Download Astronomy 15 - Homework 3 - Due Wed. April 24 1) As we`ll see

Astronomy 15 - Homework 3 - Due Wed. April 24
1) As we’ll see, stellar atmospheres are complicated enough so that stars do not radiate
exactly as blackbodies, but blackbodies are an excellent ‘zeroth approximation’ to the
radiation which stars emit. The Luminosity of a star is its total radiated power. If a star
is a sphere of radius R, its area is 4πR2 . Therefore the total power radiated by a star is
just
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L = 4πR2 σTeff
.
Teff is called the ‘effective temperature’ – it’s the temperature of a blackbody which emits
the same power per unit area as the star. Different levels in a star’s atmosphere have
different temperatures, and the effective temperature is a useful average of these. We can
learn a lot from this simple expression.
The fundamental diagram of stellar astronomy is called the Hertzsprung- Russell diagram,
or HR diagram. It’s a plot which shows the surface temperature of a star on the horizontal
axis and the luminosity on the vertical axis. For historical reasons, the temperture is
shown increasing to the left – the hottest stars are on the left-hand side of the diagram.
Many variants of the diagram exist in which the quantities plotted are not strictly the
surface temperature and the luminosity, but some closely related observed quantity. For
now, we’ll consider the so-called ‘theoretical’ HR diagram, in which the quantities plotted
are Teff and the actual luminosity of the star in ergs per second (or, if you prefer, Watts).
As it turns out, the luminosities of stars occupy a huge range – the dimmest stars are
much, much dimmer than the brightest stars. For this reason (and others, as we’ll see),
it’s convenient to plot the axes of the HR diagram as logarithms. So our diagram will
show log T on the horizontal axis – increasing leftward! – and log L on the vertical axis –
increasing upward, as is conventional.
Any point on the HR diagram has associated with it specific values of L and T , because
those are the axes of the plot.
a) Solve for R in terms of L and T .
b) Using your sheet of constants, determine the sun’s Teff from its L and R. (The answer
is in the book on p. 84 – do it yourself first).
c) Accurately draw an HR diagram with logarithmic axes. Place the sun’s luminosity and
temperature at the center. Scale the axes so that the horizontal axis spans from 100000
degrees to 2000 degrees, and so that the vertical axis spans from 10−4 to 105 times the
luminosity of the sun . . . that will compress the vertical scale greatly compared to the
horizontal scale. You can put the vertical axis in units of the sun’s luminosity L if you
like, or you can plot it in ergs/sec or Watts if you prefer. Plot a point on the diagram
representing the sun.
d) Recall that
L = 4πR2 σT 4 .
Taking logarithms this becomes
log L = 2 log R + 4 log T + constant.
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From this show that contours of constant radius on the HR diagram will be straight
lines sloping upward to the left with slope 4. On your HR diagram, draw in the line
corresponding to R = R , and then draw in lines corresponding to 10, 100, 1000, 0.1, 0.02,
and so on, times the sun’s radius.
2) Here’s a classic calculation involving blackbodies.
Consider a planet of radius rp orbiting a distance d from the sun. At this distance, the sun’s
luminosity L has been spread out over a sphere of area 4πd2 . The planet’s cross-sectional
area is πrp2 .
a) Show that the radiant power intercepted by the planet is
rp2
L 2 .
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Suppose the planet absorbs all this power and reflects none of it, that it radiates as a
blackbody over its whole surface with a single temperature (the planet might be a small,
rapidly spinning, blackened copper ball, for example – in that case heat would quickly
conduct through the whole body. Obviously, this is a little oversimplified).
b) By equating the power absorbed by the planet to the power it emits, show that it comes
to an equilibrium temperature Tp given by
Tp4 =
L
.
16πσd2
Note that the size of the planet does not matter. Plug in the expression for L in terms
of the sun’s radius and effective temperature T to find
Tp4 =
2 4
R
T
.
4d2
c) Solve for the temperature of such a body at the earth’s distance from the sun.
The excellent agreement between this and the actual temperature of the earth is somewhat
accidental – the earth absorbs only a fraction of the light it receives from the sun, but also
emits rather inefficiently.
