In 1967, Jocelyn Bell Burnell and Antony Hewish observed pulsed radio signals, o
ID: 1655500 • Letter: I
Question
In 1967, Jocelyn Bell Burnell and Antony Hewish observed pulsed radio signals, originating from the same location in the sky, arriving on Earth every 1.33 seconds. The object that emitted these signals, originally named LGM-1 ("Little Green Men") as a humorous reference to the slight possibility of the signal's alien origin, was later determined to be a rapidly rotating neutron star that shoots out a sweeping beam of radio-wave radiation like a lighthouse. These stars are called pulsars, and for every rotation of the pulsar one radio pulse is received on Earth. One of the most dramatic pulsars in the sky is PSR B0531 + 21 in the Crab Nebula, with an X-ray beam and an extremely fast period of rotation of T = 033. While pulsars are almost perfectly regular, they do slow down very gradually over time: for example, PSR B0531 + 21 slows by 38 nanoseconds every day. a. What is the PSR B0531 + 21's angular acceleration alpha? b. If alpha is constant, how many years from now will PSR B0531 + 21 stop rotating? c. Neutron stars form when a large star dies in a violent explosion called a supernova. The Crab Nebula itself is the debris generated when PSR B0531 + 21's parent star exploded. This supernova was seen in the year 1054 by early astronomers in China, Japan, and the Middle East. Assuming constant alpha, find the initial period, T_0, of PSR B0531 + 21's rotation.Explanation / Answer
(a) Here, = 2 / =2 * 3.1416 / 0.033 s= 190.4 rad/ s
1 year = 365.25 days * 24 hrs/day * 360 sec /hour =31,557,600 sec
Now, the PSR B0531+21 slows down at the rate of 38 nano sec. every day.
So, slowing rate per year = 38x10^-9x365 = 1.39x10^-5 s/year
So -
= d/dt =-2 / ^2 d/dt
=- (2 * 3.1416 / 0.033^2) s^-2 * 1.39*10^(-5)s/year * 1 year/31,557,600 sec = - 2.54 x 10^-9 rad/s^2
(b) = 0 + t =190.4 /s - 2.54x10^-9 rad /s^2 t =0
=> t = 190.4 /(2.54 x10^-09) = 7.50 x 10^10 seconds = 2377 years
(c) 0 = - t = 190.4 /s -(- 2.54x10^-09 /s^2) *(1967-1054)* 31,557,600 sec = 264 /sec
= 2 / =2 * 3.1416/264/sec=0.0238 sec