Question

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from the pulsars at regular intervals equal to the period that they rotate. Some of these pulsars rotate with periods as short as 1 ms! The Crab Pulsar, located inside the Crab Nebula in the constellation Orion, has a period currently of length 33.085 ms. It is estimated to have an equatorial radius of 15 km, an average radius for a neutron star.(a) What is the value of the centripetal acceleration of an object on the surface at the equator of the pulsar?m/s2(b) Many pulsars are observed to have periods that lengthen slightly with time, a phenomenon called "spin down." The rate of slowing of the Crab Pulsar is 3.50 multiply.gif 10-13s per second, which implies that if this rate remains constant, the Crab Pulsar will stop spinning in 9.50 multiply.gif 1010 s (about 3000 years from today). What is the tangential acceleration of an object on the equator of this neutron star?m/s2

Answer 1

(a)
`a_(c) = 5.41* 10^(9) m/s^(2)`

(b)
`a_(t) = 2.99* 10^(- 5) m/s^(2)`

Given:

Time period of Pulsar,
`T_(P) = 33.085 ms == 33.085* 10^(- 3) s`

Equatorial radius, R = 15 Km = 15000 m

Spinning time,
`t_(s) = 9.50* 10^(10)`

Solution:

(a) To calculate the value of the centripetal acceleration,
`a_(c)` on the surface of the equator, the force acting is given by the centripetal force:

`m* a_(c) = (mv_(c)^(2))/(R)`

`a_(c) = (v_(c)^(2))/(R)` (1)

where

`v_(c) = (distance covered(i.e., circumference))/( T)`

`v_(c) = (2\pi R)/(Time period, T)` (2)

Now, from (1) and (2):

`a_(c) = R\frac({2\pi )^(2)}{T^(2)}`

`a_(c) = 15000(2\pi )^(2))/((33.085* 10^(- 3))^(2))`

`a_(c) = 5.41* 10^(9) m/s^(2)`

(b) To calculate the tangential acceleration of the object :

The tangential acceleration of the object will remain constant and is given by the equation of motion as:

`v = u + a_(t)t_(s) = 0`

where

u =
`v_(c)`

`a_(t) = - (2\pi R)/(Tt_(s))`

`a_(t) = - (2\pi 15000)/(33.085* 10^(- 3)* 9.50* 10^(10))`

`a_(t) = 2.99* 10^(- 5) m/s^(2)`