1. A well-calibrated version of the Tully-Fisher (TF) relationshipcan be express
ID: 2268992 • Letter: 1
Question
1. A well-calibrated version of the Tully-Fisher (TF) relationshipcan be expressed as M1-1996-811 logo (200 km s-1 ), where Mi is the galaxy's I-band absolute magnitude and W, its velocity width, is basically twice its peak rotation speed. (a) Show that the TF relationship is equivalent to a power-law relationship of the form LI W®, where L1 is the I-band luminosity, and find the value of . (b) For the I-band TF relationship, the uncertainty in the predicted value of M is about ±0.1 magnitudes. If this relationship is used to estimate a galaxy's distance d, what percentage uncertainty results?Explanation / Answer
1. from the given data
M1 = -19.6 - 8.11*log(W/200)
a. this can be written as
-(M1 + 19.6)/8.11 = log(W/200)
W = 200*10^(-(M1 + 19.6)/8.11)
now,
absolute magnitude M is givne by
L = k*10^(nM)
where k and n are some constants and L is luminosity
hence
L/k = 10^(nM)
log(L/k) = nM
M = (1/n)log(L/k) = -19.6 - 8.11*log(W/200)
log(L/k) = -19.6n - 8.11n*log(w/200)
log(L/k) + log((W/200)^8.11n) = -19.6 n
hence
log((L/k)(W/200)^8.11n) = -19.6n
hence
k*10^(-19.6n)*200^8.11n = L*W^8.11n
or, taking 8.11n = - alpha
and other constants as A
L = AW^alpha
where A is a constant, and alpha is a constant
b. given
dM1 = 0.1
relationship between absolute magnitude and distance is
d = 10^0.2(m - M + 5)
log(d) = 0.2(m - M + 5)
d(d)/d = 0.2(dm - dM )
for dm = 0
d(d)/d = 0.2*dM = 0.02
hence relative error in measurement of d is 0.02*100 = 2%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.