Tuesday August 28
The three-semester, 10 credit hour 200-level Astronomy sequence (AST 201, 215, 225) is essentially a recapitulation of AST 101, but with a lot more detail, and a lot more physics. AST 201 covers the same concepts that we talk about in the first ~2 weeks of AST 101.
This course, AST 215, picks up from there. We will cover Newtonian gravity and celestial mechanics, the basic properties of light and the interaction of light and matter, telescopic optics, and the properties of stars (roughly the next month of AST 101).
This is as much a course about mind-set as it is a course about content. Astrophysics requires a great deal of mental flexibility, and the approach of a generalist. One needs to be familiar with essentially all of physics to do astrophysics, and one needs to develop techniques of viewing problems in the simplest possible way that still leads to a useful result.
One means toward that goal will be the use of class discussion problems. At the beginning of each main topic, I will distribute a problem to the class. I expect you to work through the problem (with hints as needed) over the duration of the topic, and discuss the work in class.
Astronomy is an Observational science. We generally cannot "do experiments" in the traditional sense ("Let's make another Sun out of pure helium to see how that's different from the one we've already got."). All we can do is come up with clever ways of looking, and of interpreting what we see. Nearly everything we know about the Universe outside the Solar System we learn by studying the light we receive. This distinguishes astronomy from other sciences, and has a lot to do with my above comments about mind-set.
A natural co-ordinate system for the sky is that defined by the celestial equator and celestial poles. One can use these reference points to construct a positional system that corresponds to the terrestrial longitude and latitude. The celestial coordinates are Right Ascension and Declination.
Another natural co-ordinate system is defined by (from a ground-based point of view) the path the Sun takes through the sky over the course of a year. This path is called The Ecliptic. From a proper heliocentric view-point, the ecliptic is the plane of the Earth's orbit around the Sun.
The ecliptic is inclined by 23.5 degrees with respect to the celestial equator. Again, from a proper heliocentric view-point, this arises because the Earth's rotation axis is tilted by 23.5 degrees away from the perpendicular to the Ecliptic plane. This is the origin of terrestrial seasons.
The planets move with respect to the background stars, but they are all confined very close to the ecliptic. In other words, all the planets orbit the Sun in very close to the same plane. Further, they all orbit the Sun in the same direction. Even further, almost all of them rotate about their spin axes in the same sense as they revolve around the Sun (Venus and Uranus are the exceptions to this last point). This is a very powerful argument that the Solar System formed coherantly from the collapse of some initial gas cloud. If planets were interstellar tramps that stars captured long after their formation, there would be no strongly preferred orbital plane. What we have learned in the last decade about planets around other stars is consistent with the general notion that planetary systems are by-products of star formation.
The apparent motions of the planets in the sky are quite complex. This observed complexity, when coupled with the philosophical prediliction to place the Earth at the center of the Universe, led to the development of a very complex system to describe planetary motion. This system is best known as the Ptolemaic model (although it's basics predate Ptolemy by several centuries).
The geocentric model required constant refinement in order to continue to match the actual observed positions of the planets. The last major refinement dates to the 14th century. The need for such refinements can be seen by considering the consequence of having a watch that runs 10 seconds slow over the course of a day. If one does nothing, then the watch will be more than an hour slow after a year.
The modern heliocentric model was first proposed by Nicolaus Copernicus at the end of the 15th century. It is conceptually MUCH simpler than the geocentric model. But Copernicus retained the Greek notion of perfect circular motion, and so his model did not represent an improvement in accuracy over the Ptolemaic model.
It is essentially impossible to disprove the geocentric model with 16th century technology. However, it became simple to do so with the invention of the telescope. The essential proof that the geocentric model could not be correct came with from Galileo's observations of the phase cycle of Venus.
