To estimate the period of the neutron star after the supernova explosion, we can use the conservation of angular momentum. The angular momentum $L$ of the star before and after the explosion can be expressed as:

$L = I \omega$

where $I$ is the moment of inertia and $\omega$ is the angular velocity.

1. Calculate the moment of inertia before the explosion: The moment of inertia for a solid sphere (which we'll approximate the star as) is given by:

   $I = \frac{2}{5} m R^2$

   where $m$ is the mass and $R$ is the radius.

2. Calculate the angular velocity before the explosion: The angular velocity $\omega$ can be calculated from the period $T$ using:

   $\omega = \frac{2\pi}{T}$

   Given that $T = 35$ days, we convert this to seconds:

   $T = 35 \times 24 \times 60 \times 60 = 3,024,000 \, \text{s}$

   Thus,

   $\omega = \frac{2\pi}{3,024,000}$

3. Calculate the new period of the neutron star: After the explosion, the neutron star has a new radius $R' = 20$ km, and the mass remains the same.

   Using conservation of angular momentum, we have:

   $I_{\text{initial}} \omega_{\text{initial}} = I_{\text{final}} \omega_{\text{final}}$

   Since the mass is constant, we can set up the equation as follows:

   $\frac{2}{5} m R^2 \omega_{\text{initial}} = \frac{2}{5} m R'^2 \omega_{\text{final}}$

   The masses $m$ cancel out, and we can rearrange to find $\omega_{\text{final}}$:

   $R^2 \omega_{\text{initial}} = R'^2 \omega_{\text{final}}$

   Therefore,

   $\omega_{\text{final}} = \omega_{\text{initial}} \left(\frac{R^2}{R'^2}\right)$

4. Substituting values: $R = 5.6 \times 10^6 \, \text{m} \quad (5.6 \times 10^3 \, \text{km})$ $R' = 20,000 \, \text{m} \quad (20 \, \text{km})$

   Now we can compute:

   $\omega_{\text{final}} = \frac{2\pi}{3,024,000} \left(\frac{(5.6 \times 10^6)^2}{(20,000)^2}\right)$

5. Calculate the new period: The period of the neutron star $T'$ is then given by:

   $T' = \frac{2\pi}{\omega_{\text{final}}}$

Now, let's compute the values step by step:

1. Calculate $\omega_{\text{initial}}$:

$\omega_{\text{initial}} = \frac{2\pi}{3,024,000} \approx 2.08 \times 10^{-6} \, \text{s}^{-1}$

2. Calculate the ratio of the radii squared:

$\frac{(5.6 \times 10^6)^2}{(20,000)^2} = \frac{(5.6^2 \times 10^{12})}{(4 \times 10^8)} = \frac{31.36 \times 10^{12}}{4 \times 10^8} = 7.84 \times 10^4$

3. Now find $\omega_{\text{final}}$:

$\omega_{\text{final}} = 2.08 \times 10^{-6} \times 7.84 \times 10^4 \approx 1.63 \times 10^{-1} \, \text{s}^{-1}$

4. Finally, calculate the new period $T'$:

$T' = \frac{2\pi}{1.63 \times 10^{-1}} \approx 38.5 \, \text{s}$

Thus, the estimated period of the neutron star is approximately 38.5 seconds.

If you have any questions or need further details, let me know! Here are some follow-up questions you might consider:

1. How does the conservation of angular momentum apply to other astronomical phenomena?
2. What are the implications of a neutron star's rapid rotation for its properties?
3. How would the period change if the mass of the star decreased?
4. What is the significance of neutron stars in astrophysics?
5. Can you explain how mass and radius relate to the characteristics of different types of stars?

Tip: When dealing with rotational dynamics, always check units and conversions, especially when transitioning from days to seconds or kilometers to meters.