knowledge of many people in the room.  I'd also like to thank the superior computational power of Wolfram Alpha for bringing to my attention that there exponents matter when calculating ratios and that the temperature of the earth is most definitely not 1270 K.



*Note: It has been brought to my attention by Professor Johnson that the masses of the earth and sun do not actually factor into this calculation at all unless we need them to derive some of our other known constants.

Tau Ceti From Palomar

by: Eric S. Mukherjee, Nathan Baskin, Daniel Lo, John Pharo, and Mee Wong-U-Railertkun

I wrote up a mix of 2 and 3 from the work sheet on LST. Hope you don't mind

Abstract

Palomar is an important observatory for the Caltech astrophysics community. Because of this, we seek to constrain the visibility of the star Tau Ceti from the Palomar Observatory over the course of the year. We determine first whether its declination allows it to be seen and then when it can be observed. We also examine the elevation from month to month. We find that Tau Ceti is observable from Palomar during certain times of the year.

Introduction

Tau Ceti is a star located in the constellation Cetus at RA = 1:44 (25.0167 degrees) and Dec = -15.9375 degrees in the southern hemisphere of the celestial sphere. The Palomar Observatory is located at 33.358 degrees north and 116.864 degrees west. It is important to understand how and where Tau Ceti can be observed from Palomar in order to study the star. To do this, we determine the elevation of Tau Ceti when it is on the meridian passing through Palomar as well as the times that it is observable optically.

Methods

First we determine what the maximum elevation of Tau Ceti is from Palomar. The elevation of a star is the angle it forms with a line connecting the centre of the earth to the observer's local horizon. We calculate this using the equation:

Where a is elevation, delta is declination, phi is the local latitude, and H is the hour angle (the difference in LST from the meridian at the time of viewing). When the star is on the meridian, it is clear that H = 0. We then get:

Since one local sidereal day is a 360 degree revolution of the earth, or the time it takes for the meridian to re-align with the object in question, it is obvious that this is the elevation any time Tau Ceti is on the meridian and also that this is the maximum elevation of Tau Ceti as observed from Palomar.

Next we notice that if we set a = 0, or say the star is at the horizon, we can find the hour angle the time the star rises. This is given by:

Dividing this by 15, we get H = 6.72 hours, which we round down to 6.5 to account for the fact that stars at the horizon are invisible to the ground based observer.

Conclusions

Plotting the elevation as a function of time for the 20th of each month of the year, we get the following graph. We choose the 20th of each month for convenience's sake as the vernal equinox is on 20 March.

This shows, as expected from considering the problem, that at LST = 1:44, or when Tau Ceti is aligned with the meridian, the elevation is always 40.71. The graph also shows the approximate times at which Tau Ceti is aligned with the meridian in UT for planning of observations. Now plotting the time (UT) of Tau Ceti's maximum elevation against the passing months, we get the following graph:

As expected, the change in time is a linear relationship to the time of year. The error bars of this graph mark the total time that Tau Ceti is above the horizon. Regions plotted in red are the times before and after sunrise and sunset, respectively. These are the times that Tau Ceti is actually visible from Palomar because interference from the sun is absent. Predictably, this range is from July to January, with maximum visibility in the October-November region where it is dark the entire time that Tau Ceti is above the horizon. This agrees with observation data that indicates Cetus is best visible in late autumn.

Acknowledgments

We thank Professor Johnson and Jackie for providing us with this problem to solve. Also, Wikipedia for the image describing horizontal coordinates and Professor Harold Geller for his notes on calculating the altitude of a star. We also thank Microsoft Excel for cooperating long enough to produce these graphs.

CCAT vs. Keck: Angular Resolution and Telescope Size

by: Eric S. Mukherjee, Nathan Baskin

This is not the actual write up I was going to do for this week. I just thought this problem was a cool look at the many factors that go into designing a telescope and decided to write it up anyway. The other one will follow.

Abstract

It's very easy to assume that a bigger telescope is always the more precise one. The question is, is this assumption always true? In this post, we examine an order of magnitude solution of problem 3 of the problem set on optics to determine whether this is the case.

Introduction

The angular resolution of a telescope is the smallest angular separation at which the telescope can resolve two distinct objects. Any angle smaller than this registers as one larger object when viewed by the telescope in question. Caltech is building CCAT, a 25m telescope which observes in the 350-850 micron wavelength. We would like to know how the angular resolution compares to that of Keck, a 10m telescope observing in the near-infrared J-band at 1.25 microns.

Methods

We want to find angular resolution, which is given by the small angle approximation:

We then convert all of our known quantities into the same units. In this case, meters. As we can see, the angular resolution has a direct dependence on the wavelength, so we need only convert the minimum wavelength for CCAT:

Immediately we can see that these differ by two orders of magnitude while the diameters, 25 m and 10 m, respectively, are on the same order of magnitude. We can ignore the factor of 1.22 since it is shared by the two equations and is close to 1 and find that

Conclusions

We find that despite the greater diameter of CCAT, there is a difference of roughly a factor of 100 between the angular resolutions of the two telescopes. This was immediately obvious from the differences between the two wavelengths, since the diameters of the two telescopes have the same order of magnitude. Solving the equation for angular resolution we get precisely:

and

We find in the end that in order to determine a telescope's power we must refer to the wavelength at which it is observing as well as the diameter and that in order to have an equivalent angular resolution at the desired wavelength, CCAT would have to be a 2500 m telescope.