Venus is observed to always be within about 43 degrees of the Sun. Given this, a geocentric model would insist that the observed phase of Venus always be a crescent (either waxing or waning). However, Galileo observed Venus to demonstrate essentially the full range of phases. He also noted that the angular size of Venus varied a great deal, reaching a minimum when Venus was nearly full, and reaching a maximum when Venus was nearly new. These observations follow naturally if Venus is orbiting the Sun, but cannot follow if Venus and the Sun orbit the Earth.
While Galileo's observations clearly rule out the geocentric model, this point was generally ignored. There are a number of sociological reasons for this. Geocentrism was Church doctrine. Also, people are generally disinclined to accept wholesale revisions of their worldview based on the results of some newfangled gizmo. Thus the work that led to the change in the generally accepted worldview had to be traditional astronomical observation. That is, work on the positions of the planets.
The essential positional data were compiled by Tycho Brahe. Brahe was the last great classical (pre-telescopic) astronomer. He spent his life observing the positions of the planets to a precision surpassing anything done before. But while he was a great observer, he was not a great mathematician. But he did have the good forture to hire a great mathematician to work for him. And that mathematician, Johannes Kepler, inherited Brahe's notebooks when Brahe died.
Kepler analysed Brahe's data, and came up with the following rules of Planetary motion, now called Kepler's Laws:
Where "P" is the period and "a" is the semi-major axis. If the period is measured in years and the semi-major axis in Astronomical Units, the relationship can be written as an equality.
The Astronomical Unit is defined as the mean distance between the Earth and the Sun. This is a very useful measuring stick, as it is possible to determine the relative distances of the planets from the Sun in terms of the Astronomical Unit (AU) by geometric considerations. Thus we knew how many AU Venus was from the Sun several centuries before we knew how many kilometers there are in an AU.
The geometric derivation of the distances to the inner planets in AU is relatively straightforward. The determination for the outer planets is more complex, and is discussed in the text. These derivations make use of the existence of several special configurations of planetary positions:
We observe the motions of the planets through the sky, and we can measure how long a given planet takes to move once around the sky. This is known as the Synodic Period of the planet. Because the Earth is *also* moving, this synodic period is not the same as the actual orbital period (or the Sidereal Period). It is possible to determine the actual period from the synodic period. The essential algebra is as follows:
For an inner planet, P2 = 1 year, and P1 follows. For an outer planet, P1 = 1 year, and P2 follows.
The fundamental unit of time we have is the Day. But how is that really defined? There are two reasonable ways to approach this:
The rotation period of the Earth is determined by measuring the interval between meridian passages of the stars. This turns out to be about 4 minutes shorter than a mean Solar day. The difference arises because the Earth is revolving around the Sun as it rotates. So if one starts a timer as the Sun crosses the local meridean (local Noon), and waits for one rotation period of the Earth (a sidereal day), one will find that the Sun hasn't quite made it back to the meridian again. The extra 4 minutes arises because the Earth has moved about 1 degree in it's orbit around the Sun since the last Solar noon (Recall that it takes 24 hour for the Earth to rotate 360 degrees. So it rotates 1 degree in about 4 minutes).
The cycle of the Moon's phases takes just under a month to complete (29.5 days. This is called a synodic month). If we measure how long the Moon takes to cycle once through the sky with respect to the stars, that tells us the Moon's orbital period around the Earth. This turns out to be about 27.3 days (or a sidereal month). The synodic and sideral months differ for the same basic reason that the solar and sidereal days differ. In the time it takes the Moon to orbit the Earth, the Earth has moved through part of its orbit around the Sun. Thus the phase cycle takes a bit longer than the orbital period.
We always see the same face of the Moon (not *quite* true - see below). This means that the Moon's rotation period is the same as it's orbital period around the Earth. This is not a coincidence. It is an example of Tidal Phase Locking.
The Moon orbits the Earth due to gravity. This causes tides. The essential point about gravity is that it obeys an inverse-square law:
This means that the oceans on the side of the Earth closest to the Moon feel a larger force of gravity due to the Moon than does the center of the Earth. And the center of the Earth feels a larger force than does the ocean on the far side of the Earth from the Moon. The result of this is that two tidal bulges are raised on the Earth. One on the side facing the Moon, and one on the side facing away from the Moon.
