CHAPTER 11 - So, angular diameter is about 1/1000...

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Astronomy –odd Homework PROBLEM 11 1./ The Radius of the Star is 10 KM. Diameter= 10 x 2 = 20. Circumference: 20 * 3.1416 = 62.838. 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 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.
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Unformatted text preview: So, angular diameter is 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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