Essential Concepts from Physics

Notes for Astronomy 1001

Compiled by Paul Woodward

What is it that we need to understand?1. How we can use Newton’s theory of gravitation to find the masses of planets, stars,

and galaxies.2. Energy conservation and some of its implications.3. How gravitational potential energy is liberated when a massive object gets smaller,

and where this energy goes.4. How mass can be converted into energy in other forms.5. How angular momentum conservation affects the rate of spin as the radius from the

rotation axis changes.6. Quantized energy levels of atoms and molecules, and the implications for spectra.7. Doppler effect: spectral line shift and/or broadening.8. Effect of temperature on spectrum.

Determining the mass of an object:1. We observe the motion of a very much less massive object that is in orbit about the

object whose mass we wish to know.2. We will apply Newton’s laws of motion and his theory of universal gravitation.3. We will simplify the problem by assuming that the object is in a roughly circular

orbit.4. To apply Newton’s laws to this case, we will need to know the relationship between

the orbital velocity, v, in the circular orbit, the radius, R, of the orbit, and the acceleration required to keep the object in circular motion (this is called the centripetal acceleration).

5. This requires an application of a little trigonometry from high school.6. First we will use simple scaling arguments, then a short derivation.

The force that produces uniform circular motion is directed from the object toward the center of the circle.You know this from your experience.

Newton’s gravitational force equation tells us how large the force is if we know the masses of the orbiting and the attracting central objects.The accelerationof the orbiting object involves only the mass of the central object, because the force is mass ×acceleration.

Determining the mass of an object (2): We will apply Newton’s law of gravitation to find out how massive Jupiter needs to

be in order to hold its largest moon Ganymede in its (nearly) circular orbit. To do this, Newton himself had to invent the calculus, and it took him 2 years. We will not be rigorous, and we have scientific calculators, so we will do it in about

10 minutes. The first problem is to determine how much force Jupiter needs to provide. This is just like a person pulling on a string to keep a ball circling around him. In half a circuit, the ball (or the moon) exactly reverses the direction of its velocity,

while keeping its speed constant.

Determining the mass of an object (3):

In precisely one half-circuit, the moon’s velocity is exactly reversed.

θ

Determining the mass of an object (4): In precisely one half-circuit, the

moon’s velocity is exactly reversed. This happens in a time equal to

(half circumference) / speed= π R / v

During this time the change in the moon’s velocity is – 2v

Hence the average acceleration is:(change in velocity) / (time interval) = – 2v2/(π R)

Things are not quite this simple, because the acceleration is applied along the radial direction, which is perpendicular to the moon’s direction of motion.

But our estimate is quite close, since the real answer is – v2/ R

θ

An Alternative Derivation (5):

• An object is in uniform motionat speed v around a circle ofradius R.

• Then the x- and y-componentsof the velocity are:vx = – v sin θvy = v cos θ

θ

θ

θ v

This alternative and more accurate derivation is for the ambitious student to review and ponder.

An Alternative Derivation (6):

• Then the x- andy-componentsof the velocityare:vx = – v sin θvy = v cos θ

θ

θ

θ

vx = – v sin θ

vy = v cos θ

v

The thing to remember is that when the orbiting object is located on the x-axis (horizontal axis), the vertical component of its motion is not changing, but the horizontal component is.

We pull along the direction of the x-axis, and this changes the x-velocity, but cannot change the y-velocity. The trick is to see how much we have to change the x-velocity as the object orbits, say, 2 degrees, and then to compute the acceleration this requires.

An Alternative Derivation (7):• Then the x- and y-components of the velocity are:

vx = – v sin θvy = v cos θ

• Circumference = 2 π R• Period of the circular motion = 2 π R / v• Acceleration = rate of change of the velocity.• When θ = 0, vy is not changing, since it has just reached its largest value and has stopped increasing

and will begin to decrease.• But when θ = 0, vx is momentarily 0 but is changing rapidly.• Acceleration = [ (value of vx for θ = + 1°) – (value of vx for θ = – 1°)] / τ• τ = (2°/360°) × (2 π R / v) = π R / (90 v)• Acceleration = [ – v sin(1°) + v sin(– 1°) ] / [π R / (90 v)]• – sin(1°) = sin(– 1°) = – 0.01745• Acceleration = (– v2/R) (0.0349 × 90 / π ) = 0.9998 (– v2/R)• In the limit that the two angle values are very close together, Acceleration = – v2/R• The acceleration is always directed from the object toward the center of the circle. (We know this,

because there was nothing special about the point on the circle where we said θ = 0.)

Determining the mass of an object (9):

8. If the mass of the large object is M then the gravitational acceleration it causes on the small object in its circular orbit of radius R is G M / R2 .

