Chapter 10-11

# Chapter 10-11 - 2 We have a^3=p^2 p=11 minutes 11...

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Astronomy –odd- numbered –Homework Chapter 10 1, M = 0.4 According to t he rule of thumb is, the main sequence : = 10 000 000 000 / M^2.5 = 10 000 000 000/(0.4)^2.5 = 100 000 000 000 year = 100 billion years 3. the stars in a globular cluster is about 0.02 stars per cubic ly and 7 stars per cubic ly 5. v = 15 km/s 1 ly = 9 x 10^12 km 1 year = 3.15 X 10^7 s d = 8.25 x 10^12 km è t = d/v = (8.25x10^12)/15 = 0.55x10^12 seconds = (0.5x10^11)/(3.15x10^7) è = 17460 years 7. we have Density = mass/ volume the volume will grow by a factor = 20^3 = 8000 So, the average density will decrease by 8000 9. 1 pc = 3.1x10^13 km = 3.1x10^16 m 1400km/s = 1.4x10^6 m/s è the supernova occurred : ( 1.35 x 3.1x10^16) / 1.4 x10^6 ~ 2.989256x10^10 seconds è = 948 years Chapter 11 1 ./ The Radius of the Star is 10 KM. Diameter= 10 x 2 = 20. Circumference: 20 * 3.1416 = 62.838.

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62.832 * 642 = 40,338. speed of light :299,997 per second Speed of the surface at neutron star's equator : 40338 / 299997 = 0.1345
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Unformatted text preview: 2./ We have : a^3=p^2 p=11 minutes 11 minutes= .0000209 years .0000209^(2/3)= 0.0007 AU =>.0007 AU= 104,718,600 meters = 104,718.6 kilometers 3./ Density of typical small neutron star: 8 * 10^16 kg/m^3 density of the Moon: 3000 kg/m^3 A neutron star of the same mass would therefore be about 30,000 times smaller. So angular diameter about 1/1000 arc-minute. 5./ Use Wien's Law relating temperature to maximum wavelength: wavelength max = 3,000,000 / T , where T is temperature in Kelvin. For T = 1,000,000 K, w-max = 3,000,000 / T = 3,000,000 / 1,000,000 7./ P 2 = a 3 / (M 1 + M 2 ) P = period of rotation in years a = distance between the objects in Aus M 1 and M 2 = mass of the first and second objects (respectively) in solar masses Average neutron star mass = 1.75 solar masses P = 13.1 / 365 = 0.03589 years a = 0.3037 AU...
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Chapter 10-11 - 2 We have a^3=p^2 p=11 minutes 11...

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