Tides also occur in solid rock, but they are much less dramatic than those occuring in oceans. The solid body tides on the Earth only produce tidal bulges with amplitudes of about 1cm. Still, this has a dramatic long term effect. The Earth spins much more rapidly than the orbital period of the moon (one day versus 27.3 days), and thus the actual tidal bulge on the Earth is always ahead of the Earth--Moon line. This causes a tidal torque: The Moon is dragging the Earth, slowing its rotation rate down. This is born out by the fossil record: The day was only 22 hours long 400 Myr ago. Eventually, this will cause the Earth's rotation rate to slow down to the orbital period of the moon.
The Earth is much more massive than the Moon, and gravity is a symmetric force, so the Earth is much more effective at slowing down the Moon than the Moon is at slowing down the Earth. In fact, the Earth has already slowed down the rotation period of the Moon to match its orbital period. This is why we always see the same face of the moon. Tidal Locking is a common phenomenon, both for moons around planets, and for stars in binary star systems.
The Earth is slowing due to the tidal torque applied by the moon. Than angular momentum has to go someplace. It goes to increase the orbital angular momentum of the Moon. The net result of that follows from some simple algebra:
If the Moon' orbital angular momentum is increasing, then the product VR must also be increasing. From this, we see that V *decreases* and R increases.
Thursday August 30
One can derive an expression for the velocity required to place an object into a circular orbit at some distance away from some mass. The calculation follows from equating the centripital and gravitational accelerations:
One can also derive an expression for the velocity required to escape the gravitational influence of a given mass (when starting from some distance away from the center of mass). This follows from considering the total energy of the system. For an object to escape from a mass, the total energy must be zero (or negative). This is why we choose to define the zero point of potential energy the way we do. The algebra looks like this:
So the circular and escape velocities differ by only a factor of 1.4. Now I am going to pull an expression out of my hat. I will derive it when we talk about binary stars. This is the expression for the velocity of a mass in an elliptical orbit:
Now, if an object is in an elliptical orbit about some mass, it is bound. Thus the total energy of the system is negative:
Recall that if one adds energy, the modulus of this expression must *decrease* and thus the semi-major axis increases. If a increases, then v must decrease (as we have already shown).
Notice that the expression we arrived at does not depend on the eccentricity of the orbit. All orbits of the same semi-major axis have the same total energy. This seems odd unless one recalls that the center of mass is not at the geometric center of the orbit *unless* the orbit is circular.
The thing that distinguishes orbits of a given a but different eccentricities is the orbital angular momentum. The angular momentum of an elliptical orbit can be written as follows:
Where theta is the angle between the radial and velocity vectors. For a circular orbit, theta = 90 degrees. Note also that theta=90 degrees at the extrema of the ellipse.
Launch Windows
If one wants to send a space probe from Earth to an inner planet (Venus, in the example worked in class), one must remove energy from the probe to do so. The minimum energy path to go from Earth to Venus is an orbit that has Earth's orbit at aphelion and Venus's orbit at perihelion, with the Sun at a focus of the ellipse. If one thinks about the geometry for a moment, this gives an orbit with a semi-major axis of
If one then uses the result quoted last week for the velocity of an elliptical orbit:
and substitutes in the appropriate numerical values, one comes to the result that a probe put into such an orbit would need an initial orbital velocity (starting from Earth!) of about 27 km/sec. But Earth's orbital speed is about 30 km/sec. So one actually has to launch the probe "backward" to get it to fall into the desired orbit.
Something else to consider about this sort of problem is that you need to time your launch carefully. The problem isn't getting to the Venus orbit. It's getting there at the spot where Venus happens to be when you get there. This is the reason why space missions have launch windows. Because the energetically favorable means of getting from one place to another only actually work once in a given synodic period.
