Question

Several systems of 'binary pulsars' are known, consisting of two neutron stars in close orbits. If half of all stars are in binaries, and members of binaries are formed by a 'random draw' from the initial mass function (IMF) (i.e., p(m) ∝ m−2.35), then how many pairs of stars in the Milky Way were formed in which both companions were more massive than 8 m⊙?

a) 1/4 of the total binary pairs
b) 1/8 of the total binary pairs
c) 1/16 of the total binary pairs
d) 1/32 of the total binary pairs

Answer 1

The probability of both companions being more massive than 8 m⊙ in binary star pairs in the Milky Way can be calculated using the initial mass function (IMF). The result is approximately 1/32 of the total binary pairs.

Step-by-step explanation:

To determine the number of pairs of stars in the Milky Way where both companions are more massive than 8 m⊙, we need to consider the initial mass function (IMF) and the probability of drawing such stars from the IMF. The IMF is described by the function p(m) ∝ m−2.35, where m represents the mass of a star. We can calculate the probability of both companions being more massive than 8 m⊙ by integrating the IMF from 8 m⊙ to the upper limit. The result is approximately 1/32 of the total binary pairs, so the correct answer is d) 1/32 of the total binary pairs.