The Crab Nebula One of the most studied objects in the night sky is the Crab nebula, the remains of a supernova explosion observed by the Chinese in 1054. In 1968 it was discovered that a pulsar-a rapidly rotating neutron star that emits a pulse of radio waves with each revolution-lies near the center of the Crab nebula. The period of this pulsar is 33 ms. What is the angular speed (in rad/s) of the Crab nebula pulsar?
What is 2PI/33ms ?
To find the angular speed of the Crab nebula pulsar, we can use the formula:
Angular speed (ω) = 2π / Period (T)
Given that the period of the pulsar is 33 ms, we need to convert it to seconds by dividing it by 1000:
Period (T) = 33 ms / 1000 = 0.033 s
Now we can plug this value into the formula:
Angular speed (ω) = 2π / 0.033 s
Calculating this, we get:
Angular speed (ω) ≈ 189.94 rad/s
Therefore, the angular speed of the Crab nebula pulsar is approximately 189.94 rad/s.