Transcript Lect07-2-4-09

Essential Concepts from Physics
Notes for Week 3 of Astronomy 1001
Compiled by Paul Woodward
February 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.
θ
Determining the mass of an object (3):
 In precisely one half-circuit, the moon’s velocity is
exactly reversed.
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
θ
= πR/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
An Alternative Derivation (5):
•
•
An object is in uniform motion
at speed v around a circle of
radius R.
Then the x- and y-components
of the velocity are:
vx = – v sin θ
vy = v cos θ
θ
θ
θ
v
An Alternative Derivation (6):
•
Then the x- and
y-components
of the velocity
are:
vx = – v sin θ
vy = v cos θ
θ
v
vy = v cos θ
vx = – v sin θ
θ
θ
An Alternative Derivation (7):
• Then the x- and y-components of the velocity are:
vx = – v sin θ
vy = 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 circular
orbit via this gravitational
acceleration thus
requires that
v 2 / 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 = (2p R / v)2 = ( 4p2 / 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, a, 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 a .
Compute the radius of the moon’s orbit about Jupiter via
R  D sin a  D a
Then if we approximate the moon’s orbit by a circle, we find
M = ( 4p2 / 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 that
g = 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☼ = ( 4p2 / 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, 2p 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 = a 3
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 [4p2 / G (m1 + m2)]
5. When one object, with mass M, is much more massive than the
other, we have
P2 = (4 p2 / 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 p2 / 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 p2 / 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.
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.
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.
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.
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.
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.
This galaxy looks pretty
symmetrical, and pretty
thin.
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 has two
companions, that we could
use to help determine its mass
The Andromeda Galaxy, which is probably quite like our own.
Its nuclear bulge is about 12,000 light years across.
These two companions of our
own galaxy should have motions
that we could relate to our
galaxy’s mass.
The
Magellanic
Clouds
as viewed
from
Australia
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.