Subfield: Mechanics

The rotating neutron star. 

A 'millisecond pulsar' is a source of radiation in the universe that emits very short pulses with a period of one to several milliseconds. This radiation is in the radio range of wavelengths; and a suitable radio receiver can be used to detect the separate pulses and thereby to measure the period with great accuracy.

These radio pulses originate from the surface of a particular sort of star, the so-called neutron star. These stars are very compact: they have a mass of the same order of magnitude as that of the sun, but their radius is only a few tens of kilometers. They spin very quickly. Because of the fast rotation, a neutron star is slightly flattened (oblate). Assume the axial cross-section of the surface to be an ellipse with almost equal axes. Let $r_{p}$ be the polar and $r_{e}$ the equatorial radii; and let us define the flattening factor by:

$$
\epsilon=\frac{\left(r_{e}-r_{p}\right)}{r_{p}}
$$



| a mass of | $2.0 \cdot 10^{30} \mathrm{~kg}$, |
| :--- | :--- |
| an average radius of | $1.0 \cdot 10^{4} \mathrm{~m}$, |
| and a rotation period of | $2.0 \cdot 10^{-2} \mathrm{~s}$. |

Question: a - Calculate the flattening factor, given that the gravitational constant is $6.67 \times 10^{-11}$ N.m $\cdot \mathrm{kg}^{-2}$.

Solution:
a - 1st method

For equilibrium we have $\mathrm{F}_{\mathrm{c}}=\mathrm{F}_{\mathrm{g}}+\mathrm{N}$ where $\mathrm{N}$ is normal to the surface.

Resolving into horizontal and vertical components, we find:

<img_4429>

$$
\begin{aligned}
& F_{g} \cdot \cos (\phi)=F_{c}+N \cdot \sin (\alpha) \\
& \quad F_{g} \cdot \sin (\phi)=N \cdot \cos (\alpha)
\end{aligned} \rightarrow F_{g} \cdot \cos (\phi)=F_{c}+F_{g} \cdot \sin (\phi) \cdot \operatorname{tg}(\alpha)
$$

From:

$$
F_{g}=\frac{G \cdot M}{r^{2}}, \quad F_{c}=\omega^{2} \cdot r, x=r \cdot \cos (\phi), y=r \cdot \sin (\phi) \text { en } \operatorname{tg}(\alpha)=\frac{d y}{d x}
$$

we find:

$$
y \cdot d y+\left(1-\frac{\omega^{2} \cdot r^{3}}{G \cdot M}\right) \cdot x \cdot d x=0
$$

where:

$$
\frac{\omega^{2} \cdot r^{3}}{G \cdot M} \approx 7 \cdot 10^{-4}
$$

This means that, although $r$ depends on $x$ and $y$, the change in the factor in front of $x d x$ is so slight that we can take it to be constant. The solution of Eq. (1) is then an ellipse:



$$
\frac{x^{2}}{r_{e}^{2}}+\frac{y^{2}}{r_{p}^{2}}=1 \rightarrow \frac{r_{p}}{r_{e}}=\sqrt{1-\frac{\omega^{2} \cdot r^{3}}{G \cdot M}} \approx 1-\frac{\omega^{2} \cdot r^{3}}{2 \cdot G \cdot M}
$$

and from this it follows that:

$$
\epsilon=\frac{r_{e}-r_{p}}{r_{e}}=\frac{\omega^{2} \cdot r^{3}}{2 \cdot G \cdot M} \approx 3,7 \cdot 10^{-4}
$$
2nd method 

For a point mass of $1 \mathrm{~kg}$ on the surface,

$$
U_{p o t}=-\frac{G \cdot M}{r} \quad U_{k i n}=\frac{1}{2} \cdot \omega^{2} \cdot r^{2} \cdot \cos ^{2}(\phi)
$$

The form of the surface is such that $U_{\text {pot }}-U_{\text {kin }}=$ constant. For the equator $(\Phi=0$, $\left.r=r_{e}\right)$ and for the pole $\left(\Phi=\pi / 2, r=r_{p}\right)$ we have:

$$
\frac{G \cdot M}{r_{p}}=\frac{G \cdot M}{r_{e}}+\frac{1}{2} \cdot \omega^{2} \cdot r_{e}^{2} \rightarrow \frac{r_{e}}{r_{p}}=1+\frac{\omega^{2} \cdot r_{e}^{3}}{2 \cdot G \cdot M}
$$

Thus:

$$
\epsilon=\frac{r_{e}-r_{p}}{r_{e}}=\frac{1+\frac{\omega^{2} \cdot r_{e}^{3}}{2 \cdot G \cdot M}-1}{1+\frac{\omega^{2} \cdot r_{e}^{3}}{2 \cdot G \cdot M}} \approx \frac{\omega^{2} \cdot r_{e}^{3}}{2 \cdot G \cdot M} \approx 3,7 \cdot 10^{-4}
$$

Final answer: $3.7 \times 10^{-4}$