A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of T = 0.033 s that is increasing at the rate of
1.26×10−5s/y
. (a) What is the pulsar’s angular acceleration α? (b) If α is constant, how many years from now will the pulsar stop rotating? (c) The pulsar originated in a supernova explosion seen in the year 1054. Assuming constant α, find the initial T.

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Olajidey Olajidey · Answer 1 · 2020-03-14T03:58:52+01:00

Answer:

α [tex]= -2.303 \times 10^-^9rad/s^2[/tex]

b)t = 2619.8 years

c) T(i) = 0.0208536s

Explanation:

Given that ,

The period of pulsar T = 0.033 s

The period increase at a rate of [tex]\frac{dT}{dt}[/tex] = 1.26×10⁻⁵s/y

a) Pulsar angular speed is

ω = θ / T = 2π / T

θ = one revolution of pulsar

T = period of pulsar

Pulsar angular acceleration is given by

[tex]\alpha = \frac{dw}{dt}[/tex]

[tex]\alpha = -\frac{2\pi }{ T^2}\frac{dT}{dt}[/tex]

[tex]\alpha = -\frac{2\pi }{0.033^2} \frac{1.26 \times 10^-^5}{60\times60\times24\times365.25}[/tex]

[tex]= -2.303 \times 10^-^9rad/s^2[/tex]

b) ω₀ = θ / T

0 = 190.4 - 2.303 × 10⁻⁹

[tex]t = \frac{190.4 }{2.303 \times10^-^9} \\= 8.27 \times10^1^0s \\= 2619.8y[/tex]

c) 2018 - 1054

= 964years

The pulsar is originated in a supernova explosion 964 years ago.

then the initial period of pulsar

[tex]T_i = T - 964 \times \frac{dT}{dt} = 0.033 - 964 \times 1.26 \times 10^-^5 \\= 0.0208536s[/tex]

T(i) = 0.0208536s