Gravitational Slingshots
One often hears about space probes using gravity boosts or gravity assists or such. These are gravitational analogs of the perfect elastic collisions one learns about in Physics 221. The story there is that if one collides a billiard ball with a stone wall, the billiard ball recoils with a velocity equal and opposite to its incoming velocity. If, instead of a brick wall, the target is a freight train full of bricks travelling at some velocity u, then a billiard ball with an initial velocity v will collide, and recoil with an final velocity v + 2u.
The situation with gravitating bodies is very similar. The main distinction is in what is meant by a "collision". An actual physical crash isn't what we are talking about. Instead, a gravitational collision is an interaction in which one or both bodies exchange a significant amount of gravitational energy.
Planets are much more massive than spaceprobes, so the planets act in the role of the freight train full of bricks. If one can send a probe past a planet, one can gain kinetic energy from the planet by such means. This turns out to be a MUCH more efficient way to send probes to the outer Solar System than direct rocket burns. The Cassini mission to Saturn used both Venus and the Earth as gravity boosts on its way out. And the Voyager 2 probe used gravity boosts at each of the outer planets on its way through the outer Solar System.
Consider a circular orbit. Beginning with the general expression for gravitational energy, one can show the following:
This turns out to be an entirely general result for any bound system in gravitational equilibrium. As such it is an absurdly useful tool. Given that, it is well worth the time to plow through the general derivation. I summarize the derivation below:
Now, we go back to the original statement, and use this first bit of insight:
Now let's consider what we have on the right-hand side:
Where I've left out some steps that involve rearranging the radial and force vector terms. Now consider the following:
So here is what we've got:
Recall that we want to apply this to equilibrium systems. Given that, we want to take the time-average of the above. But for a system in equilibrium the moment of inertial is constant. Thus all of the derivatives of the moment of inertia are zero. As a result we have the following:
This result is used in many, many astrophysical contexts.
Tuesday September 4
Consider a simple two-body system in which the masses are bound to one another, and on which there are no net external forces. For such a system, the net acceleration must be zero, and the total momentum must be conserved. If one defines a simple 3-D coordinate system, in which the two masses are at postions r1 and r2, one can use momentum conservation to make the following statement:
Now make the following definition for the radial vector between the masses, and solve for the two radial vectors r1 and r2 as follows:
Finally, we define the reduced mass as follows:
The reason for this definition is that it allows us to recast the full two-body problem (m1 and m2) into a one-body problem with an "object" with a mass equal to the reduced mass orbits an "object" with mass M=m1+m2. Some of the resulting algebra is summarized below:
Isaac Newton first examined inertial or unaccelerated motion. This led him to what are now called Newton's Laws of Motion:
And the second law can be written algebraically as
From here, Newton went on to consider the behavior of falling objects. As falling objects accelerate downward, they MUST be acted on by a force. But, as noted by Galileo, the rate of acceleration does not depend on the mass of the falling object. This is inconsistent with Newton's second law UNLESS the force of gravity depends on the mass of the falling object.
Newton realized that his third law of motion (the reaction law) meant that any two massive bodies must exert equal gravitational forces on each other. So the force of gravity between any two objects has to depend on the masses of both objects.
Newton asked the question "Is this force on the Moon the same as the force of gravity we feel on the surface of the Earth?"
Now, the Moon is in a roughly circular orbit around the Earth. Circular motion also requires a force. This is because any change in velocity requires an acceleration. And velocity isn't the same as speed. Two objects that are moving at the same speed (say two cars on the highway, both moving 50 mph), but in opposite directions (one car moving north, and the other south) do NOT have the same velocity (in this example the velocities differ by 100 mph). To change the direction that an object is moving in requires a force, even if the object's SPEED remains constant.
