The clear recursive relation is that for every 5th term, the pattern increases by a factor of 10^2 which makes sense given that \(10^{0.4(5)} = 10^2\). This process repeats until infinity with the same constants leading the recursive pattern.
Saturday, October 3, 2015
Blog #14, Worksheet #5.1, Distance Ladder, Problem #1
1) Many students consider the astronomical magnitude system for measuring stellar fluxes baffling and
a bit scary. With this exercise, I’d like to convince you otherwise. Magnitudes are defined such
that, for Star One and Star Two with fluxes \(F_1\) and \(F_2\), \[\frac{F_1}{F_2} = 10^{0.4 \cdot (m_1 - m_2)}\]Where \(m_1\) is the magnitude of Star One and \(m_2\) is the magnitude of Star Two. Notice that, to two
sig figs, \(10^{0.4}\) is just 2.5. As a result \[\frac{F_1}{F_2} = 2.5^{(m_1 - m_2)}\] Note also that a larger magnitude corresponds to a fainter star. Using this information, fill out the
following table, where ∆m = m2 - m1, and identify the simple (approximate) recursion relation:
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Good job Simon! To be more specific, your entry under 2,7,... should be more like 6.25, 625,...etc. But you're definitely right to an order of magnitude :) 5/5
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