91Ó°ÊÓ

The radius of star is,

$$\begin{gathered} \mathrm{r}=20\mathrm{km} \\ =20×{10}^3\mathrm{m} \end{gathered}$$

The period of the star is T = 1.00 s

A neutron star is rotating once at every second. The particle is placed on the star’s equator; hence it covers the distance as circumference of circle within time T. We can use the formula of speed and centripetal acceleration and find the effect of faster rotation of the neutron star on the speed and centripetal acceleration of the particle.

Formula:

The speed of a particle on a star’s equator is,

$v=\frac{2\mathrm{Ï€°ù}}{T}$ …(¾±)

The centripetal acceleration is,

$a=\frac{v^2}{r}$ …(¾±¾±)

Use equation (i) to calculate the speed of the particle.

$$\begin{gathered} \mathrm{v}=\frac{2\mathrm{Ï„Ï„°ù}}{\mathrm{T}} \\ =\frac{2\mathrm{Ï„Ï„}×20×{10}^3\mathrm{m}}{1.00\mathrm{s}} \\ =1.3×{10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the particle on the star’s equator is $1.3×{10}^5\mathrm{m}/\mathrm{s}$.

Use equation (ii) to calculate the magnitude of the particle’s centripetal acceleration.

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}} \\ =\frac{{(1.3×{10}^5\mathrm{m}/\mathrm{s})}^2}{20×{10}^3\mathrm{m}} \\ =7.9×{10}^5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$$7.9×{10}^5\mathrm{m}/{\mathrm{s}}^2$

Therefore, the magnitude of the particle’s centripetal acceleration is $7.9×{10}^5\mathrm{m}/{\mathrm{s}}^2$.

If the rotation of neutron star increases, its period decreases. Hence, the speed of a particle on star’s equator increases. The magnitude of the particle’s centripetal acceleration depends on speed of the particle; hence,this increasing also increases centripetal acceleration.