AST201 Assignment 2 Solution 1 a Puzzling Binary

AST201 Assignment 2 Solution

1 A puzzling binary

PSR B2303+46 is a radio pulsar in an eccentric binary with a 12-day orbit. Prof. Shri Kulkarni (of the California Institute of Technology) and one of your instructors (MHvK) have found that the other star in the binary is a massive white dwarf. Given that the binary is eccentric, we know that the last thing that happened was the supernova explosion that formed the pulsar. Thus, the pulsar in this binary was formed later than the white dwarf. (a) [5 points] Look up for what initial masses a star is expected to end its life by forming a neutron star, and for what initial masses it is expected to leave a white dwarf. Which is expected to live longer, the ones that leave neutron stars or those that leave white dwarfs? Given this, and the fact that stars in binaries are born at the same time, what is puzzling about the above binary?Feel free to answer this question in point form.

Solution. From the book (e.g., p. 574), stars with masses above 8 M end as neutron stars [1 point], stars with masses below as white dwarfs [1 point]. Stars forming a neutron star live shorter [1 point], since they are more massive [1 point]. Thus, in the binary one would have expected the neutron star to form ﬁrst rather than second [1 point]. Many people said 4-8M was the range for neutron star from the NASA website. This is a case where a seemingly respective source is not correct. The textbook makes clear distinction between high, intermediate and low mass star. No marks were deducted for this answer. 0.5 marks were deducted for giving a reasonable but incomplete mass range. Also many people were confused and wrote down the mass of a neutron star and a white dwarf. The Chandrasekhar limit of a white dwarf has nothing to do with this question. (b) [6 points] How could the paradox be resolved? I.e., what kind of life could lead to a binary system like the one we observe? Your answer should give a brief, point-form description of the major stages of the preceding life of such a binary (Hint: read Section 17.4 of the text book.) Solution. In a binary, it is possible to transfer mass, leading to an apparently more evolved, less massive object around a less evolved, more massive one . A brief outline would be: start with two stars of unequal mass, one close to but below 8 M ; the more massive one evolves and transfers most of its envelope, increasing the mass of its companion to beyond 8 M ; the originally more massive star ends its life as a white dwarf ; the originally less massive star is now more massive than 8 M and thus ends its life in a supernova explosion, leaving a neutron star. Most people were able to get that mass transfer was required. Up to 4 marks were awarded for explaining mass transfer in a binary as explain in the textbook. Many people failed to mention that the white dwarf was formed from the originally more massive star. 2 additional marks were awarded for mentioning the correct progenitors for the white dwarf and the pulsar. 1 mark was deducted if the explanation was too long.

1 2 Stars in clusters and the mystery of the Blue Strag- glers

In class, we discussed the reasons star clusters are extremely useful to astronomers: all stars are at the same distance and all stars have the same age. As a result, stars in a cluster only appear at well-deﬁned locations in an H-R diagramme. As we will see, however, some exceptions are found.

(a) [6 points] In a few sentences, summarize why in a cluster of stars one expects to see stars on the main sequence only up to a certain ‘turn-oﬀ mass.’ What would you expect more massive stars to look like? (List at least four possibilities.) Solution. The cluster was formed some certain amount of time ago [1 point], and only stars that have main-sequence lives longer than this age can still be around [1 point]. Stars with masses above the turn-oﬀ mass could be seen as subgiants, red giants, horizontal branch stars, AGB stars, planetary nebulae or white dwarfs [1 point each, up to 4]. Supergiants and hypergiants are not valid answers, as they do not exist in globular clusters. However neutron stars and black holes are seen in globular clusters, therefore they are valid answers and will be given credits.

(b) [3 points] In Fig. 1, the observed H-R diagram is shown of the globular cluster NGC 288. You see a clear turn-off away from the main-sequence to the giant branch at V −I = 0.6 and V = 18.5. However, there are also a number of exceptions: these are the so-called Blue Stragglers. Explain in one or two sentences why, given your answer to (a), it is surprising to find these stars? Solution. From their location, these stars appear to be main-sequence stars [1 point], but with masses beyond the turn-off mass [1 point]. With such high masses, their main- sequence life times are shorter than the age of the cluster, and thus they should have ended their main-sequence phase [1 point].

(c) [5 points] Research blue stragglers and summarize concisely (no more than a third of a page) at least two possible explanations for their existence. (Hint: One possibility should be obvious from question 1!) What common ingredient do all explanations share? Solution. Possibilities include: (1) Mass transfer in a binary. In this case the binary still remains a binary even after a blue straggler is formed. (2) Merger between two stars (after a collision, or in a binary). In this case two stars merge and form one blue straggler. (3) Extra mixing due to rapid rotation. In this case there is only one star involved in forming a blue straggler. [2 points per valid explanation]. All solutions share that additional fuel is brought to the core [1 point].

(d) [5 points] In Fig. 1, you also see that while most of the stars along the main sequence form a very narrow band, there are some that are slightly brighter and/or redder.

2 What could these be? (In order to verify your answer, you may ﬁnd it useful to know that a factor two in brightness equals 0.75 units along the vertical axis.) Solution. Binaries [1 point] that cannot be resolved in the image [1 point]. The summed light will exceed that of one star, and hence the object will appear above the main sequence [1 point]. For equal-mass binaries, it will be a factor two [2 points]; for unequal mass, the binary would appear less than a factor two brighter, but its color would be redder [2 points, bonus].