23P2Q08

8a)

\displaystyle T=\frac{6.0}{10}=0.60\text{ s}

\displaystyle \omega =\frac{2\pi }{T}=\frac{2\pi }{0.60}=10.5\text{ rad }{{\text{s}}^{-1}}

8b)

The pulsar and the parent star must have the same angular momentum.

Since the pulsar has a much smaller radius, it must have a much larger angular velocity.

8c)

Binding energy per nucleon peaks near iron-56.

The fusion of iron to form even larger nuclei requires energy input (instead of releases energy).

8di)

\displaystyle \begin{aligned}   g&=\frac{GM}{{{R}^{2}}} \\  &=\frac{(6.67\times {{10}^{-11}})(1.4\times 2.0\times {{10}^{30}})}{{{10000}^{2}}} \\  &=1.87\times {{10}^{12}}\text{ N k}{{\text{g}}^{-1}} \end{aligned}

8dii)

\displaystyle \begin{aligned}  \Delta g&=GM(\frac{1}{{{r}_{1}}^{2}}-\frac{1}{{{r}_{2}}^{2}}) \\ & =(6.67\times {{10}^{-11}})(1.4\times 2.0\times {{10}^{30}})(\frac{1}{{{10000}^{2}}}-\frac{1}{{{10001}^{2}}}) \ & =(6.67\times {{10}^{-11}})(1.4\times 2.0\times {{10}^{30}})(1.9997\times {{10}^{-12}}) \\ & =3.73\times {{10}^{8}}\text{ N k}{{\text{g}}^{-1}} \end{aligned}

8diii)

It becomes longer than 3.0 m.

The difference in gravitational force across the two ends of the rod is so large that the intermolecular forces of the rod is not able to keep the entire rod accelerating at the same rate.

8ei)

\displaystyle \begin{aligned}   v&=r\omega  \\ & =(100000)(\frac{2\pi }{2.0\times {{10}^{-3}}}) \\ & =3.14\times {{10}^{8}}\text{ m }{{\text{s}}^{-1}} \\ & >c \end{aligned}

The speed of the surface cannot exceed the speed of light.

8eii)

The pulsar disintegrates if, at a point on the surface of the pulsar, the gravitational acceleration is equal to the centripetal acceleration.

\displaystyle \begin{aligned}   \frac{GM}{{{r}^{2}}}&=r{{\omega }^{2}} \\  \frac{G(\frac{4}{3}\pi {{r}^{3}}\rho )}{{{r}^{2}}}&=r{{(\frac{2\pi }{T})}^{2}} \\  \rho &=\frac{3\pi }{G{{T}^{2}}} \end{aligned}

8eiii)

\displaystyle \rho =\frac{3\pi }{G{{T}^{2}}}=\frac{3\pi }{(6.67\times {{10}^{-11}}){{(0.60)}^{2}}}=3.93\times {{10}^{11}}\text{ kg }{{\text{m}}^{-3}}

8f)

\displaystyle \text{age}=\frac{T}{2\Delta T}=\frac{33.5\times {{10}^{-3}}}{2[(38\times {{10}^{-9}})\div (24\times 60\times 60)]}=3.808\times {{10}^{10}}\text{ s}=1210\text{ years}

8g)

\displaystyle \begin{aligned}  {{E}_{photon}}&=\frac{hc}{\lambda } \\  (400\times {{10}^{12}})(1.60\times {{10}^{-19}})&=\frac{(6.63\times {{10}^{-34}})(3.00\times {{10}^{8}})}{\lambda } \\  \lambda &=3.11\times {{10}^{-21}}\text{ m} \end{aligned}

8h)

With a larger collecting area, the SKA can collect a larger power for the same intensity of radio signals, allowing it to detect lower intensity pulsars.

2 thoughts on “23P2Q08”

  1. Pingback: A-Level 2023 – xmPhysics
  2. Pingback: A-Level 2024 – xmPhysics

Leave a comment