• Keeping the small object in its circularorbit via this gravitationalacceleration thusrequires that

v2 / R = G M / R2

θ

We have used simple trigonometry and the definition of acceleration as the time rate of change of velocity, together with uniform circular motion, to find that the acceleration must be -v2/R. Newton’s law of gravity tells us this acceleration must be -GM/R2, so these two expressions must be equal. The unknown M is now determined, if we measure G in the lab, and we get the orbital period and R from observation.

Determining the mass of an object (10):• Multiplying through by R in this relation gives:

v2 = G M / R• The square of the orbital period, P, the time to complete one full orbit, that

enters Kepler’s third law is therefore given by

P2 = (2π R / v)2 = ( 4π2 / G M ) R3

• Here we see Kepler’s third law, derived from Newton’s laws.But now the constant of proportionality between P2 and R3 can, through the constant of universal gravitation, G, tell us the mass of the large mass, M.

• It is a key fact that we can measure G by doing experiments with gravitating masses in laboratories here on earth. Then we can use this same value of G to compute the mass of a planet.

• Saying that G is the same on earth and everywhere else and for all time is an assumption. We can check that it is reasonable if its results are consistent.

Determining the mass of an object (11):9. Weighing Jupiter by observing the motions of its moons.• Measure the period P of the orbit of a moon of Jupiter.• Measure the greatest angular separation, α, between the moon and Jupiter

during an orbit.• From our knowledge of the orbits of the earth and Jupiter about the sun,

compute the distance, D, to Jupiter at the time of the measurement of the angle α .

• Compute the radius of the moon’s orbit about Jupiter viaR = D sin α ≅ D α

• Then if we approximate the moon’s orbit by a circle, we find

M = ( 4π2 / G ) ( R3 / P2 )

Galileo’s observations of Jupiter’s moons.

Jupiter with its four Galilean satellites

Determining the mass of an object (12):9. Weighing the earth.• Measure the acceleration, g, on an object

near the earth’s surface caused by the earth’s gravity.

• Measure the radius of the earth, R⊕

• In 230 B.C. the Greek astronomer Eratosthenes measured the size of the earth by noting how the elevation of the north polar star changed as the result of travel over the earth’s surface either north or south.

• Newton’s law of gravity then implies thatg = G M⊕ / R⊕

2

• Knowing G from laboratory experiments and R⊕ from the above method, we can compute the mass of the earth fromM⊕ = g R⊕

2 / G

Obviously, those ancient Greeks knew the earth was round. They even measured its radius.

Determining the mass of an object (13):

This is a short aside for the ambitious or curious student.The famous mathematician Gauss, born about the time of the American revolution, proved that the gravitational force from a spherically symmetric object like the earth is precisely the same as the force from a point mass of the same total mass located at the object’s center. This is just amazing. You need calculus to see it.

Determining the mass of an object (13):

Second blurb on Gauss’s Law.

Determining the mass of an object (13):9. Weighing the sun.• Newton’s laws imply that: M☼ = ( 4π2 / G ) ( R3 / P2)⊕• Measure the radius of the earth’s orbit, R .• This was first done in the 1720s, using an effect known as the aberration of starlight.• The earth’s motion in its orbit about the sun causes light from distant stars located

above or below the earth’s orbital plane to appear to come from locations displaced slightly in the direction of the earth’s motion.

• Over the course of a year, such stars appear to move in tiny ellipses on the sky, whose size and position allow us to compute the velocity, v, of the earth’s orbital motion.

• The circumference, 2π R, of the orbit is just v P.

Light from distant stars appears to come from a direction that is affected by the motion of the earth in its orbit around the sun.

The observation of the aberration of starlight was the first direct and clear proof that the earth in fact is moving relative to the frame of reference of the distant stars.It is a larger effect than that of stellar parallax, and it was observed before the parallax of any star was observed and measured.

At this Web site, you can find the story of how we first really proved that the earth actually moves.In looking for parallax of stars, the aberration of starlight, a much larger effect of the earth’s motion, was discovered by Bradley instead.The speed of light, determined about 50 years earlier, allowed us to measure the speed of the earth’s orbital motion about the sun, and from this the radius of the earth’s orbit.

Although at age 5 I did not believe the earth rotates, for much the same reasons stated in Courtney Seligman’s text, it is very hard to believe that Ptolemy, by around 200 AD, did not believe that the earth rotates. Without the earth rotating, his model of the solar system would truly have been a mess.

The objections of the ancients to the earth’s rotation can be viewed by modern people as not laughable at all.