Newton reasoned that the force should follow an Inverse- Square Law. This is just a reflection of the geometry of a sphere. The surface area of a sphere increases as the square of the radius. Take the example of a light source of some fixed luminosity. If you get twice as far away from it, it appears only a quarter as bright.
Newton reasoned that gravity should behave in this way also. Taken together, this led Newton to the following formulation for the law of gravity:
The m1, M2 are the two masses involved, the R1,2 is the seperation between the masses. The G is a constant (the Gravitational Constant).
Newton knew this law was consistent with his Laws of Motion, and with the observed behavior of falling objects near the Earth's surface (local gravity). Now we return to his question: "Is the force that keeps the moon in a constantly accelerating (nearly circular) orbit around the Earth the same as the force of gravity we feel on the surface of the Earth?"
Among the consequences of Newton's Laws of Motion are rules that govern objects in circular motion (and you'll need to take a physics class if you want to see the derivation of this). So he knew that any object in circular motion must have an acceleration that depends on its speed and the radius of the circle involved:
Newton knew the distance to the Moon (this was known from parallax measurements), and he knew the moon's orbital speed (from its distance and its orbital period). So he knew the acceleration the Moon HAS to be experiencing to be in the orbit it is in:
Newton knew the size of the Earth, and he knew that the distance between the Earth and the Moon was about 60 Earth radii. Given this, he calculated
Or about 1/3600 of the acceleration due to gravity at the Earth's surface. The acceleration of gravity on the Earth's surface is about 9.8 m/sec/sec. This means the acceleration on the Moon due to the Earths gravity should be about 2.7x10^-3 m/sec/sec. This is exactly the acceleration that the Moon must be experiencing in order to stay in a circular orbit around the Earth.
It is possible to derive Kepler's Laws of Planetary Motion from Newton's Laws of Motion and Gravity. The text presents a detailed, vector-form derivation of all three of Kepler's laws. I will not repeat the technical derivation of the first law. Instead, I will assume that orbits are conic sections, and make a statistical appeal for the existence of elliptical instead of circular orbits. Why? It isn't because circular orbits are not allowed. It's because a circular orbit is a special case. For any circular orbit, there are an infinite number of elliptical orbits with the same starting point. So if you pick one, at random, it will be elliptical.
I will now skip ahead to Kepler's third law. And I will make the further simplification that planetary orbits are circular. (They are pretty close to circular, and the algebra is easier. One gets the same result if one wades through the elliptical orbit algebra). The approach is similar to that discussed above for the orbit of the moon. One simply equates the gravitational and centrifugal forces of an object in a circular orbit, and turns the algebra crank:
Back to the 2nd law now. Here I will do the solution for the elliptical-orbit case, and in scaler form. The text provides the vector-form solution.
Consider an object in an elliptical orbit. The total angular momentum of the orbit must be conserved. The scale form for the angular momentum is as follows:
Where m is the object mass, v is its instantaneous speed at some time, r is the distance from the Sun to the object, and theta is the angle between the radius and velocity vectors. In some time delta-t, the object sweeps out a triangle of area delta-A, with one side equal to r, and with the perpendicular to that side h. The trig is as follows:
Because both L and m are conserved, the areal time derivative is a constant, and the proof is complete.
The Solar System is approximately a two-body system, with the Sun and Jupiter as the two bodies. One can write down the sum of their orbital angular momenta in terms of their masses, their orbital speeds, and their seprations from the center of mass point of the orbit. If they have some total seperation d, one can then step through the algebra as follows:
One can then put numbers into this expression to determine a value for the orbital angular momentum. I gave a number in class. But I had a different agenda when I launched into this problem. Consider the following:
That is, the angular momentum of Jupiter's orbit is some three orders of magnitude larger than that of the Solar orbit. If we also consider the rotational angular momenta of the two objects, the upshot is that the angular momentum of Jupiter's orbit is still about an order of magnitude larger than that of the Solar rotation. We live in a planetary system in which most of the angular momentum is in the planetary orbits. Observations of other stars indicate that this in not Universally the case. Many stars rotate much more rapidly than the Sun. These are likely to be cases in which no Jupiter-mass planets formed.
