Essential Concepts from Physics

Notes for Week 3 of Astronomy 1001Compiled by Paul WoodwardFebruary 4, 2009

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.

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

θv

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

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

An Alternative Derivation (8):

• 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

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, that enters Kepler’s third law is therefore given by

P2 = (2 R / v)2 = ( 42 / 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 value of G to compute the mass of a planet.

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, , 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 .

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

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

M = ( 42 / 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 from M = g R

2 / G

Determining the mass of an object (13):

9. Weighing the sun.

• Newton’s laws imply that: M☼ = ( 42 / 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, 2 R, of the orbit is just v P.

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 star light, 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.

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 [42 / G (m1 + m2)]

5. When one object, with mass M, is much more massive than the other, we have P2 = (4 2 / 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 2 / 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 what the “main sequence” is and how we do this later in the course).

4. Assuming that the star, with mass M, is much more massive than the planet, we have P2 = (4 2 / 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.

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.