astronomy math question

卓霖

2013-09-17 10:49 pm

1)You are looking at one star for a year and notice that it appears to shift back and forth, a condition nown as stellar parallax. From the shifting, you determine that the maximum angular separation from the initially observed position is 0.026 arcseconds. What is the distance in parsecs to this star? What is this distance in AU?

2)The nuclear bomb called an “H-bomb” released 54 TeraJoules of energy to destroy an entire city. How much matter (in grams) was used?

3)The Hipparchos satellite designed to measure the parallax motion of the stars measures a star that has a full angular shift over 6 months of 0.008 arcseconds. It is determined to have a brightness of 6.013 x 10-10 W/m2, and its wavelength of maximum emission of continuous spectrum is found to be at 743.59 nm. How many times larger than the Sun is this star?

回答 (1)

天同

2013-09-18 12:58 am

✔ 最佳答案

1. 1 arcsec = 1/3600 degrees
Disatnce = 1/tan(0.026/2x3600) AU = 1.587 x 10^7 AU = 1.587x10^7/206265 pc
= 76.9 pc
(Take 1 pc = 206 265 AU)

2. 1 TeraJoule = 10^12 J
Using E = mc^2
where E is the energy in Joule
m is the mass in kg
c is the speed of light (= 3 x10^8 m/s)

Hence, mass vanished = 54 x 10^12/(3x10^8)^2 kg = 6 x 10^-4 kg = 0.6 g

3. 0.008 arcsec = 0.008/3600 degrees = 2.22 x 10^-6 degrees = 3.879x10^-8 rad
Distance of star from earth
= 2/3.879x10^-8 AU = 5.156x10^7 AU = 7.734x10^18 m

Luminosity of the star = 6.013x10^-10 x [4 x pi x (7.734x10^18)^2] w
= 4.52x10^29 w

Using Wein's Displacement Law, temperature of star T is
T = 2.9x10^-3/(743.59x10^-9) K = 3900 K

Hence, 4.52x10^29 = (4.pi.R^2).(5.67x10^-8).(3900)^4
where R is the radius of the star, and (5.67x10^-8 w/m^2.K^4 is Stefan's constant)
solve for R gives R = 5.24x10^10 m

Because the radius of the sun is 6.96x10^8 m, the radius of the star is 5.24x10^10/6.96x10^8 = 75 times longer than that of the sun. The star is therefore 420,000 times larger in volume than the sun.

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