2023 H2 Physics Paper 2 Q8 — Pulsars: How Fast Can a Star Spin?

The model, live

Radio beam / pulsesGravityLimit brokenWithin the limit

Part (a)

2 marks

Use the radio trace to find the angular velocity of the first pulsar discovered.

Approach

Each pulse is one sweep of the beam past the Earth, so the time between pulses is the period of rotation. Measure across as many pulses as possible: the first is at $0.40\,\text{s}$ and the thirteenth at $7.6\,\text{s}$ , twelve intervals.

T = \frac{7.6 - 0.40}{12} = 0.60\,\text{s} \qquad \omega = \frac{2\pi}{T} = \frac{2\pi}{0.60} = 10.5\,\text{rad s}^{-1}

angular velocity $= 10\,\text{rad s}^{-1}$

Watch

The pulse heights vary a lot. Use the positions of the peaks, not their heights, and don't skip the small ones.

Part (b)

1 mark

Given that angular momentum $L = \tfrac25 mr^2\omega$ is conserved, suggest why a pulsar spins much faster than the star it formed from.

As the core collapses its radius decreases enormously while its mass stays about the same. Since $L = \tfrac25 mr^2\omega$ is conserved, $\omega \propto 1/r^2$ , so the angular velocity increases greatly.

Note

The effect is enormous. A core shrinking from the size of the Sun ( $7\times10^{5}\,\text{km}$ ) to $10\,\text{km}$ shrinks its radius by a factor of $7\times10^{4}$ , so its spin speeds up by about $5\times10^{9}$ .

Part (c)

2 marks

Explain why the iron in the pulsar's crust did not undergo fusion.

Approach

Fusion releases energy only if the product has a higher binding energy per nucleon than the nuclei that fuse. Iron sits at the peak of the binding-energy-per-nucleon curve.

Iron nuclei have the highest binding energy per nucleon. Fusing iron into heavier nuclei would give products with a lower binding energy per nucleon, so the process would absorb energy rather than release it. Fusion therefore cannot continue beyond iron.

Part (d)(i)

2 marks

The pulsar has mass $1.4$ solar masses (the Sun is $2.0\times10^{30}\,\text{kg}$ ) and radius $10\,\text{km}$ . Find the gravitational field strength at its surface.

M = 1.4(2.0\times10^{30}) = 2.8\times10^{30}\,\text{kg}

g = \frac{GM}{R^2} = \frac{(6.67\times10^{-11})(2.8\times10^{30})}{(1.0\times10^{4})^2} = 1.87\times10^{12}\,\text{N kg}^{-1}

$g = 1.9\times10^{12}\,\text{N kg}^{-1}$

Part (d)(ii)

1 mark

Find the difference $\Delta g$ between the surface and a point $1.0\,\text{m}$ above it.

Approach

Subtracting two values of $GM/r^2$ that agree to four figures loses most of the precision on a calculator. Either keep all the digits, or use the gradient: $\dfrac{dg}{dr} = -\dfrac{2GM}{r^3} = -\dfrac{2g}{r}$ .

\Delta g = GM\left(\frac{1}{(10\,000)^2} - \frac{1}{(10\,001)^2}\right) \approx \frac{2g}{R}(1.0) = \frac{2(1.87\times10^{12})}{1.0\times10^{4}} = 3.7\times10^{8}\,\text{N kg}^{-1}

$\Delta g = 3.7\times10^{8}\,\text{N kg}^{-1}$

Note

That difference over one metre is about forty million times the whole of $g$ at the Earth's surface.

Part (d)(iii)

1 mark

A $1.0\,\text{m}$ steel rod falls vertically towards the pulsar from $3.0\,\text{m}$ above its surface. Suggest what happens to its length, with a reason.

The rod is stretched (and pulled apart). Its lower end is in a much stronger gravitational field than its upper end, so the lower end has a much greater acceleration and pulls away from the upper end.

Part (e)(i)

2 marks

Explain, with a calculation, why a pulsar of radius $100\,\text{km}$ cannot have a period of $2.0\,\text{ms}$ .

Approach

The passage gives two limits. Test the first: the speed of a point on the equator.

v = \frac{2\pi r}{T} = \frac{2\pi(1.0\times10^{5})}{2.0\times10^{-3}} = 3.1\times10^{8}\,\text{m s}^{-1}

The surface at the equator would move at $v = 2\pi r/T = 3.1\times10^{8}\,\text{m s}^{-1}$ , faster than the speed of light ( $3.0\times10^{8}\,\text{m s}^{-1}$ ), which is not possible.

Part (e)(ii)

4 marks

Show that the minimum average density of a spherical pulsar that does not disintegrate is $\rho = \dfrac{3\pi}{GT^2}$ .