The earth’s rotation should cause a wind, and it does. We all live in Minnesota and are aware that weather comes to us most days from the west. We will see when we discuss the weather on Venus, Earth, and Jupiter that this is caused by the earth’s rotation. In fact, the phenomenon of the trade winds is a result of the earth’s rotation together with the sun’s greater heating of the regions near the equator. So the ancients were right. There should be a wind set up, and there is.

The second objection to the earth’s rotation cited by Seligman is that if you throw a ball into the air, it should not come back down in the same location.

Again, the ancients were correct.They just didn’t throw the ball up high enough to see the effect.But we do that today by firing rockets into the sky.They do not come down where we sent them off, even if we fire

them straight upward.This is indeed caused by the earth’s rotation.So the ancients once again were right.And we too are right – the earth rotates!

Just take a ride on the space shuttle and watch it turn.There is absolutely no question about it.

Seligman also says that Tycho Brahe rejected the concept of the earth’s motion, because he saw no parallax of any star.

This is correct.But it does not follow that he did not accept Copernicus’ theory.He generated a variation of Copernicus’ model, in which all the

planets except earth orbit the sun, and then the sun, carrying all those planets with it, orbits the earth.

This model is in a certain sense actually correct.In the coordinate system in which the earth is always located at the

origin of coordinates (the center of the universe), this is the correct actual observable motion of the sun and planets.

This is not a useful coordinate system to use, but there is nothing wrong with it. In fact, one could say that it is a “natural” coordinate system for us.

But Brahe was wrong. Parallax is observed for stars, but you need a telescope to see it, and Brahe died before the telescope was invented.

Seligman tells the story of how, armed with the telescope, Bradley sought to observe the parallax of stars and hence to prove that the earth moves (relative to the frame of reference of the “fixed” stars) in its orbit about the sun.

Partly because he lived in England, which is awfully far north, and partly because his observations were not as accurate as we now can do, Bradley discovered a different, completely unanticipated result of the earth’s orbital motion – the aberration of starlight.

This is a very much larger effect, caused by the finite speed of light. It dwarfs the effect of parallax. It is also seen for the stars most easily observed from England, namely those near the north celestial pole.

The story from Seligman’s Web page is continued on the next slides.

Although the details of this story are fascinating, you need only know what the effect of the aberration of starlight is, how it is caused, and that it was measured before parallax of stars, because it is so much larger.

You should also know how the speed of light was measured using the eclipses of the Galilean moons of Jupiter (see following slides).Together these allow us to weigh the sun.

Aside on the Speed of Light:1. In order to weigh the sun using the earth as a test particle and applying Newton’s law of

gravity, we need to know the radius of the earth’s orbit.2. We can get this from the aberration of starlight, but to do so we need to know the speed of

light.3. In the late 19th century the speed of light was studied in a series of very famous experiments,

but it had been estimated 2 centuries earlier using the eclipses of the moons of Jupiter.4. Ole Rømer, a Dane, realized that the predictions for the eclipses of Jupiter’s moons were

either early or late depending upon whether Jupiter was near or far from the earth at the time.

5. He used an inaccurate determination of the radius of the earth’s orbit to deduce the speed of light from this effect, and his value was in use for 2 centuries.

6. His was the first proof that light does not get where it’s going instantaneously, a point that had been argued for many years on philosophical grounds without arriving at any conclusion.

7. What was his interest in the timing of eclipses of Jupiter’s moons? He was building tables of these eclipse times that could be used by world explorers to determine their longitude when far from any land. To determine longitude, you need a good clock, and the Jupiter system served this function beautifully before quartz crystal watches.

It takes light from Jupiter longer to reach us at the earth when we are located on the opposite side of the sun from Jupiter. If we know the radius of the earth’s orbit about the sun, we can then figure out what the speed of light must be. The delay is quite a few minutes, so our clocks do not need to be perfect to make this measurement.

Aside on the Speed of Light:1. It might seem that Rømer’s determination of the speed of light is

circular in our context, since he used the radius of the earth’s orbit to get it, and we want to use it with the observed aberration of starlight to get the radius of the earth’s orbit.

2. BUT, we have 2 independent numbers we need to know – namely the speed of light and the radius of the earth’s orbit – and we have two independent phenomena that pin them down – namely the measured size of the apparent elliptical paths on the sky of stars near the north celestial pole and the measured discrepancies in the observation and prediction of the eclipses of Jupiter’s moons.

3. THEREFORE, we may use both of these together to determine the two things we want to know – namely the radius of the earth’s orbit and the speed of light.