3) How big is the sky? A Schmidt telescope is designed to take photographs with an
especially wide field of view. The most famous Schmidt, the Oschin Schmidt at Palomar
Mountain in California, has a 1.2-meter diameter aperture and a focal length of 3.07 meters.
a) What is its image scale in arcseconds per millimeter?
b) The plates it takes are 14 inches (35.6 cm) on a side. How many degrees does this
correspond to in the sky? How many square degrees of sky is this? (Yes, you just square
it.)
c) This leads on to the subject of solid angle. A solid angle is much like a regular angle, but
with one more dimension; it’s the appropriate unit for “area” in the sky. One standard unit
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for solid angle is the steradian (‘stereo radian’), which is analagous to the more familiar
radian measure for angles in a plane. To develop that analogy, consider a circle of radius
R, and an arc of length l; the angle subtended by that arc at the center of the circle is
simply l/R radians, so the whole circle contains 2π radians. Note that radians are unitfree (dimensionless), since they are ultimately derived from a length divided by another
length. Likewise, if you have an area A on the surface of a sphere of radius R, the solid
angle subtended by that area at the center of the sphere is given by A/R 2 steradians. Note
that this is also a unit-free (dimensionless) ratio, since the area is divided by another area
(R2 ). Now: How many steradians in the whole celestial sphere?
d) Steradians are not always convenient measures of area; in (b), you expressed the area
of the plate in square degrees. One can compute the number of square degrees in the
whole sky by considering the surface area of a sphere which is 360 units (meters? miles?
light years?) in circumference; that’s isomorphic (to use fancy math talk) to the celestial
sphere. So now compute the solid angle of the whole celestial sphere in square degrees.
e) Suppose one were to photograph the whole sky with a Schmidt camera like the Oschin.
(It has been done! The Oschin Schmidt has a couple of twins in the South, one in Australia
and one in Chile; working together, they have done the whole celestial sphere.) Assuming
one could do it without significant overlap between plates, how many plates would it take?
And finally, suppose one were to make a giant celestial globe by taking these plates and
pasting them onto the surface of a sphere. What would the radius of this sphere be?
(Gotcha!)
4) A Taylor series is a handy mathematical tool for exploring the behavior of a function in
a limited region. Here’s the basic idea. Suppose you have a function f (x), with derivatives
f 0 (x), f 00 (x), and so on (the primes denote 1st, 2nd, and so on derivatives). Suppose
furthermore that (a) you know the value of the function and the values of all the derivatives
at some x = x0 and (b) the function obeys certain conditions of continuity which we need
not belabor here; most functions one meets in physics obey these conditions. The magical
thing about the Taylor series is that given this information, you also know the value of the
function at all x 6= x0 ! The relevant formula is
f 00 (x0 ) 2 f 000 (x0 ) 3
a +
a + ...
f (x0 + a) = f (x0 ) + f (x0 )a +
2!
3!
0
This offers a way to compute, say, a sine function; at θ = 0 one finds for f (θ) = sin(θ) that
f 0 (0) = 1, f 00 (0) = 0, f 000 (0) = −1, and so on, so that
sin(θ) = θ + 0 −
θ3
θ5
+0+
+ ...,
3!
5!
where θ is in radians.
Now, another practical use of this comes from the fact that, in many circumstances, the
high-order terms get very small very fast. If a < 1 this will be true, and the factorial
in the denominator also does its part. This makes the Taylor series a powerful tool for
determining the behavior of a function in the vicinity of some known point. Let’s see how
to do this.
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a) Look at the right-hand side of equation 3.13 in the book - the relativistic formula for
the Döppler shift. Let x = v/c. Show that the derivative of this mess for x = 0 is equal to
one.
b) Now use a Taylor series to derive equation 3.14 from equation 3.13. Note that for this
you need only use the first term of the Taylor series, so you’re practically there already;
also, be sure to note the definition of z in the intervening text. (As Shu notes, one can get
the same result from a binomial expansion, but a Taylor series is more generally useful).
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