- Attachment: Hubble’s composite image
- This is the phase that the Cat’s Eye Nebula is in. A planetary nebula lasts about 10,000 years before the dissipation gets too faint to see. The Cat’s Eye, NGC-6543, is one of the most unusual planetary nebula because it has 11 rings or shells created by consecutive outbursts. 10,000 years is not long in astronomical terms so we would not expect to see very many. However, some 5,000 planetary have been catalogued in our own galaxy.
- The Cat’s Eye is located in the Constellation Draco the Dragon 3,300 lightyears away. It can be viewed with a 6-inch backyard telescope. The Nebula is located directly above the North Ecliptic Pole, 90 degrees perpendicular from the plane of the Solar System. The Hubble Space Telescope has made some careful measurements of the nebula. To learn how old it is one measurement was made of the outer most visible shell. The edge of the shell was 16 arc seconds from the star remnant at the center. The expansion was measured at 16 kilometers per second, or 36,000 miles per hour.
- To calculate the radius of the outer shell we use trigonometry. The distance to the nebula is the adjacent side of a right triangle and the radius of the shell is the opposite side. The tangent of the angle is the ratio of the opposite to adjacent sides. For very small angles the tangent is equal to the angle itself in radians.
- There are 2pi radians in the full circumference of every circle. 2pi radians = 360 degrees.
------------------------- Therefore, 1 radian = 57.3 degrees.
------------------------- And, 1 arc degree = 0.175 radians
------------------------- 1 arc minute = 0.00029 radians
------------------------- 1 arc second = 0.000048 radians.
------------------------ Tangent = Radius of the shell / Distance to the Nebula
------------------------ Radius of the shell = Tangent * Distance to the Nebula
------------------------ Radius of the shell = (tangent 16 arc seconds) * (4.8*10^-5 radians / arc second ) * (3300 lightyears )
----------------------- Radius of the shell = 0.25 lightyears.
- So the shell has been traveling 36,000 miles per hour and has traveled 0.25 lightyears distance. How long did it take?
--------------------- Time = Distance / Velocity.
-------------------- Time = 2.5 lightyears / 16 km / sec
-------------------- There are 9.3 * 10^9 kilometers in a lightyear
------------------- There are 3.1 * 10^7 seconds in a year.
-------------------- Time = 0. 25 LY * 9.5*10^12 km/LY / (16 km / sec) * (3.1 *10^7 sec / year)
------------------- Time = 4,800 years.
- The Cat’s Eye Planetary Nebula started expanding 4,800 years ago. This was about the time when the earliest Egyptians devised the 365 day calendar.
- The gases in the outer shells have expanded to 2.4 lightyears. Their total diameter in the sky is 300 milliarcseconds. These 11 shells began expanding 48,000 years ago and are equally spaced at 4,800 year intervals. This whole process must have started 48,000 years ago when “homo sapiens sapiens” subspecies was the only surviving humanoid on the planet. Cro-Magnon man was making stone tools and living in caves.
------------------------ Radius of the shell = (tangent 150 arc seconds) * (4.8*10^-5 radians / arc second ) * (3300 lightyears )
----------------------- Radius of the shell = 2.4 lightyears.
- The center star’s surface temperature is 80,000 C compared to our Sun’s 6,100 C. It is 10,000 times more luminous than our Sun. The star is creating a solar wind that is blowing material off at the rate of 20 trillion tons per second. The velocity of this wind is 1,000 miles per second , or 10 times faster than the material in the expanding bubbles. This immense solar wind slams into the slower bubbles of gas creating the rings we see surrounding the star. The nebula has a density of 5,000 particles per cubic centimeter and a temperature that varies from 7,000 to 15,000 Kelvin. This results in 11 empty bubbles appearing to have been created at 4,800 year intervals
.
- The dying star is ejecting 20 trillion tons of matter every second ( 2*10^16 kilograms / second). The original star had a 5 Solar Mass.
------------- Mass = 2*10^16 kg/sec * 48,000 years * 3.1*10^7 sec/year
------------- Mass = 2.8 *10^28 kilograms
------------------ 3 Solar Mass is 150*10^9 kilograms
---------------- Percentage = 0.28 *10^29 / 150*10^29 kg
-------------- Percentage = 0.18%
- Less than 2/10th of the star’s mass has been ejected into the planetary nebula in 48,000 years.
- The hot temperature gas by itself could not have created the powerful X-rays being emitted. Something else is going on? We still have more to learn from this 48,000 year old debris. An announcement will be made soon , stay tuned.
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(1) When the White Dwarf remnant star that is less than 1.4 Solar Mass, like this one in the center of the Cat’s Eye, is able to accrete mass from an outside source to where it exceeds 1.4 Solar Mass then it explodes as a Type 1a Supernova, a thermonuclear explosion. See Review # 1318 Kepler’s Supernova to learn more about these type of star’s death.
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