Determining the mass of an object (summary):1. We observe the motion of a very much less massive object that is in orbit about the object whose

mass we wish to know.2. We estimate, one way or another, the distance to the object.3. Kepler’s third law of planetary motion (which falls into this category) states that the square of the

period, measured in years, equals the cube of the average distance, measured in Astronomical Units (AU), or

P2 = a3

4. Using Newton’s law of gravitation, we can generalize Kepler’s third law, so that it holds for any two masses orbiting about each other with period P and average separation a,

P2 = a3 [4π2 / G (m1 + m2)]5. When one object, with mass M, is much more massive than the other, we have

P2 = (4 π2 / GM) a3

6. Since only the mass, M, is unknown in this equation, we may solve for it in terms of the other quantities:

M = (4 π2 / G) (a3/P2)

Application to Planets in other Solar Systems:

1. We observe periodic changes in the velocity of a star close to us.

2. We can measure the distance to the star from, for example, its parallax.

3. We can estimate the mass of the star from its surface temperature and luminosity, if the star is on the main sequence. (We will see why we can do this later on in this course.)

4. Assuming that the star, with mass M, is much more massive than the planet, we have P2 = (4 π2 / GM) a3

5. This tells us the orbital radius of the planet about its star.

6. Now, from the amplitude of the velocity variation of the star, we can determine the mass of the planet.

The first image of a planet orbiting another Sun-like star.This is a composite image, made using a 270-inch telescope’s

adaptive optics, of the star 1RSX J160929.1-210524, about 500 light years away, with its planet. The planet is about 7 to 12

times the mass of Jupiter & orbits about 30 Gmiles from its star.

A fish-eye view of the Milky Way from Australia

The direction of the center of our galaxy

is fairly obvious here, but it can be determined much

more accurately by observing the

speeds of approach or recession from us of individual

stars in the Milky Way.

We can use individual stars in the Milky Way as our small-mass

orbiters and thus measure the mass of the galaxy inside their orbits.

In infrared light, our galaxy looks a lot flatter.

To get a rough measure of the mass from velocities of orbiting stars, we need only assume circular orbits and an axisymmetric distribution of mass. We will

also assume all orbits lie within a single plane.

The Milky Way, photographed in diffuse infrared radiation by the COBE satellite.

We get the velocities of the stars from the Doppler effect, which is discussed later this week.

A galaxy is a system in which the mass is distributed throughout the plane in which our test objects (stars) orbit.

This makes it much more difficult to apply Newton’s law of gravity than it was in the case of the solar system or for moons orbiting Jupiter.

To apply Newton’s law in this case we need to use the calculus, which is beyond the scope of this course.

The procedure is nevertheless fairly simple.

We build a simple model of the galaxy that permits us to compute the orbital velocities of stars at different radii from the center.

In such a model, we might have perfect circular orbits for stars in a perfectly flat disk in which the density of stars changes only with radius from the center and not with angle around the disk.

We might also add a spherically distributed nuclear bulge or halo.

Once we have or model, we can compute the orbital velocities of the disk stars at all radii.

We do this and compare the result with the actual velocity observed.

They will not agree.

We then change the parameters of our model, such as the degree of central concentration of the disk mass or the relative amount of halo and disk material, and keep improving the agreement with the observations until it is acceptable.

If no acceptable set of parameters is found, we will have to develop a different, perhaps more elaborate model.

This process works very well, and we use it all the time to determine the masses of galaxies and the distributions of mass within galaxies.

For disk galaxies, commonly made model assumptions are:

1) The galactic disk is thin and flat.

2) The orbits of stars in the disk are circles centered on the galactic center.

3) If we go around one of these circular orbits, the local density of stars at each point will be roughly the same.

4) There may be a nuclear bulge centered on the galactic center and with a spherical or oblate spheroidal distribution of stars within it.

Based on these assumptions we can compute the orbital velocities at all radii in the disk.

Also, these assumptions imply that velocities of stars that we measure for such a system when seen edge on are in fact the orbital velocities.

We have built our model of our own galaxy partly by looking at other ones in order to get a better perspective. This one is M83, 14 million

light years away from us.

Clearly, if our galaxy is like M83, shown here, our

assumption of mass spread out evenly around concentric rings

is in trouble.

Happily, what we see in this picture is the distribution of

light in M83, not the distribution of mass.

NGC 891, a spiral galaxy in the constellation Andromeda that is seen edge on, probably looks pretty much the way our galaxy would look when viewed from outside and edge on.

This galaxy looks pretty symmetrical, and pretty

thin.

The Andromeda Galaxy, which is probably quite like our own.Its nuclear bulge is about 12,000 light years across.

This galaxy has two companions, that we could

use to help determine its mass

The Magellanic

Cloudsas viewed

from Australia

These two companions of our own galaxy should have motions

that we could relate to our galaxy’s mass.

The Large Magellanic Cloud,

a satellite of our own Milky Way

We could use the velocities of stars in

this irregular galaxy to help to determine its mass, but we would

have to generalize our method.