Approach

The second limit: at the equator, gravity must be able to supply the centripetal acceleration. At the limit they are equal. Then replace $M$ by density times volume and $\omega$ by $2\pi/T$ ; the radius cancels.

At the limit, for a point on the equator,

r\omega^2 = \frac{GM}{r^2}

With $M = \rho \cdot \tfrac43\pi r^3$ and $\omega = \dfrac{2\pi}{T}$ :

r\left(\frac{2\pi}{T}\right)^2 = \frac{G\rho\,\tfrac43\pi r^3}{r^2} = \tfrac43 \pi G\rho\, r

\frac{4\pi^2}{T^2} = \tfrac43\pi G\rho \quad\Longrightarrow\quad \rho = \frac{3\pi}{GT^2}

A denser pulsar has more gravity per unit of radius and holds together; so this is the minimum density.

Setting the centripetal acceleration equal to the gravitational field strength at the surface, $r\omega^2 = GM/r^2$ , with $M = \tfrac43\pi r^3\rho$ and $\omega = 2\pi/T$ , gives $\dfrac{4\pi^2}{T^2} = \tfrac43\pi G\rho$ , so $\rho = \dfrac{3\pi}{GT^2}$ .

Part (e)(iii)

1 mark

Use the expression to find $\rho$ for the first pulsar discovered.

\rho = \frac{3\pi}{GT^2} = \frac{3\pi}{(6.67\times10^{-11})(0.60)^2} = 3.9\times10^{11}\,\text{kg m}^{-3}

$\rho = 3.9\times10^{11}\,\text{kg m}^{-3}$

To check your answer:

The pulsar's actual average density is $\dfrac{2.8\times10^{30}}{\tfrac43\pi(1.0\times10^{4})^3} = 6.7\times10^{17}\,\text{kg m}^{-3}$ , about a million times the minimum. It is in no danger of flying apart.

Part (f)

1 mark

The Crab Pulsar has period $33.5\,\text{ms}$ , increasing by $38\,\text{ns}$ per day. Use $\text{age} = \dfrac{\text{period}}{2\times\text{rate of change of period}}$ to find its age in years.

\text{rate of change of period} = \frac{38\times10^{-9}\,\text{s}}{86\,400\,\text{s}} = 4.40\times10^{-13}

\text{age} = \frac{33.5\times10^{-3}}{2(4.40\times10^{-13})} = 3.81\times10^{10}\,\text{s} = \frac{3.81\times10^{10}}{3.16\times10^{7}} = 1.2\times10^{3}\,\text{years}

age $= 1.2\times10^{3}$ years

Note

The supernova that made the Crab Nebula was recorded by astronomers in 1054, about a thousand years ago. The estimate is close.

Part (g)

2 marks

Find the wavelength of the highest-energy photons from the Crab Pulsar, $400\,\text{TeV}$ .

E = (400\times10^{12})(1.60\times10^{-19}) = 6.4\times10^{-5}\,\text{J}

\lambda = \frac{hc}{E} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{6.4\times10^{-5}} = 3.1\times10^{-21}\,\text{m}

wavelength $= 3.1\times10^{-21}\,\text{m}$

Watch

T is tera, $10^{12}$ . Convert eV to J with $1.60\times10^{-19}$ before using $hc/E$ .

Part (h)

1 mark

Suggest why the SKA's larger collecting area will let it detect pulsars that the Parkes Telescope cannot.

The power collected from a faint or distant pulsar is proportional to the collecting area. The SKA's area, $1.0\,\text{km}^2$ , is about 300 times that of Parkes ( $\pi \times 32^2 = 3.2\times10^{3}\,\text{m}^2$ ), so it receives enough power for weak signals to be detected above the background noise.

What this question was really testing

Reading data well. Period from many pulses, not one.
Using an unfamiliar equation the question gives. $L = \tfrac25 mr^2\omega$ conserved, so $\omega \propto 1/r^2$ .
Steep gravity. $\Delta g \approx (2g/r)\Delta r$ , and what a large gradient does to an extended object.
Turning a sentence into an inequality. "Centripetal acceleration must not exceed $g$ " becomes $r\omega^2 \le GM/r^2$ , and $r$ cancels.
Unit care. ns per day, TeV, km², all converted before use.

A dead star spinning in fractions of a second, and the two things that stop it spinning faster

Part (a)

Part (b)

Part (c)

Part (d)(i)

Part (d)(ii)

Part (d)(iii)

Part (e)(i)

Part (e)(ii)

Part (e)(iii)

Part (f)

Part (g)

Part (h)

What this question was really testing

This is how we mark, and how we teach.