Suppose the Sun contracts (collapses) to a pulsar. Estimate the minimum radius of the pulsar and its period of rotation. The period of revolution of the Sun about its axis is 25.38 days (1 day = 24 hours).

Answer

15 km ; `10^(-3)` s

Explanation

Applying the law of conservation of momentum we can find the relation between the pulsar radius and its period of revolution. We have
`I_(o.)omega_(o.)=Iomega, " or " 2/5MR_(o.)^2(2pi)/(T_(o.))=2/5MR_(o.)^2(2pi)/T`
Hence
`R_(o.)^2/T_(o.)=R^2/T`
To prevent the escape of matter as the speed of rotation is increased the force of gravity should exceed the centrifugal force, i.e. `momega^2RltR_(gr) ` Hence `(4pi^2mR)/(T^2)lt(gammamM_(o.))/(R^2)`
Therefore the second relation sought between the pulsar radius and its period of revolution is of the form
p`R^3/T^2lt(gammaM_(o.))/(4pi^2)`
Eliminating the pulsar period from both equations, we obtain
`Rge(4pi^2R_(o.)^1)/(gammaM_(o.)T_(o.)^2)=(4pi^2xx7^4xx10^(32))/(6.67xx10^(-11) xx2xx10^(30) xx2.2^(2) xx10^(12)) ~~15km`
`T ~~ R^2T _(o.)//R_(o.)^2 ~~10^(-3)8`