Common Astronomy Calculations on the TI-30XA Matter-Energy Equivalence Suppose you wish to compute the energy liberated when 1 kg of matter is converted to pure energy. To do this you use Einstein’s’ equation 2mcE = where smc /1038×= - the speed of light in free space ()()1628109/1031 ×=×× smkg joules On your TI-30XA (data entry is in normal font, computation keys are underlined, result is bold) 1 × 3 EE 8 x2 = 9. 16 Note: 9. 16 is how the TI-30XA represents 9.0 × 1016 in it’s normal mode. Notice that we are not concerned about carrying the units (kg, m/s, etc.) through the calculation and there is no provision for keeping track of units on the TI-30 even if we wanted to. It is the number that matters most. You should know, however, that the result of this calculation is in joules since it is energy. Let’s try doing the same calculation with 1 gram of matter instead of 1 kilogram. Recall that a gram is 1/1000th of a kilogram or 10-3 kg. ()()13283109/103101 ×=×××−smkg joules 1 EE 3 +/- × 3 EE 8 x2 = 9. 13 Suppose you wish to compute a mass deficit. The example given in class was from the proton-proton cycle. 4 × 1 hydrogen = 4 × 1.673 × 10-27 kg = 6.692 × 10-27 kg - 1 helium = 1 × 6.645 × 10-27 kg = 6.645 × 10-27 kg ________________________________________________________ = 0.047 × 10-27 kg On your TI-30XA: 6.693 EE 27 +/- - 6.645 EE 27+/- = 4.7 –29 Which is the same as 0.047 × 10-27 kg.Distance Suppose you wish to compute the distance to a star with a measured parallax half angle of p = .75 arc seconds: 3.175.01sec==spard parsecs (~ 4.3 LY) On your TI-30XA: .75 1/x 1.3333333 Wien’s Law Suppose you wish to compute the temperature of our sun from the color of the photosphere. Our sun radiates most strongly at λmax = 500nm. Wien’s Law yields: KnmTK60005001036=×= On your TI-30XA: 3 EE 6 ÷ 500 = 6000 The Stefan-Boltzman Law Suppose you wish to compute the luminosity of an object given its temperature and surface area. Recall that this may be accomplished with the Stefan-Boltzman law: 4TEσ= , where E = L/A, σ = 5.67 × 10-8 W/m2 K4 (Stefan-Boltzman constant). More conveniently written: 244 RTLπσ=for an isotropic radiator (spherical). Our sun has a radius of 7 × 108 km and a surface temperature of about 6000 K. ()()()()262848104107460001067.5 ×≈××=−πL 5.67 EE 8 +/- × 6000 yx 4 × 4 × π × 7 EE 8 x2 = 4.5247 26 Dénouement These examples should be sufficient to get you through the calculations you need for astronomy class. Some practice with your calculators will insure success. As always, feel free to ask for help if you need