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One of the very first objects observed with the newly launched Hubble Space Telescope was SN 1987A. At the position where the massive blue star Sk -69 202 once was, we now see the remains of the exploded star surrounded by a ring. The angular diameter of this ring is so small that it was not observable from the ground where the atmosphere blurs the image. This ring was one of the most exciting discoveries made with the Hubble Space Telescope. A very important paper used the image of the ring, combined with the strength of special emission lines, to determine the diameter and distance to the ring and SN 1987A. For this assignment, you will take their data and calculate the numbers for yourself.

The ring does not look circular to us because it
is tilted at an angle toward our line of sight. This angle is called the
inclination angle and is labeled ** i**. A quick review of some
trigonometry may be necessary here:

No matter how the ring is tilted, the long axis
we see (the *major *axis, or **a**) will always be the same length, but
as we tilt the ring further and further the *minor *axis, or **b**,
becomes smaller and smaller. The manner in which the ring around SN 1987A is
tilted is such that the top (upper right corner of photo in Figure 1) is closer
to us than the bottom.

When the supernova went off, it released a bright flash of visible and ionizing
UV and x-ray radiation. After about 3 months, the x-rays and UV radiation
reached the ring of gas causing the ionized atoms in the ring to emit light.
While the radiation from the supernova illuminated the entire ring
simultaneously (because the supernova is at the center of the ring and the
radiation emits in all directions at the same speed), the entire ring **did not
**"light up" simultaneously from our point of view (as you saw in
the video in class). The light from the top of the ring, which is tilted toward
us, did not have as far to travel as that from the bottom of the ring, which is
tilted away from us. The interval between the time when we saw the light from
the top of the ring and the time when the entire ring (including the bottom part
of the ring) was illuminated represents the *light travel time* between the
near side of the ring and the far side of the ring. In our trigonometric diagram
above, this distance is represented as **L _{t}**. This represents the
difference in light travel time,

Now that you understand the geometry of the ring and have reviewed some trigonometric relationships, you can determine important information about the ring.

**PRINT THIS OUT OR USE A SEPARATE SHEET OF
PAPER TO SHOW YOUR WORK! YOU MUST SHOW YOUR WORK WHERE APPROPRIATE TO GET FULL
CREDIT**

Questions:

**1)** At sometime after the supernova explosion, we
began to see the ring emission. Call this time **t _{o}**. At a later
time,

**t**** _{o
}=__________**+/-(error)

**2)**
The difference, **t**_{max}**-**
**t _{o}**, is the light travel time from the far edge of the ring to
the near edge. This time can be converted to a distance by multiplying by the
speed of light ,

How many km is this distance? **L _{t}**=

**3)**
Refer back to our trigonometric
diagram, Figure 1. Note that the distance you just determined is not the
diameter of the ring, **a**, but is actually the distance marked as **L**,
the length along our line of sight. Now you know **L _{t}** and that Q

The major axis, **a**, of the ring in km
is______________.

How many parsecs is this? (1pc = 3.1 x 10^{13}
km)_________________

**4)** Now
we can calculate the distance of SN 1987A. In astronomy, where typical objects
are at great distances, there is a very important and simple relationship
between an object's angular
diameter, b, (how
large an angle in the sky it subtends) and its actual diameter (**a**)
and distance away from us (**d**). This is:

If b is expressed in units of *radians*
(remember, there are 180^{o} in p radians) then **d** will have the
same units as **a**. For instance, if b=.01 radians and **a **=10 cm, then
**d** = 10cm/.01=1000cm. You know the actual diameter, **a**, of the ring
from part 3. Now you need to determine the angular diameter of the ring in
radians. The angular size of the photo in Figure 3 is 1.15 x 10^{-5}
by 1.15 x 10^{-5} radians. **READ CAREFULLY****:**
you must measure what fraction the **MAJOR AXIS** of the ring is of the photo
height and convert this to radians. If you don't understand this procedure read
this.

The angular diameter of the ring major axis **in
radians**

The distance to the ring is (**d** = **a**/b):**_______________**
light years (there are 9.46x10^{12 }km

in a light year).

**5)**
The Blast Wave! If you have ever seen
those movies from the 1950s showing the buildings that are blown up by atom
bombs, you may remember that the radiation from the explosion makes the
buildings begin to burn almost immediately. Then a few moments later, the
buildings are completely leveled when the powerful blast of supersonic air hits
it. The blast wave travels very, very fast, but not nearly as fast as the
radiation. SN1987A was very much like an atomic bomb (in the same way that the
Earth is like grain of sand): the flash of radiation reached the ring a few
months after the explosion; but the blast wave will reach the ring after a
greater delay. We are going to estimate that delay and predict when the blast
wave will hit the ring.

First we must estimate the speed of the blast
wave. We do this by measuring the Doppler
shift of absorption lines in the spectrum of the supernova. Below is a
spectrum about one day after the explosion. The supernova is expanding so
absorption lines due to the atoms in its atmosphere are **blue-shifted. **If
**L **** _{o} **is the wavelength of an
absorption line in the laboratory and

(FYI 2)

where **c** is the
speed of light (300,000 km/s). Look at figure 4, below. The dip just to
the right of 6000 angstroms is a blue-shifted absorption line from Hydrogen
called Ha (H-alpha). Its "rest" (laboratory measured) wavelength
is **L **** _{o}** =
6563 angstroms. Find

Figure 4

**D****L****
**= ____________+/-(error)**____________**angstroms.

**v**_{blast}=
__________________km/s

**6)**
Assuming that the blast wave travels at constant velocity, you can calculate the
time for the blast wave to reach the ring by dividing the radius of the ring
(half its diameter, or **a**/2) by **v**_{gas}.
Given that the blast wave began in 1987 when the supernova went off (in OUR time
reference), in what year will we see the blast wave
hit the ring? (1 yr = 3.16x10^{7} seconds)

________________

**FYI 1**: This equation only works when b
<< 1, as it is in this case. If you become good at estimating
smallish angles you can use this simple relationship to figure out the distances
to objects, like the distance to an airplane flying overhead. Example: Very
approximately, your index finger is .01 radians across when held at arms length.
So if a six foot tall person is just covered up by your index finger turned
sideways at arms length, you know that they are about 6/0.01=600
(the same as 6*100=600) feet away. Pretty neat, huh?

**FYI 2**:The "**D**"
in the equation is the Greek letter "delta". In math this is
traditionally used as an adjective to mean "the change in something".
Now when you hear some scientist talking about "delta-lambda" you know
he is talking about a red shift or blue shift, and if you hear
"delta-v" you know that refers to a change in velocity.)

**Question 4 supplement**:
In question four, you are asked to measure the angular size (in radians) of the
ring around SN 1987A. You are given the angular size of the piece of sky of the
picture of the ring and you have to estimate the angular size of the ring itself
which is smaller than that of the whole picture. This is like being shown a
dollar bill and being told it is 15 centimeters long and having to estimate how
many centimeters long the picture of George Washington is, but your ruler only
has inches on it. Go back to the question.