Thursday September 6
Decomposition of the velocities along an elliptical orbit
This is problem 2.3a in the textbook.
One is given the geometric equations for an ellipse, and is to assume the validity of Kepler's second law. From this, one is asked to calculate the radial and angular components of the velocity of a mass on an orbit around the Sun. The relevent geometric equations are as follows:
The problem is to determine the component velocities in terms of only the variable theta, and assuming a,
P and e are known. One begins by taking the time derivative of the radial equation:
Taking the formulation for Kepler's second law from the text, one procedes as follows:
From here, we need to evaluate an expression for L, the angular momentum, in terms of stuff we know:
Now we take this expression, and fold it into the expression for d(theta)/dt from above. Once that is accomplished, we can find the expression for vr directly, and that for v-theta in one further step:
Relative forces of Solar and Terrestrial gravity on the Moon
We typically speak of the moon "orbiting the Earth". That this can be misleading is demonstrated by the following:
That is, the Sun exerts about twice the gravitational force on the Moon that the Earth does. The moon is really orbiting the Sun, with a perterbation from the Earth.
Tidal torque effects the Earth and Moon rotations
The tidal torque from the Moon is slowing down the Earth's rotation. But the total angular momentum of the Earth-Moon system is conserved. That means the angular momentum the Earth loses has to go into the Lunar orbit. The upshot of this is that the Moon is moving further away from the Earth with time. One can begin with a statement of the total angular momentum of the system, and then take time derivatives to see how the system of equations has to behave.
One could also evaluate the asymptotic behavior by combining the constraints of the conservation of angular momentum (as discussed in class) with the conservation of energy (I skipped that bit). You should be able to reconstruct the algebra.
Local Escape Velocities and Interstellar Overdrive
The escape velocity from the surface of Earth is about 11 km/sec. One can also evaluate the local escape velocity from the Solar System. It turns out to be about three times larger. So in order to launch an object from Earth's surface and have it escape the Solar System, we need to give it an initial velocity of about 40 km/sec. I worked out the energetics of this for a custom spamprobe, but slipped up in my on-the-fly unit conversion. The actual energy required is a million times what I said (km to m conversion forgotten). So that's a few 100 million Joules. But that's just for asymptotic freedom. A long trip even to the nearest stars.
From here I moved on to consider Solar Sails. Photons carry both energy and momentum. So shining a light on something is equivalent to pushing on it (slightly). Down here, under all this atmosphere, the effect is very small. It's measurable, but doing so is not easy. In outer space, one could build large, thin sails, supported by low-mass carbon composite frames, and use them to move material away from the Sun. This is another excellent idea for interplanetary travel, but it is lacking for interstellar voyages. This is because the pressure drops rapidly with distance from the light source.
A variant on the Solar-Sail idea that has been discussed is to use high-intensity lasers instead of sunlight. This would produce a much greater pressure and result in a much larger acceleration. The problem is that it would cost a HUGE amount of energy (= $ !) to operate such a facility at a level that would be useful for interstellar travel.
As an example, consider again a probe with a total mass equal to that of a can of Spam (340 grams). If one were to accelerate such a spamprobe to 0.2c, that would require about 6e14 Joules. Quite a lot. But much less than what a real Starship would need. It would take about 20 years for such a probe to reach alpha Cen. So, including the light travel time for the signal to get back to us, that's about 24 years from launch to receiving data.
That isn't outlandish by the standards of current interplanetary probe timescales. The annoying part is that there isn't any high-intensity laser in the alpha Cen system to slow the probe back down when it gets there. So we'd get a fly-by. Our spamprobe would cross the inner 1000 AU of the alpha Cen system in about 5 days. It would cross the inner 40 AU (where all the planets are in our own Solar System) in about 5 hours.
