Solutions to Intermediate Astronomy Problems (1)

Recall from the blog on intermediate astronomy problems:

1) For the example problem given, if the distance to Cepheid A = 10 pc, find the distance to Cepheid B.


Solution

We already know Cepheid (A) is 2.512 times brighter than the shorter period one, B.
Thus, B(A) = 2.512 B(B).

We also know: d(A) = 10 pc

From the inverse square law for light:

(d1/d2)^2 = B2/ B1

or, in this case, letting d2 = d(A), so B2 = B(A):

[d(B)/d(A)]^2 = 2.512

and:

[d(B)/10 pc]^2 = 2.512

[d(B)/10 pc] = [2.512]^½ = 1.58

Then: d(B) = 1.58 (10 pc) = 15. 8 pc or 51.5 Ly


2) One Cepheid is found to have a period of 6.32 days and a mean apparent magnitude of +4.5. Another Cepheid in the same constellation is found to have a period of 15.9 days and a mean apparent magnitude of +7. If the P-L relationship for both has an uncertainty of +/- 0.35 m, find the approximate distance for each one.

Solution

For the Cepheid with Period(P) = 6.32 days, log P = log(6.32) = 0.8

For the Cepheid with Period(P) = 15.9 days, log P = log (15.9) = 1.2

Using the P-L graph (see last blog) the intrinsic brioghtness associated with a period of log P = 0.8 is M_v= -3.2, and for log P = 1.2 it is (-4). Assuming M_v= abs. magnitude, and using the distance modulus equation:

(m - M) = 5 log r - 5, we have:

for Cepheid (1):

log r1 = (m - M + 5)/ 5 = [{+4.5 - (-3.2) + 5]/5 = 12.7/5 = 2.54

and: antilog_10 (2.54) = 346.7 so r1 = 346.7 pc

for Cepheid (2):

log r2 = [+7 - (-4)+5 ]/5 = 16/5 = 3.2

and antilog _10(3.2) = 1584.9

so: r2 = 1584.9 pc

Now, let r1 and r2 have probable errors of e_r1 and e_r2, respectively. Then:

e_r1/r1 = -0.46 (delta M) and e_r2/r2 = -0.46 (delta M)

Finally:

e_r1 = (346.7 pc)(0.46)(+/-0.35) = +/- 55.6 pc

and

e_r2 = (1584.9 pc)(0.46)(+/-0.35) = +/- 255.2 pc


3) Two Cepheids, Alpha and Beta are observed to have the same period of 10 days. at maximum brightness A has an apparent magnitude of +3.0 and B has an apparent magnitude of +8.0. If the distance of A (associated with a cluster) is known to be 60 pc, how far away is B?

Solution

The magnitude difference between Alpha and Beta is (8 - 3) = 5 magnitudes, which corresponds to a brightness ratio of 100. (E.g. (2.515)^5 = 100) so Alpha is apparently 100x brighter than B. According to the inverse square law of light, the brightness of a light source diminishes as the square of the distance.

Accordingly:

[d(Beta)/d(Alpha)]^2 = 100 = B(alpha)/ B(Beta)

and


[d(Beta)/d(Alpha)] = [100]^½ = 10

and: d(Beta) = 10 {d(Alpha)]

So, Beta must be ten times more distant than Cepheid Alpha.



4) A CXC astronomy student plots a light curve for the Cepheid Zeta Geminorum as shown (Fig. 2). Using this curve and the P-L graph, estimate the brightness of this Cepheid in terms of absolute visual magnitude.

Solution


We bring up the student's sketch of his light curve again (top).

The period will be found as the time between two exactly similar phases, or 'peak to peak' on the curve. This is the time from day 7, to day 18 or (18 d - 7 d) = 11 d.

Take the log of the period: log (11) = 1.04

and now use the P-L graph to obtain the corresponding absolute visual magnitude:

We obtain M_v = (-3.8) approx.